Alternately, I would've just transformed/rotated the x and y axes by an angle of 45° about the origin using the rotational matrix. Then using the standard method of computing volumes of revolution ,the problem becomes easy.
@stefanschroder46943 жыл бұрын
That's a ingenious idea. This makes sense👌
@emanuellandeholm56573 жыл бұрын
That was my first thought. That parabola is still a "nice" function in a rotated coordinate system (x' = y - x, y' = y + x). Trivial substitution.
@quintonpierre3 жыл бұрын
Yeah, it is actually the same thing (just less convoluted). The change of variable is (u,v)=1/sqrt(2) (y-x , x+y) and you get the curve by solving sqrt(2) (u+v)=(v-u)^2. Then you can apply the usual technique on the interval u in [0,sqrt(2)].
@emanuellandeholm56573 жыл бұрын
@@quintonpierre Thing is, it becomes multivariable when you let u = (y-x) and v = (y + x). Then you have to deal with Jacobians, which you did implicitly.
@jakobunfried26693 жыл бұрын
@@emanuellandeholm5657 things are nicer if you include a factor 1/sqrt(2) in the new variables. otherwise you need to think about scaling factors in the integral limit and differential
@mohitkrjain93963 жыл бұрын
8:40 the y-coordinate should have been (4*xi+1-sqrt(8*xi+1))/2
@mathiasfjsne88543 жыл бұрын
Yes i noticed the same thing
@thomashoffmann88573 жыл бұрын
It's compensated around 13:20 when Delta - y is calculated for the distance (?)
@sweetcornwhiskey3 жыл бұрын
@@thomashoffmann8857 No that's where it changes the problem entirely to make it far easier to solve than it otherwise would have been. This error persisted throughout the video, and I'm not sure whether the answer that he got is incorrect or whether he just pulled it from mathematica without checking it through the calculations.
@Dysan723 жыл бұрын
@@sweetcornwhiskey No, the value (4*xi+1-sqrt(8*xi+1))/2 us the value that makes the problem easier to solve. xi - (2*xi+1-sqrt(8*xi+1))/2 = (-1+sqrt(8*xi+1))/2 which removes the xi term so the two parts would not be the same.
@andrewfischer-garbutt28673 жыл бұрын
Yeah that honestly bothered me a lot lol.
@SpencerTwiddy3 жыл бұрын
11:04 - should be sqrt(2*(deltaX)^2) = sqrt(2)*deltaX
@joshspektor43893 жыл бұрын
I used a rotation matrix to get a parametric equation for the curve: x=sqrt(2)/2*(t^2 + t) y=sqrt(2)/2*(t^2 - t) Then you can integrate pi*y(x)^2 over x from 0 to sqrt(2), which after making a u-substitution of x in terms of t turns into the integral over t from 0 to 1 (that’s the only t interval on which y is negative) of pi*sqrt(2)/2*(t^2 - t)^2*(sqrt(2)*t+sqrt(2)/2), which works out nicely to the pi/(30*sqrt(2)).
@pierre-marcshinkaretzky88513 жыл бұрын
Much simpler method. The volume is an onion made of cones which slant height is h(x) = x - x^2. The radius is just h(x) / sqrt(2) (simple isocele 45° triangle) Thus, the surface of this cone is s(x) = pi * h(x)^2 / sqrt(2) =pi * (x - x^2) ^2 / sqrt(2) = pi * (x^4 - 2x^3 + x^2) / sqrt(2) The thickness of each cone is dx Then V = / s(x) dx in interval (0,1) V = pi * (F(1) - F(0)) / sqrt(2) with primitive F(x) = x^5 / 5 - x^4 / 2 + x^3/3 F(0) = 0 F(1) = 1/30 " 1/5 - 1/2 + 1/3 = 1/30 " V = pi / (30* sqrt(2) ) CQFD NOTE : Démonstration rapide de la surface d'un cône de longueur latérale L et de rayon R. Un cône déplié fait un secteur d'angle Alpha et sa surface est le ratio Alpha/ 2*pi du cercle de rayon L (pi*L^2) Sa surface est donc S = pi*L^2 * Alpha/ 2*pi = L^2* Alpha/ 2 La circonférence de la base du cône fait 2*pi*R mais aussi Alpha*L Donc Alpha = 2*pi*R /L donc Alpha/ 2 = pi*R /L Finalement S = L^2 * pi*R/L = pi*R *L la surface d'un cône de longueur latérale L et de rayon R vaut S = pi*R *L Conic onion peels characteristics : PRIMO : Each cone has a constant angle of 45° with its axis (base radius and height are equals) SECUNDO : dx is the thickness of the peel because x axis is perpendicular to the cone generating line Then s(x) is something simple proportional to the square of the slant height and slant height is also simple y(top of the conic peel ) - y(parabola matching lower point of the peel)
@mohammedkhalili11543 жыл бұрын
Pretty much any other method is simpler than his
@Zarunias3 жыл бұрын
My way for this kind of problems is to remember: The volume of a rotated area is the area times the distance the center of gravity traveled in this rotation. So I break it down to several steps. Step 1: Calculate the area of this [1/6]. Step 2: Calculate the volume of it if rotated by the x-axis [2*Pi/15]. Step 3: Calculate the volume if rotated by the y-axis [Pi/6]. Step 4: Use the numbers to calculate the coordinates of the center of gravity [1/2; 2/5]. Step 5: Calculate the distance between the center of gravity and the line of actual rotation [sqrt(2)/20]. Step 6: With these numbers, calculate the volume [Pi*sqrt(2)/60]. All done with simple integrations, and moreover: Without thinking.
@debdeepmajumder91363 жыл бұрын
Pappus theorem.
@scipionedelferro3 жыл бұрын
+1 I was going to comment on the same idea: Calculate the center of gravity and use the Pappo-Guldin Theorem. Depending of the problem, calculating the center of gravity can be done in different way, for a parabola segment is known since Archimedes times, sometime symmetries can help, or there is a general formula in worse cases.
@abenedict853 жыл бұрын
The point is that there is not ONE way to arrive at a correct answer. That the answer is correct because there is agreement of the solution INDEPENDENT of the method of its derivation. You can integrate disks or revolve curves, it's the same outcome. If you only ever confine yourself to revolving curves, you may not include integrating disks into your thinking, and will be subsequently stopped because there are no appropriate curves to revolve. Remember, everything looks like a nail when all you have is a hammer. But if you never learn to use a screwdriver, screws just appear as poor performing nails.
@jursamaj2 жыл бұрын
@@abenedict85 For this, I prefer integrating cones. (Take the segment from x^2 to x and revolve it around the x=y line.)
@primsiren1740 Жыл бұрын
Can you explain how you calculated the centre of gravity in step 4 please? I don't understand what function you are rotating about the axes to get the volumes in parts 2 and 3
@demenion35213 жыл бұрын
it is a very satisfying feeling to just find rotated coordinates and ending up at pretty much exactly the same integral and hence final value
@davidbizzozero34583 жыл бұрын
There is a simpler way to do this problem since we have the line y=x which is a nice 45 degree angle. The trick is to take a small rectangle (width dx) and note that when you rotate it about y=x you get a lateral surface of a cone with volume: V_conearea = pi*r*L*dx. The slant height "L" is easy to calculate since it's the difference of the curves: L = x-x^2. Next, the radius "r" is a bit more complicated since that line is at an angle. However, since we know that for that 45 degree cone h = r = L/sqrt(2), we know that r = (x-x^2)/sqrt(2). Now integrate x from 0 to 1! V = integral(pi*r*L dx,0,1) = integral(pi*(x-x^2)^2/sqrt(2) dx,0,1) = pi/(30*sqrt(2)) I think this is a "cooler" way to do this problem since we don't need to do any rotation, or use washers or shells! Instead, we use lateral area of cones with thickness dx! You can do this "cone" trick with any angle but that just makes finding the cone radius "r" from the slant height "L" involve an arc trigonometric function. Have fun!
@realedna3 жыл бұрын
|\ |\ \ | \ \ | \ So the infinitesimal volume (V_conearea) is the volume between those 2 slanted lines rotated around the vertical line (vertically extruded lateral cone area), where the vertical line represents y=x.
@HolgerGoesKiwi3 жыл бұрын
I am a math teacher in Germany. First of all thanks a lot for the idea of this challenge. It is not necessarily very hard, yet a good training in arithmetics! Here are my two cents: The orthogonal distance from the line y=x to the parabola is: sqrt(2xi^2 + 6xi + 1 -2(2xi + 1)sqrt(0.25 + 2xi)) Finally the whole integration leads to the volume of V= 1/60 * sqrt(2) * pi. Hope that's the right answer, but it seems at least plausible... Thanks again for this great channel!
@algorithminc.88503 жыл бұрын
Wow - this takes me back to the 80's. I was fortunate enough to have a high-school teacher who once worked for NACA (later became NASA) - named Roselle D'Orazio. She was a war-effort mathematician and worked on the X-1 too. A brilliant lady - a kind grandmotherly type who could run circles around most anyone with math. Most fun I ever had in a class. She did bring up such a topic, as extra out-of-interest work. She touched on Hamilton's work and quaternions, among other ways of addressing these types of problems. If the USA would remove the politics from teaching, and get back to attracting (and keeping) teachers like this with backgrounds like this who really love the subjects ... this country would be accelerating forward ...
@UltraMaXAtAXX3 жыл бұрын
Except districts are worried about test scores. I love my subject but my district may have a hard time keeping me.
@whatelseison89703 жыл бұрын
@@UltraMaXAtAXX How backasswards is it that funding should depend on test scores. Low test scores should indicate students and teachers need _more_ funding not less.
@TheCloudyoshi3 жыл бұрын
I spent like 2.5 hours on this and thought it was going to require a bunch of trig sub and whatnot... and then I realized I used 2*pi*r instead of pi*r^2 for the disc volume 😂 which meant I just had an extra square root in there unnecessarily complicating the heck out of my integral
@udic013 жыл бұрын
8:42 the y coordinate should be =1/2* (4Xi+1+.... That's why the magic happens at 13:00... Classic michael...
@thomashoffmann88573 жыл бұрын
The right result shows that you made an even number of errors 😉
@goodplacetostop29733 жыл бұрын
0:03 Is it skipped because they don’t have enough time in the semester or becaue it’s too hard? 🤔 21:07 Good Place To Stop
@skylardeslypere99093 жыл бұрын
I think a combination of both. In general, surfaces of revolution are kind of skimmed over. They just give the integral formula without giving it too much thought, which results in not knowing how to do these, since there's not a standard formula for this problem.
@jgray27183 жыл бұрын
As someone who has taught college calculus a few times I can tell you that the semester tends to be pretty packed. This would work well in something like calculus BC in high school where there's a fair amount of time and you can do some neat projects, but if you tried this in a college calculus class (probably calc 2, but some people do solids of rotation at the end of calc 1 in a semester system; I don't remember where everything goes in quarters) you'd be pressed for time and the students wouldn't get as much out of it as you might want. This would make for a cool extra credit project, though, and if you were looking for an undergrad paper this would probably be an interesting one. It does kind of get trivialized by rotation matrices, though. Another reason I would want to skip this is that I generally think of solids of rotation as a somewhat gentle introduction to applications of integration. They're easy to visualize - the slices for the Riemann sum are literal slices - rather than trying to "slice" time or some other abstract concept. It goes well alongside the classics like lifting objects with ropes of non-zero mass and pumping water out of various troughs and such. If you then make the algebra complicated it defeats the purpose of having a gentle introduction as the students get lost in the algebra and don't comprehend the cool stuff you're showing them.
@jgray27183 жыл бұрын
@@skylardeslypere9909 It's funny, I never even remember the formulas myself because I always want to draw everything. I think my students want or expect the formula but I just keep drawing stuff; the formula is in the book, they can look it up if they want to. I'm just going to keep unrolling cylinders and stacking discs. I mean, who cares about the formula anyway? The whole point of having solids of revolution as part of the introduction to Riemann sums is to help gain intuition about what a Riemann sum really is by having a visualizable application, so why ever worry about the formula? That just skips all the visualizing and jumps to the answer, which defeats the purpose. Also, I'm pretty sure (though I haven't done it) that you could derive a general formula for rotations about lines not square to the axes by using a rotation matrix to rotate your axes, do all the calculations in the new coordinate system, and then another to rotate them back. My guess is that you can probably just run the existing formulas for discs and shells through those rotation matrices to get the formulas straight away. You could probably also do it with standard algebra (like Michael did in the video), but that feels like it would be harder. So I strongly suspect there's a reasonably-derived standard formula, but again, why bother if the whole point is to think about the problem and how Riemann sums work? Now that I think of it, this might make a good project in a linear algebra class when talking about either rotation matrices or change of basis, as this is a really good use of a change of basis matrix. I think I might steal this for exactly that purpose. Thanks for making me think on this topic Skylar, I think my students will really like this :-)
@skylardeslypere99093 жыл бұрын
@@jgray2718 You mentioned that it gets trivialized by rotation matrices, and I was thinking something along those lines too while watching the video. Is it actually easier to find the volume of the surface of revolution around any line in the plane, if you apply a change of base? That is, translate your coordinate system so the origin is on that line, and then rotate it so it becomes one of the axes.
@jgray27183 жыл бұрын
@@skylardeslypere9909 In general, yes it's probably easier. Applying a rotation matrix is easy, then you have a standard solid of revolution, which is also easy. The algebra Michael did seemed much harder to me. "Trivialized" was probably an overstatement on my part, but it's not difficult, just a little bit longer than the usual solid of revolution. This particular problem was also chosen specifically to have "easy" algebra since it was rotated around y = x. I'd think a rotation matrix would be the way to go for sure if you wanted to rotate around y = 2x. God help you if you want a general formula for rotating about y = ax + b, though. I'd have to consider how to do that, but I think you're into affine algebra rather than linear algebra at that point, so maybe the approach Michael took in the video would actually be easier in that case. I don't even know what an affine rotation matrix looks like, now that I think of it. Probably a slide then a spin?
@gurkiratsingh7tha9933 жыл бұрын
This is something new, this is the first video about this new idea I have ever seen in KZbin. Thank you for this amazing video
@johnsaunders15273 жыл бұрын
I worked through this when I found it in the James Stewart Calculus book. What a great problem!
@digxx3 жыл бұрын
Rotating the old coordinate system X by 45° the coordinates for (x,x^2) in the new system X' become (x',y')=R*(x,y)=1/sqrt(2)*(x^2+x,x^2-x) where R=1/sqrt(2)*[ (1,-1) , (1,1) ] is the rotation matrix by -45° (rotating the coordinate system by 45° is equivalent to rotating the old coordinates by -45°). Hence y' becomes the radius of the rotating body and V=Pi*Int(y'^2 * dx',x=0..1)=Pi/2*Int( (x^2-x)^2 * d(x^2+x)/sqrt(2) , x=0..1) = Pi/(2*sqrt(2))*Int( (x^4 - 2x^3 + x^2)*(1+2x)*dx,x=0..1)=Pi*sqrt(2)/60.
@etoolster3993 жыл бұрын
Thank you, Mr. Penn. This video just taught me how to do "u substitution". My calculus class (in 1977) didn't get this far. Hey, I learned something today! Thanks again!
@AirAdventurer1943 жыл бұрын
This was just a wonderful video! I know I would have really struggled with that if a student in a Calc II class had asked it as a question and I didn't have any notes prepared. I thought for sure the square root integrals were going to need to be done with trig substitutions; it's always a little nice when they can just be done with u-substitutions (I guess you'd need squares under the radical for trig substitutions).
@AirAdventurer1943 жыл бұрын
(P.S. Your "Mr. Rogers mannerisms" are really nice; I should try to emulate that more)
@petersievert68303 жыл бұрын
12:48 Actually the magic easily derives from the fact that the slope of the line QP being -1 has the absolute value 1, doesn't it?
@hozelda3 жыл бұрын
Instead of relying on that magic, you can avoid the sqrt altogether by not solving for x in terms of xi. Instead turn x^2+x=2xi into: xi=x(x+1)/2 dxi=(x+1/2)dx and substitute accordingly so x is integration variable (and bounds remain at 0 and 1). After much reduction, end up with: sqrt(2)pi/4 integral from 0 to 1 dx [2x^5 - 3x^4 + x^2] which is very easy to integrate.
@hozelda3 жыл бұрын
To clarify, the "much reduction" part involves many but easy reductions since we are simplifying sums and multiplications of small polynomials (no radicals).
@ConradAdamsMrJUMBO3 жыл бұрын
I guess the next challenge would be to find the volume of the solid of revolution formed by rotating the region bounded by y=x and y=x^2 about the parabola y=x^2.
@jakobunfried26693 жыл бұрын
what would it even mean to rotate sth about a curve? afaik that doesnt make sense. take a random point in the shaded area. can you tell me unambiguously where it would rotate to?
@derendohoda38913 жыл бұрын
@@jakobunfried2669 Hmm... how about using the curvature of a point on the parabola to generate a circle of inversion and then invert the point on the line?
@jakobunfried26693 жыл бұрын
@@derendohoda3891 only 90% sure i understand what you mean, but assuming i do: a point on the line could then be on several different such circles and it would be unclear, where it should rotate to
@williamwolfs48193 жыл бұрын
@@jakobunfried2669 there's only one circle perpendicular to the curvature with all points equidistant from a particular point of the curve. The team problem we would run into is overlapping circles -- a naive integral would double count some volumes when rotating about a curve
@dizzyd73153 жыл бұрын
Maybe next year
@jakobunfried26693 жыл бұрын
i found it much more intuitive to do a double integral over all rings and then you can even let the curve be an arbitrary function f(x). the radius of the ring that intersects the shaded surface at (x, y) is (x - y)/sqrt(2). integrating the circumference of these rings over the shaded area (x from 0 to 1, y from f(x) to x) gives a general formula V = π/sqrt(2) \int_0^1 dx (x - f(x))^2 valid for a function with f(0)=0, f(1)=1, f(x)
@ellingsondavid3 жыл бұрын
Had teachers cover this in '96 and after. They not only did y=x, but a lot of other values for the slope and we had to calculate by hand. I loved my TI-92 that would do this!
@stabbysmurf3 жыл бұрын
My way was "simple" but fraught with tiny mistakes: I wrote A+B = , where r = 1/sqrt(2). Then just write the integral of (pi*B^2)dA. This simplifies greatly because 2rA = x+x^2 and 2rB=x-x^2, resulting in an integral of powers of x from 0..1.
@jorgepresto88233 жыл бұрын
I believe there is a more easy way to solve this problem. The distance between any point of the parabola and the line y=x in [0,1] is (by basic geometry courses) d(x)=(x-x*x)/sqrt(2). Because it is going to be rotate around y=x the length is sqrt(2)*(xi+1-xi), then the volume will be pi*integral[0,1] d2(x) sqrt(2) dx. Solving this integral the result is pi/30sqrt(2)
@MathTutor13 жыл бұрын
I normally teach this topic for honors classes, specially the revolution about the line y = mx. Revolution about y=mx+b can be little more involved. Thank you for doing these.
@MasterHigure3 жыл бұрын
How is revolution around y=mx+b more involved? You just move everything down by b and you're back at y=mx, don't you? Well, technically, that one simple thing makes it a little more involved. But you know what I mean.
@Mmmm1ch43l3 жыл бұрын
Why?
@davidebracalicioci38033 жыл бұрын
I applied a different method, in principle it should be more general. Let us assume to have a straight line t, with equation y=mx+q in the plane, and a function y=f(x) to rotate w.r.t. to the direction defined by the straight line by an angle of 2pi. Then, if we take a general point P=(s,f(s)) on the graph of the function f, the distance between point P and the straight line is given by the classical formula d(P,t)=|ms+q-f(s)|/sqrt(1+m^2). So the volume of the rotation of the graph of bounded by the interval [a,b] is simply the sum of all the circles of radius d(P,t) all along the interval [a,b]. Obviously the sum becomes the following integral \int_a^b \pi*d(P(s),t)^2ds In the case proposed in this video, a=0, b=1, f(x)=x^2, m=1, q=0 and the integral becomes \int_0^1 \pi/2*(s-s^2)^2ds=\pi/2*(x^3/3-x^4/2+x^5/5)|_0^1=\pi/2*(1/3-1/2+1/5)=\pi/2*1/30=\pi/60 The result is not the same, I do not have the square root of 2 but just 2 at the denominator. I hope you enjoy this solution.
@Goku_is_my_idol3 жыл бұрын
Very cool approach and easier than in the video. I would like to point out that until the line "sum becomes the following integral" everything is correct. The problem comes after that. If u take an area of πr² then u need to multiply the differential element perpendicular to the given area. So multiplying "ds" simply gives the wrong solution. Differential element should be along the line ms+q acc. to notations u have taken which then gives the differential element as ds.√(1+m²). (From pythagoras theorem directly; taking a triangle of ds base and dy =mds as altitude) The entire integral then becomes Integration from 0 to 1[π(s-s²)²/√2 ds] I think this will give the correct value now.
@davidebracalicioci38033 жыл бұрын
@@Goku_is_my_idol Yes, you're right, very smart fix. Thank you
@kruksog3 жыл бұрын
Probably not that exciting, but this was definitely covered in my calc 2 course. I remember because it was one of the only problems I skipped on the final (not because I couldn't do it, but I recognized right away it would take too much of my time, because I wasn't as prepared for that problem as I was the rest of the material.)
@martinnyberg92953 жыл бұрын
00:06 That skipping must have started recently. 🤔I did many of these from the Swokowski (?) textbook in 1993 at LSU. And the book was purple. 😁👍🏻 (I also think that I have taught these problems in a class or two, but that was many moons ago.)
@hakerfamily3 жыл бұрын
The distance from a point (x,x^2) up to the point (x,x) is L=x^2-x. The shortest distance from (x,x^2) to the line x=y is thus r=(x^2-x)/sqrt(2) since there is a 45-45-90 triangle involved. Rotating the vertical line segment from (x,x^2) to (x,x) around the x=y line gives a cone with slant height L and radius r. The volume can then be found from the surface area of a cone: pi r L, i.e. \int_0^1 pi /sqrt(2)* (x-x^2)^2 = pi/ (30 *sqrt(2)). Disks, shells, washers... and cones!
@Goku_is_my_idol3 жыл бұрын
Genius
@Vladimir_Pavlov3 жыл бұрын
Great idea! Only the distance between the point of the parabola (x,x^2) and the axis of rotation y=x is equal to L=(x-x^2)/√2, 0≤ x≤ 1. When the parabola rotates, this segment forms a cross-section of the figure perpendicular to the axis of rotation - a circle with an area of π L^2. If dl is an element of the axis of the geometric body , then dl=√2dx. V=∫(from 0 to 1)π*[(x-x^2)^2/2]√2dx=π√2/60.
@Vladimir_Pavlov3 жыл бұрын
We calculate the volume of the geometric body as the sum of the volumes of elementary cylinders of differentially small height. In this sense, the cone has nothing to do with it.
@hakerfamily3 жыл бұрын
@@Vladimir_Pavlov You are right of course, r=(x-x^2)/sqrt(2), since (x^2-x)/sqrt(2) is negative. I think what you call L was r for me. Splitting the volume up into disks as you suggest is very good and simpler. I do like the idea of splitting it into cones though!
@Vladimir_Pavlov3 жыл бұрын
It should be noted that this approach to solving the problem makes it easy to get a result if the axis of rotation is an arbitrary straight line passing through the origin, y=tga *x, a≠π/2.)
@koenth23593 жыл бұрын
I used a slightly different approach, I set P(x) as the projection of the point A = (x, f(x)) onto the axis (the line x=y). P(x) can be found as P(x)=(t,t) setting t=x-c=f(x)+c. This solves as c= ½(x-x²), so that P(x)= (½(x+x²), ½(x+x²)). Now let s(x) be the distance of P(x) from the origin: s(x) = ½√2 (x+x²), and let r(x) be the distance from P to A: A-P = (x- ½(x+x²) , x²- ½(x+x²)) = (½(x-x²) , ½(x²-x)) and r(x)=|A-P|= ½√2(x-x²). We now find ∫ πr²ds = ∫ πr² • ds/dx dx = ∫ π ½(x-x²)² • ½√2 (1+2x) dx = ¼π√2 ∫ (x-x²)² •(1+2x) dx = ¼π√2 ∫ x^2 -3x^4 + 2x^5 dx = ¼π√2 [1/3 x^3 - 3/5 x^5 + 1/3 x^6] Because x runs from 0 to 1, the last expression must be evaluated at the boundaries 0 and 1 this gives a volume of π√2/60
@vornamenachname80013 жыл бұрын
alternative solution: generate weight field which is 0 at the revolving axis and linearly goes to 1pi when it reaches distance 1 from the revolving axis, integrate over x. y equals the integral of the weight field beween the revolving axis and other function
@polyhistorphilomath2 жыл бұрын
Using Archimedes’ geometric proof of the parabolic quadrature, the relative height and width of each triangle is known and the position of each center can also be calculated. So the centroid can be calculated as the weighted sum of the centers. Then take the foot/elevation, r, of the centroid from y=x. This is 1/(10*2^(1/2)). Multiply 2πr by 1/6 (per Pappus) an lid the volume obtains.
@pappaflammyboi57993 жыл бұрын
Let Sₖ be the set containing 2 and 5 and the first _k_ primes that end in 7. For example, S₃ = {2, 5, 7, 17, 37}. Define a _k-Ruff_ number to be one that is not divisible by any element in Sₖ. If Nₖ is the product of the numbers in Sₖ then define _F(k)_ to be the sum of all _k-Ruff_ numbers less than Nₖ that have the last digit 7. You are given _F(3)_ = 76101452. Find _F(97),_ give your answer modulo 1,000,000,007.
@zdrastvutye3 жыл бұрын
i need to write, that the perpendicular point xl,yl on y=x is the solution of 2 linear equations. after this is done, r(x)=sqr((x-xl)^2+(y-yl)^2). then there is the guldin rule or a for next loop to add parts with different r(x) together. finally, a part of the volume is dv=pi*r(x)^2*dl l=sqr(xs^2+ys^2) so dl=l/n (n=number after "to" in a for next loop)
@Vladimir_Pavlov3 жыл бұрын
If for a parabola y=x^2 the axis of rotation is a straight line y=tga *x (a ≠π /2), then the required volume of the body of rotation will be equal to V=(π/30) *cosa *(tga)^5. At a = 45 degrees, we get an answer for a this case.
@dpatulea3 жыл бұрын
Nice! Never did this before! Thanks!
@misterdubity30733 жыл бұрын
I used a series of cone shaped shells each having slant height of x-x^2 and radius sqrt(2)/2 of that. Integrated 0 to 1 dx. I got the same answer. First I started the same way as in the video but stopped short of calculating that diagonal distance. Then I tried rotating axes but couldn't figure that bit out.
@dinhotheone3 жыл бұрын
Same, although I think the cones have the same radius and height of r(x)=(x-x^2)/sqrt(2), we want the shells so I did Vshell(x)=pi/3*r(x)^3-pi/3*(r(x)-sqrt(2)dx)^3 I.e. volume outer cone minus volume of inner cone. Note you need sqrt(2)dx because the cone height lies on the diagonal line rather than aligned with the x axis. After cancelling and dropping terms with dx^2 or dx^3, you get Vshell(x)=pi/sqrt(2)(x-x^2)^2 dx and when you integrate this from 0 to 1 you get pi/(30*sqrt(2))
@insouciantFox3 жыл бұрын
My calc teacher did this exact problem as an extra-curricular example, but we weren't tested on it.
@zdrastvutye3 жыл бұрын
hi! the difficulty would be greater if the axis y=x was not in the xy plane. but so, the rotated points are, if xl and yl are perpendicular points: xr=x+(xl-x)*cos(w) yr=y+(yl-y)*cos(w) zr=sqr((x-xl)^2+(y-yl)^2)*sin(w) have fun using ba$ic instead of calculus!
@cicciobombo74963 жыл бұрын
i just got the parametric form of the rotated function and used the parametric version of the volume integral from 0 to 1 :D
@stephenhousman69753 жыл бұрын
11:10 Small mistake: sqrt(2 * delta x) does not equal sqrt(2) * delta x
@cc3loki3 жыл бұрын
It was delta x squared inside the square root, he just forgot
@Iridiumalchemist Жыл бұрын
If you just subtract x from y = x^2 then you the object bounded above by the axis. This is equivalent and EASIER than the rotation matrix. This problem is simply rotation y = x^2 - x about the x-axis between 0 and 1.
@pendragon76003 жыл бұрын
this can be trivialized by a change of basis, but it's nice to see this alternative method. Also yeah, not sure what's up with calc teachers skipping this. My class didn't cover it, as far as I can remember.
@Oscar16180333 жыл бұрын
Got to the same result using the integral of surface area of cones General case with that method gives V=pi/sqrt(m^2+1) * integral[(mx+q-f(x))^2]
@devonshrestha32793 жыл бұрын
Small mistake 8:40 it should be 4xi in the y coordinate of the blue point, not 2xi.
@duncanw99013 жыл бұрын
At least one Mr. Turnbaugh in Bentonville, AR didn't skip it! Spent about 3 weeks on it, and like 4 people in the class ended up being comfortable with it. But we did it.
@scorch25able3 жыл бұрын
I am sure there are others who have commented that the rotation of axis method would be simpler to implement, but this is also a good video to demonstrate the use of the Riemann Sum. I wonder how this can be solved using Lebesgue Integral...
@FineDesignVideos3 жыл бұрын
I used Lebesgue integration, and it felt much neater to me! It worked by splitting the volume into hollow cylinders with the central axis being y=x. The radius r cylinder would have width dr and volume 2 pi r len(r) dr, where len(r) is the length of the cylinder of radius r. To find len(r), we just need to find the points such that y = x^2 and y = x - r/sqrt(2) (since that is the parallel line to y=x at distance r from it). The x values of the points are the solutions to x^2 - x + r/sqrt(2) = 0. The difference between the x values is sqrt(b^2 - 4ac), which in this case is sqrt(1 - 2sqrt(2)r). The distance between the points is then sqrt(2)sqrt(1-2sqrt(2)r) (because the slope of the line is 1). So our answer is the integral of 2sqrt(2) pi r sqrt(1-2sqrt(2)r) dr, where r goes from 0 to... 1/2sqrt(2), since that's where len(r) becomes 0. Substitute t=1-2sqrt(2)r, so r = (1-t)/2sqrt(2) and dr = -dt/2sqrt(2). Now we're solving integral of 2sqrt(2) pi (1-t)/2sqrt(2) sqrt(t) (-dt/2sqrt(2)), which is: -pi/2sqrt(2) times integral of sqrt(t)-tsqrt(t) dt, where t goes from 1 to 0. The final integral to solve is super easy, and you get -pi/2sqrt(2) times [2/5 - 2/3], which is 2pi/15sqrt(2), which is different from the answer given in the video. :o I have no idea where I went wrong. I had a silly calculation mistake in my first attempt and it magically matched the video answer with that mistake.
@buxeessingh25713 жыл бұрын
I covered this once. The class had a very hard time following.
@Roescoe3 жыл бұрын
This situation happened to me on accident when our prof made an error in the homework. Me and another guy couldn't solve it so we asked, but he just fixed the error instead of telling us how to solve it.
@The1RandomFool3 жыл бұрын
I think I've done one similar to this before. The method I chose was rotate the curve pi/4, then use the normal method around the x-axis.
@GreenMeansGOF3 жыл бұрын
That how I would do it.
@JoQeZzZ3 жыл бұрын
Without doing it, you would need to change the integration to 0 to sqrt(2) no?
@MasterHigure3 жыл бұрын
Not entirely trivial to rotate the parabola. Not impossible. But I'm not sure it's easier than this video.
@drhubblebubble73 жыл бұрын
I always love pausing the video and trying this out first.
@matiasjoaquinbustamantevej32783 жыл бұрын
You could also use Pappus, you only need to find the area of the curve, his center of gravity and the distince of that center to the axis of rotation (y=x in this case). I had these problem (but with a different curve) in my Calc 2 final exam
@Bulbulim943 жыл бұрын
rotating gives y=x^2 --> f(x)=x+(1-sqrt(1+4x*sqrt(2)))/sqrt(2), now calc pi*integral (f^2)dx
@rezabidar10423 жыл бұрын
Hi Michael, there is an easier way to solve this: if OP represents the segment from the origin to P(x,y=x^2), and t is the angle between OP and x-axis, then radius of revolution is r=OP*sin(pi/4-t)=sqrt(2)/2*OP*(cost-sint)=sqrt(2)/2(x-y). Then V=integral_0^1 pi*[(sqrt(2)/2(x-x^2))]^2*sqrt(2)dx = pi*sqrt(2)/60
@MarieAnne.3 жыл бұрын
I did the same, but differentiated with respect to t. Like you, I get w = √2 dx But when solving for slope from (x, x) to (t, t²) = −1, I solve for x in terms of t instead of t in terms of x (x−t²)/(x−t) = −1 xᵢ− t² = −x + t 2x = t² + t x = 1/2 (t² + t) Differentiting both sides we get: dx = 1/2 (2t + 1) dt Note that points (t, t²) go from (0,0) to (1,1), so t is between 0 and 1 Now we calculare r², where r is distance between (x, x) = (1/2 (t² + t), 1/2 (t² + t)) and (t, t²) r² = [1/2 (t² + t) − t]² + [1/2 (t² + t) − t²]² = [1/2 (t² − t)]² + [1/2 (t − t²)]² = 1/4 (t² − t)² + 1/4 (t − t²)² = 1/2 (t⁴ − 2t³ + t²) Now we can calculate volume of a disk with radius r and width w v = πr²w = π · 1/2 (t⁴ − 2t³ + t²) · √2 dx = π · 1/2 (t⁴ − 2t³ + t²) · √2/2 (2t + 1) dt Integrating over all disks, we get volume of rotation V = π/(2√2) ∫ [0, 1] (t⁴ − 2t³ + t²) (2t + 1) dt = π/(2√2) ∫ [0, 1] (2t⁵ − 3t⁴ + t²) dt = π/(2√2) (t⁶/3 − 3t⁵/5 + t³/3) | [0, 1] = π/(2√2) (1/3 − 3/5 + 1/3) = π/(2√2) (1/15) V = π/(30√2)
@paul213532 жыл бұрын
At 3:09 the interval [0,1] is sudivided into n+1 subintervals, not n subintervals. Maybe he had in mind to dtart with x1 as his first point, not x0. Anyway is is an error.
@ekisvioleolivaradamos67013 жыл бұрын
Very informative... Amazing content ... Tnx prof...
@TaladrisKpop3 жыл бұрын
This is actually the standard situation (covered for example in section 5.1 in Stewart's Calculus) of a solid whose volume is computed as the integral of the cross-section area. Volumes of solids obtained by rotating a region bounded by graphs of functions about the x-axis (V=\int_a^b \pi f(x)^2 dx) is a special case of it.
@adrianignacioirias3 жыл бұрын
No! The better example for method is in page 588 ( chapter 9 ) 4th Edition. (Discovery Project). Adrián Irias
@TaladrisKpop3 жыл бұрын
@@adrianignacioirias I don't have the 4th edition but I guess you are talking about the Discover project "Rotation about a slant line" in the chapter about Applications of Integration. You don't need this for the example of the video. And, if you read the whole chapter, you'd better use the Theorem of Pappus to compute the volume.
@hunterlouscher92453 жыл бұрын
Had a lot more fun finding the length of the line perpendicular to the rotation axis between its intersections with the curves, and getting area with pi r^2 (squaring the square root), and integrating the resulting simple polynomial over x.
@yoav6133 жыл бұрын
Yes i solved it this way,it was nice
@koenth23593 жыл бұрын
Me too, check out my solve!
@newwaveinfantry83622 жыл бұрын
Could you not have simply subtracted x from x^2 (or vise versa) and done a horizontal rotation on that? (of course that rotation would be from 0 to sqrt(2) since that's the length of the line)
@arnoldbailey75503 жыл бұрын
Going about it the hard way. Set the slope as your x value and calculate the area under the curve. Convoluted methods take too much time. Good exercise for fun.
@walterbrown86943 жыл бұрын
Rotation of coordinate systems was covered on one my senior year math courses leading to my electrical engineering degree. Course was called "Multidimensional Calculus". The material in the course I found to be quite helpful in my employment for a defense contractor in missile and weapons systems targeting/engagement. (Been retired for over 30 years, and still remember some of that stuff)
@billyoung81183 жыл бұрын
You will be hard pressed to find a better mathematician than an EE grad. I earned my BSEE studying computer chip design. Graduated in the tech bubble burst. No jobs anywhere, world-wide. Ended up doing statistics for the insurance industry.
@milesrout3 жыл бұрын
@@billyoung8118 A better mathematician than an EE grad is pretty easy to find if you've ever heard of a 'mathematician'.
@MultiAndAnd3 жыл бұрын
Lebesgue measure is inavariant under the action of isomtries, so I would rotate around e_1 instead
@MrAshtordek3 жыл бұрын
I solved this way differently (desmos demo: desmos com/calculator/fci6ffmtdx) by parametrizing the distance between the two functions with a line orthogonal two x=y. I will call the intercepts k_i. First I considered that the function for the orthogonal line must be: -x + 2x_0 = y Then I found the intercepts with the two other functions at x_0. For x=y its almost by definition: k_1 = (x_0, x_0) and for x^2=y its a little more complicated, but basically you can solve the equation -x + 2x_0 = x^2 and you get: k_2 = ((sqrt(1+8*x)-1)/2,((sqrt(1+8*x)-1)/2)^2) Which looks a bit terrifying, but doesn't cause any major problems down the way. Then I considered that the radius of the volume, r, at x_0 must be given by the function: r(x) = ||k_1 - k_2|| Where we are a bit lucky because actually both terms in the norm are equal (not intuitive in my opinion). However that simplifies the equation A LOT: r(x) = sqrt(2) * ((sqrt(1+8x)-1)/2 - x) Which we can use to calculate the area of a circle-slice: c(x) = r(x)^2 * pi = (sqrt(2) * ((sqrt(1+8x)-1)/2 - x))^2 * pi = 2 * pi * ((sqrt(1+8x)-1)/2 - x)^2 Then we can construct a integral using the formula for the area of a circle-slice to calculate the volume: V = int_0^1{c(x) dx} = int_0^1{2pi * ((sqrt(1+8x)-1)/2 - x) ^2 dx} = pi/60 ~ 0.52 So, I must say I am pretty confused. Weirdly this is "off" from your result by a factor of exactly sqrt(2), so that is probably not a coincidence, however I for the life of me cannot fathom it right now.
@nuranichandra21773 жыл бұрын
Convoluted but elegant. Around 11.20 mark there is a slight error in computation related to W.
@douglasmagowan27093 жыл бұрын
Man, I did this a much simpler way. Rather than finding the distance from the point (x_i,x_i) to the corresponding point on the parabola, I found the distance from the the point (x_i, x_i^2) to the corresponding point on the line. This quickly gave pi sqrt 2 ∫ ((x - x^2)/sqrt 2)^2 dx
@byronwatkins25653 жыл бұрын
I think you could rotate the two curves '-45 deg' and evaluate the regular way -- effectively, this is what resulted.
@polyhistorphilomath2 жыл бұрын
I see comments regarding forward rotation of the axes. … I am not impressed by the counterintuitive nature of that idea. Rotate the curves instead as you suggest. Det(J) or det(g) =1 so there is no need to do any heroic math. The standard parametric statement of the volume integral definition includes the derivative of x with respect to a parameter t. y’^2(dx’/dx)dx can all be expressed in terms of the original variable x, so the bounds of integration are unchanged.
@nathanisbored3 жыл бұрын
im guessing starting out by rotating the picture so that y=x is flat would have made most of the steps turn out simpler, but it would have been harder to get info about the transformed parabola. but i guess having a rotated parabola defining the curve is better than having a rotating axis defining the solid, and probably involves a lot less writing. maybe thats why they dont do slanted examples, because you can transform a slanted example into a standard one at the start, albeit it would be good to know how that process plays out.
@AntoshaPushkin3 жыл бұрын
Wouldn't it be easier to rotate everything by -45 degrees, i.e. go from (x, y) to (u, v) where u = (x+y)/sqrt(2), v = (x-y)/sqrt(2) (maybe need to change some plus to minus or vice versa), express the line y = x^2 in terms of u and v, get it in a form v = f(u) on the interval from 0 to 1, and then calculate ∫πf(u)^2 du from 0 to 1?
@Vladimir_Pavlov3 жыл бұрын
When rotating the coordinate system by 45 degrees , it is necessary to calculate π*∫[from 0 to√2 ]{(u+1/√2)-√(√8 u+1/2)}^2 du =√2/60 When the coordinate system rotates by (-45)degrees will have to calculate the same integral. Note that you will have to integrate from 0 to √2.
@AntoshaPushkin3 жыл бұрын
@@Vladimir_Pavlov right, thank you
@MathFromAlphaToOmega3 жыл бұрын
What about taking a surface of revolution around a curve? The radii of the circles would be the perpendicular distance between the two curves. Maybe that's a silly thing to do, but it would make for an interesting problem.
@Akheronlevrai3 жыл бұрын
Le volume étant simplement connexe et la symétrie xy, pourquoi s'embêter à prendre des rectangles d'intégration penchés, les verticaux auraient aussi bien fonctionnés dans ce cas précis. Un exemple plus compliqué avec un sinus "rapide" borné par un sinus x lent" par rapport à y=1/2x aurait montré la généralité de la méthode présentée
@NonTwinBrothers3 жыл бұрын
Oh man, I JUST so happened to derive the formula for a 45º rotated parabola a year ago. I never would've thought that would come in handy
@MasterHigure3 жыл бұрын
Both r and w are obviously hypotenuses in right, isosceles triangles (you already established their slopes of 1 and -1). That would save a lot of writing. Just take one of the legs, multiply by √2 and call it a day.
@odysseus96723 жыл бұрын
"I think you can work out a general formula for this.." Yep, and it's really easy: the volume is the integral of the cross sectional area d h. Go forth and calculate. :)
@zdrastvutye3 жыл бұрын
21:06 das pi taucht erst auf, wenn man die guldinsche regel anwendet
@goedelite2 жыл бұрын
Math arguments should not be interrupted by ads. An ad at the beginning is enough for youtube and Mr Penn to be well compensated. We taxpayers, who paid for the development of the interned receive no royalties at all for the risk of our taxes. We should not have to tolerate the greed of youtube. The is part of the corruption of our government that permits such greed.
@lenskihe3 жыл бұрын
When I took AP Calc in High School, we were supposed to come up with problems and put together our own "final examn". My group put this problem on it and our teacher wasn't able to solve it 😉
@wizard73143 жыл бұрын
Teachers 'skip' this because it's obviously just a co-ordinate change away from regular revolution. What would be more interesting is if you rotated every point on the 'slanted' curve around the X axis, but the centre of rotation was given by the other curve.
@grathwohldominik69843 жыл бұрын
The area of the depicted figure is A = \int_0^1 x dx - \int_0^1 x^2 dx = 1/6 A rectangle with width a and length sqrt(2) with the same area: a * sqrt(2) = 1/6 => a = 1/(6*sqrt(2)) Rotating this rectangle gives the solution: pi*a^2*sqrt(2) = pi * sqrt(2)/72 =pi/(36*sqrt(2)) However Michael obtained pi/(30*sqrt(2)). Can someone point me to my error?
@grathwohldominik69843 жыл бұрын
I was thinking all the time what may be wrong with such an easy approach and I found the explanation: Rotating figures with the same cross-section do not lead to bodies with the same volumina, e.g. a rectangle A_r=a*h_r has the same area as a right-triangle A_t=½*a*h_t if h_r=½*h_t. Now I introduce r=h_t that the formulas are looking more familiar: A_r=a*½*r, A_t=½*a*r. If I rotate the rectangle around a, I obtain a cylinder with V_cy=¼*pi*a*r^2 and if I rotate the right triangle around a, I obtain a cone with V_co=1/3*pi*a*r^2
@willyh.r.12163 жыл бұрын
Math is the best, when well explained like this.
@bobbwc70113 жыл бұрын
21:06 My German math teacher in those situations 20 years ago: *Deduction of points!! Pi/(30sqrt2) what? Potatoes? Pancakes? It's "volume units"!* (which very often downgraded me from a "1" ("very good", "A") to a "2" ("good", "B+"). Different times though. Today, grading in German secondary schools is less strict and less pedantic.
@barsercan72033 жыл бұрын
That's a revolutionary problem
@FinderTheIcewing3 жыл бұрын
I did learn this in calc 2...and the exact example.
@Mryeo53543 жыл бұрын
Honestly an easier way is to use pappus's Centroid theorem.
@williamwolfs48193 жыл бұрын
Yes, but that's not something a second semester calculus student would have access to when first studying solids of revolution
@1ucasvb3 жыл бұрын
@@williamwolfs4819 It's a shame because Pappus' centroid theorem is easily understood earlier, and it is FAR more general than solids of revolution so restricting it to that is a disservice to the theorem.
@assassin016203 жыл бұрын
@@1ucasvb can you explain to me how to use it in this specific problem?
@txikitofandango3 жыл бұрын
I would have partitioned the solid into cones and integrated over the lateral areas of all the cones
@dr.palsonp.h.d8153 жыл бұрын
problem would have been much easier if you used the inverse, and did things in terms of yi, yi+1 instead of xi, xi+1 (horizontal rieman)
@9WEAVER93 жыл бұрын
this was always left as an exercise to the readers in our courses.
@marsgal423 жыл бұрын
"Gnarly equations with a side of Pappus to go!" "Coming right up!"
@BikeArea Жыл бұрын
Recognize as he steps on the marked spot at 0:51 to ensure the following magic knock becomes seamless. 🤓
@random199110043 жыл бұрын
mmmm imagine as an extra advanced problem if the line being rotated around was not one of the boundaries (i.e. not y = x ) , but it would have to be parallel to y = x to stop it intersecting the region. Consider rotating around y = x + 1 ( or y = x + c for c not equal to 0 ).
@random199110043 жыл бұрын
mmm actually, it doesn't even need to be constrained to have the same slope as y = x. It just needs to be such that it does not intersect the region between y = x and y = x^2.
@onceuponfewtime3 жыл бұрын
11:09 I don't get the reason why is sqrt(2*delta x) = sqrt(2) * delta x
@dr.palsonp.h.d8153 жыл бұрын
Your definition was that x0 = 0 ... xn = 1, but your sum had index i = 1 ... n
@alexey.c3 жыл бұрын
It looks like the radius formula has wrong sign: it is negative if x = 1/2.
@AnkhArcRod3 жыл бұрын
I think coordinate transformation is the easiest way to deal with this...but sure, there are other methods too.
@robertmonroe9728 Жыл бұрын
Rotate 45 and use volume formula
@obidahali2 жыл бұрын
What would happen if we rotated the axes, would we get the same result? ....Thanks
@a.osethkin553 жыл бұрын
y = x - x^2 with rotation => simpler solve..
@mcwulf253 жыл бұрын
Have you tried rotating the axes? Maybe you get a complicated equation for the parabola.
@frentz73 жыл бұрын
Rotate the axes and then set up circular cross-sections perpendicular to the horizontal axis like usual?
@jessejordache18693 жыл бұрын
Couldn't you also take the area of the 45/45/90 triangle between 0 and f(x) = g(x), and then just integrate y=x^2 within the same bounds? You'd Have to fiddle with it a bit; like you might have to take the complement of the integral within the quadrant before you rotated it around the linear function, but i'm just curious. It's too late for me to tackle right now the full "slanted revolution" technique: i'm sleepy, but i don"t see why my kludge wouldn"t work in this case at least.
@blackrasputin33563 жыл бұрын
I think my teacher had us do this in AP calc in high school, but it's been a while so I'm not 100% sure.
@mickmenn23 жыл бұрын
Because it just standard revolution volume with extra geometric steps. ;)