Volume of revolution slant axis

In this video, I calculate the volume of a solid of revolution, but this time where the solid is rotated about the line y = x instead of the usual horizontal or vertical lines in calculus! This requires us to rotate the plane by 45 degrees, which in turn requires a sprinkle of linear algebra. This again should show how powerful linear algebra is, it allows us to solve impossible calculus problems. Enjoy!
Note: My apologies that I didn't calculate the integral at the end. Here’s how to calculate the integral: • Square rooty squared i... The answer is pi/120

[@drpeyam]

Here’s how to calculate the integral: kzbin.info/www/bejne/g17WYWN7nLZqmqM The answer is pi/120

[@frozenmoon998]

It's sad that the video is private and was unseen until now. Great that you shared that extra spice in there.

[@brunolevilevi5054]

Hey Dr Peyam, I was curious, instead of rotating the plane can't you use the distance from point to line formula and get a new function? from a point (x0,y0) or in other words (x0, 2^(-1/4)sqrt(x0) ) the distance from that point to the line y = x is abs(x0-2^(-1/4)sqrt(x0)) / (sqrt(2)). Since that is the radius, I just integrate pi * radius squared from 0 to 1/sqrt(2) . The problem is I got a different answer, but I'm not sure if I made an error in calculations or if the logic is flawed. Is that also a valid method to get the volume or is there a problem with it? Edit: I got the bounds of integration wrong, now I believe its right

[@cameronkhanpour3002]

Hey!!!! I remember I tried asking this question on r/calculus when I was learning about volumes of revolution but people seemed to misunderstand my question :( thank you so much for fulfilling my curiosity!

[@mcfloh55]

Ungeahnte Perspektiven eröffnen sich dank der lesefreundlichen Frontal-Kameraeinstellung - unbedingt für alle kommenden Videos so beibehalten. :)))

[@AndDiracisHisProphet]

Mein Ironiedetektor funktioniert grade nicht ganz richtig. Meinst du das ernst? Ich finde, natürlich kann Dr. Peyams hübscher Rücken auch entzücken, aber ich kann so nicht sehen was er gerade schreibt...

[@AndDiracisHisProphet]

@Poo Guy also ironie

[@williamadams137]

How about rotating about a parabola?

[@GreenMeansGOF]

Because of this video, I finally understand why the rotation substitutions require the inverse matrix(clockwise) as opposed to the counterclockwise matrix.

[@serkanergun0]

Is it possible to solve this problem by calculating the distance of the point (x,y) to the line x=y as r=|x-y| / sqrt(2) and then using r as the radius of the circle and integrate with respect to x as int( (x - f(x))^2 * pi / 2 dx) ?

[@zajoliroA]

You could also find the area and center of mass of the region. Then find the distance from the center of mass to y=x and multiply that distance by the area and by 2pi.

[@Hey-io4xx]

It does work indeed and is even easier than the method shown in the video, at least I think so, thanks for that. But I'm just not sure WHY it works, could someone maybe explain?

[@colorx6030]

@@Hey-io4xx Not sure if this is really it but there's a certain theorem in Engineering that looks a lot like the solution of OP. It's called Pappus-Guldinus Theorem if you're interested.

[@Rundas69420]

Combining calculus and linear algebra is the same level of awesomeness as combining pork and beef to form an even better patty for your burger (or the 2 most important ingredients in your favorite dish).

[@thedoublehelix5661]

Functional analysis

[@БогданЛисянський]

Great video! P.S. This is a guy with olympiad from the last gaussian integral video. I want to say that I wrote it not bad. I've done 9 of 10 and all answers were correct. Thank you for inspiring me for math.

[@bulldawg4498]

I, too, solved the problem theoretically before watching the video but didn't think the algebra would be so messy ... Again, another outstanding video by the good doctor ...

[@mohammedal-haddad2652]

That's a cool volume of rotation problem. Thank you very much.

[@crazye7132]

I love problems where you need to change the basis

[@tomatrix7525]

This was cool! Rotation matrices are used the whole time in 3d video games to make stuff rotate

[@عطاالدهامشة-ع8ح]

if xy=2,rotate about y=-x from x=√2 to 2 find volume

[@drpeyam]

Same technique

[@عطاالدهامشة-ع8ح]

@@drpeyam if you have time answer my question i need the answer very necessary with best wishes

[@drpeyam]

Just watch the video

[@ТимофейЧерников-щ2х]

Solution is unnecessary complicated, but I loved the problem. Good video!

[@triton62674]

how would you do it?

[@ТимофейЧерников-щ2х]

Unfortunately I can't post pictures here, but I will try to explain it in plain text. Main idea is to understand that distance from point with coordinates (x0, y0) and line y=x is (y0-x0)/sqrt(2). You can derive it using linear algebra but it's easier to do with simple drawing and remembering the fact that altitude from right angle of an isosceless right triangle to it's hypotenuse is length of it's leg divided by sqrt(2). Well, after you got that, just integrate ( (f(x)-x)/sqrt(2) )^2 * pi * sqrt(2) (multiplied by sqrt(2) because of the slope of line y=x) from 0 to 2^-0.5, where f(x) is 2^-0.25 and you got pi/120. Okay once I wrote that down I realized it may not seem clear how I did it. I just want to say that method from video is probably most efficient when dealing with some hard problems rather with simple as in video.

[@triton62674]

@@ТимофейЧерников-щ2х could you pls upload to imgur and share thanks

[@ТимофейЧерников-щ2х]

imgur.com/gallery/cmz831p And about multiplying by the 2^0.5 in the end: you have to do this because shape that you get from (f(x)-x)*2^-0.5 is not only rotated but also stretched horizontally by a factor of 2^-0.5 I don't know how to properly explain it you can just plot both functions and see it yourself

[@MrRyanroberson1]

I would try the problem, but an instant glance at the video description means that I would end up doing the same thing. Or maybe not. I'll start anyways: y = -4\2 * 2\x, rotating such that the line y=x becomes y' = 0, y' = (y-x)/2\2; x' = (y+x)/2\2, this both conserves the area of the plane and sets the old line y=x (y-x)/2\2 = 0 to be y'=0. Solving for x and y given x' and y' gives us 2\2 * (y' + x') = 2y therefore (y'+x')/2\2 = y, and (x'-y')/2\2 = x. Plugging back in, we get (y'+x')/2\2 = -4\2 * 2\((x'-y')/2\2) = 2\((x'-y')/2), therefore we get that y' + x' = 2\(x'-y'), which is somewhat difficult to isolate y in... Substituting the blank space in the search box on wolfram alpha with the equation gives y = -x ± sqrt(8x+1)/2 - 1/2. Finally, since it's revolving around the x axis, it's actually just asking for the integral of pi * y^2 dx. Plugging in... pi * int(x^2 ±x sqrt(8x+1) + (8x+1)/4 - x ±sqrt(8x+1)/-2 + 1/4)dx, we can solve some of the integral first... pi (x^3 /3 + x^2 + x/4 -x^2/2 + x/4 ± int(x sqrt(8x+1) - sqrt(8x+1)/2)) = pi (x^3 /3 + x^2/2 + x/2) ±.5 int((2x-1)sqrt(8x+1)). Perhaps a U sub would do it... u = 8x+1, du = 8dx. The integral part is 1/64 int((u-5)sqrt(u))du which is solvable: 1/64 * ((2/5)u^(5/2) - (10/3)u^(3/2)) = 1/64 * ((2/5)(8x+1)^(5/2) - (10/3)(8x+1)^(3/2)). All of this is fine but i get a strong feeling that ± means i'm very wrong

[@heliocentric1756]

I solved it simply by disk method, where R=(f(x)-x)cos(45), and the infinitesimal of integration= (dx)/cos(45). In general if the line is y=kx, and the angle it makes with x-axis is a, then R=(f(x)-kx)cos(a), and the infinitesimal of integration= (dx)/cos(a)

[@qxtr5853]

Isn't the radius of a slice of the body just the distance from a Point (x,f) to a corresponding Point ((x+f)/2,(x+f)/2) on the y=x axis? If you take an Intergal for the rotation shape with that formula and multiply it with sqrt(2) to account for the different thickness of dx in that perspective, I think you should get the right result without any LinA.

[@VortexGamesYT]

Marvelous solution, Dr 3.1415 am. Pappus theorem it's other way.

[@MsSlash89]

Hi doctor πm, thanks for the video as usual. I was wondering, however, how does one formalize rotation of a curve or a surface around an axis. What does it look like in higher dimensions? Wouldn’t it be nice to make an extravaganza about it?

[@funkycude57]

Doctor peyyaaaaaaaam

[@math4u769]

We can also solve the problem by second theorem of pappus which gives the same result: pi/120.

[@josephdomingo4413]

Thank you sir :) Really having a tough time, because of this video, my stress has lessen 😊

[@Nevermind445]

couldnt you have just said y= *thingy* - x ? that would have moved the line to be horisontal

[@ben11628]

Is it possible to rotate the function horizontally on the diagonal?

[@berenjervin]

This works well if you are rotating around a linear function like Y=X. Can you do one where you rotate around a non linear axis? e.g. rotating y=cos(x) around y=x^3?

[@drpeyam]

But that would be strange, imagine rotating f(x) = 0 about y = x^2, what would that look like? There would be some overlap in the volume

[@wahyuhidayat7042]

@Dr peyam...if a^2+b^2+c^2+d^2=36. The highest value of 2a+3b+6c+24d ? The answer= 137 7/8 but I can't solve this question how to do it?

[@drpeyam]

Lagrange multipliers :)

[@wahyuhidayat7042]

@@drpeyam thank you

[@neilgerace355]

Thanks! Very interesting

[@wooyoungkim2925]

It is great !!

[@zivssps]

Actually, there is a direct formula. Integral of pi * cos(theta) * (f(x) - tan(theta)*x)^2 between a and b, where f(x) is the function that rotates, theta is the positive angle between the line and the x axis, and a and b are the x value of the intersection between the function and the line.

[@drpeyam]

How did you get that?

[@zivssps]

@@drpeyam A lot of calculations. You can check that is true. Basically, I tried to evaluate the volume like we evaluate normally across the x axis. Maybe I can upload my calculations tomorrow.

[@benjaminarias5193]

Cool!!!

[@NoName-eh8fz]

Is this a reupload? Or why is the video with the integral already online? Btw very interesting solution!! :)

[@drpeyam]

It was unlisted for a long time, but now it’s officially published

[@NoName-eh8fz]

@@drpeyam Ok thank you!

[@txikitofandango]

I did it by "cone method", integrating over all the areas of cones. The surface area of a cone = pi * radius * slant height. The slant height is just f(x) - x. The radius is 1/sqrt(2) times the slant height.

[@txikitofandango]

Of course, I got the wrong answer with this method. I got pi/15. So...

[@txikitofandango]

I think the problem is, what's the volume of a very thin sheet draped over a cone? Is it just the area of the cone times dx?

[@sodahead0]

One way to check for volume is do the calculus manually, then design the slanted axis revolution in CATIA. Take a measurement of the volume in CATIA and evaluate the integral. If the results are close, then the formula is good. One thing is CAD software is very math-based, and it is a good way to check derivations.

[@eliasarguello9961]

is it possible to find the volume of a region that is bounded by a function f(x), sin(x), and two x values where f(x) intersects sin(x) with no intersection between the two x values, and then rotate f(x) with sin(x) being the axis of rotation?

[@husklyman]

Volume about y=mx+b

[@drpeyam]

I think you just shift it

[@ozaman-buzaman9300]

Tan( α)=m find alfa and rotate to his angle