Triple Integrals in Cartesian Coordinates | Volume between Surfaces

[@madhumithatk8681]

Finally , I've found someone who explains in the most beautiful way :-) ...YAAYY!! SUCCESS:))

[@Junker_1]

Like always the visual style helps a ton. You already have helped me a lot in understanding these things. So thank you.

[@DrTrefor]

Happy to help!

[@giovannianimbono17]

Pls Add subtitles for deaf people, it’s a little hard to catch all the words you’re pronouncing. Anyway great video!

[@sujalkapoor4845]

You can use googles live caption thingy, if that helps

[@udoberk6647]

for anyone wondering, the volume in question is 9/4 * pi

[@harry_page]

Phew, I worked it out and got that number

[@SachinKumar-dy4hh]

do we polar coordinates here to find this out

[@SachinKumar-dy4hh]

i did it with polar coordinates and its coming 9pi/4. is there another way where i wont have to use polar coordinates

[@udoberk6647]

Yes, however polar is by faaaar the easiest. Always adapt your coordinate system to fit the type of problem at hand!

[@edwarddi3833]

the idea you are explaining is so clear. thanks man!

[@AndyOrtiz93]

Bro’s the CEO of Calculus. Got a perfect score on my last exam thanks to these videos!

[@DrTrefor]

Amazing!

[@isaac5990]

This is a criminally underviewed video

[@ogunsadebenjaminadeiyin2729]

If you were my maths teacher🏆, I would never skip class😂😂😂😂

[@DrTrefor]

And I’d give you an A+

[@Christian-mn8dh]

@@DrTrefor for what ?? (;;

[@klemen6502]

Yes you would

[@joe_mama92]

@@Christian-mn8dh for oiling him

[@gshomie9908]

Cause he'd be understanding everything​@@Christian-mn8dh

[@StaticBlaster]

I'm sure you can do quadruple, quintuple, n-tuple integration if needed. I'm sure in string theory, they use 11 integrations or other fancy functions that do the same thing.

[@himalpandey09]

Wow

[@ilias-4252]

For anyone trying to compute this by using "tricks" here is an easy way: Instead of adding up infinitely small volumes, think about adding up infinite disks centered on the z axes. We can see this is possible by substituting r^2=x^2+y^2 and getting f1=r^2 and f2=3-r^2. Next, we know that each of these circles will have area of πr^2, so our integral will look something like integral from a(z) to b(z) of πr^2 dz since our radius will always depend on how far up we are. Now, from z=0 to z=3/2, the radius of our circles will be given by f1=z=r^2 and our bounds for z will be z=0 and z=3/2, and from z=3/2 to z=3 , our radius will be given by f2=z=3-r^2 and our bounds will be z=3/2 and z=3. To skip some calculations we can see that these 2 integrals are exactly the same, since our volume is completely symmetrical above and under our circle of intersection, so we can just calculate the 1st one and double the result. Integral from 0 to 3/2 of πr^2 dz= int πz dz = π (z^2/2) evaluated at 3/2 and 0 = π*9/8. So the answer is 2* (9π/8)=9π/4

[@devjyotiroy4741]

The best explanation in the entire internet ! ❤

[@DrTrefor]

Thank you!

[@devjyotiroy4741]

@@DrTrefor if you can please make a video on physical representation of green's theorem and stoke's theorem ! Interlinking with closed line integral and closed surface integral ❤ love and support ❤

[@ogunsadebenjaminadeiyin2729]

@@devjyotiroy4741 I would really love to see this too.

[@erikawanner7355]

I could never figure out how to even set up triple integrals when I took calc 3! This is amazing!

[@Lexyvil]

This helped a lot! Thanks! I just found it a bit confusing why the dz integral goes from blue to red instead of red to blue when the red paraboloid is below blue, but it really helps visualize how to set it up.

[@flippert0]

Lol, a couple of hours ago, I pondered exactly about the question, what a triple integral actually means. Now this video pops up. Thanks, Dr. Bazett!

[@henryharmon3656]

Any possibility of videos on the big topics and theorems at the end of vector calculus? Line integrals, surface integrals, divergence and curl, Green's Theorem and so on? I'm thinking of using your videos as the basis for a flipped calculus course. Thanks, Trefor.

[@DrTrefor]

Have a whole playlist on vector calc!!

[@henryharmon3656]

@@DrTrefor Great! I’ll check it out.

[@mlop2484]

thanks for helping science!!

[@nadia-sy8cn]

thank you for the best teaching ever

[@olehborys1462]

Like your approach to visualize all your words on graph. Well done!

[@benjaminyellin5095]

Quick question: at 1:39, how would I approach the problem if the drawing of the graph was not given? Like how would I understand that the entire volume is constrained within the intersection circle?

[@nott_applicable]

Few things, the graphs of x^2+y^2 are pretty easy to recognize so just like 2d imagine just shifting them up/down and try to imagine what it would look like. For example I assume you know what x^2 looks like, now flip it so it becomes -x^2, now lets shift it up by 3 and boom we have our function in 2d. Now take another function lets say t^2, can you imagine what these two would look like on the same graph? This sure wouldn't work for harder graphs but for a lot of the simpler ones it does. alternatively you could try to find the intersection on these two functions and look what values the functions take within that intersection. Here for example the intersection is a circle and if you plug in some point inside of this circle in both f1 and f2 you will find that the one is above the other, that combined with some basic analysis to see that they're 3d parabola and you could conclude that it would look somewhat like this. If all else fails you could in worst case also just calculate some points/extrema and from that try to find how the functions are shaped. A lot of it is intuition but often, especially on tests when you don't have a 3d plotter available, they will use common functions.

[@akhildundra7834]

I think you are the greatest mathematician

[@SAMARTHSAMANT]

Fantastic ! Love from india !!

[@dumdum6178]

Can I use polar coordinates instead of x and y?? Edit: nvm i got using polar too.

[@sergiolucas38]

Nice video, the colours and the images are really outstanding :)

[@ANJA-mj1to]

Great to see triple integral of the cones of maxiumum intesity for input of the two interfering sources on the x-axcis

[@AsmitSou]

You are awesome... i learned how to imagine higher mathematics

[@nishugill9477]

Sir ,very helpful video... 🤗🤗Thanku so much... 🙏🙏😊

[@alfieplant6927]

What function would this integral be in respect to. Would it just be 1 since you are finding a finite region?

[@Jofetg]

Great video with clear explanation

[@AyaNi1214]

my master gave us subject to explain in class when I searched it on KZbin I didn't get it even i saw your videos ....thanks 🌹

[@dipalichakrawarti6847]

Very well explained!!!!!!😇😇😇😇😇😇😇

[@riffster]

Best explanation This far 👏🏽

[@somiaelshemy9666]

that was very clear, Thank You!

[@solaire7046]

It would have been easier to follow if the visual representation had its axis labeled.

[@sathyavageesan9507]

THANK YOU SO MUCH THIS HELPED A LOT

[@robinmc142]

How did you get 3/2 rather than 2? Don't you set z=0 for some reason then solve for y?

[@surajv1986]

Thanks, Sir, I had a bit of difficulty in understanding the middle integral limits, May I ask how did you obtain the values of g1(x) and g2(x) for the limits i.e.?

[@agabaedwin7090]

Yeahhh...me too I had the same problem

[@adarsh5997]

hey man i have a hard time changing the order of integration in some questions they ask to integrate by changing the original order

[@paulnokleberg5188]

I converted to cylindrical coordinates, performed the integration, and got 18pi/8, which is approximately 7.07. By looking at the volume and visualizing a 1×1×1 cube based on the coordinate system, it looks like about 7 blocks of that size might fit inside. Is my answer correct? Thanks for a great video!

[@nonamex3052]

i watched my professor video like 4 times and i didnt get it but 7 min here and i have a very good idea. might as well not go to uni and just watch youtube lol. thanks dr

[@video_camera]

If x^2+y^2=3/2, why is the y limit going from -sqrt(3/2-x^2) to sqrt(3/2-x^2) but the x limit isn't going from -sqrt(3/2-y^2) to sqrt(3/2-y^2)? Why is the y gone?

[@NEHAYADAV-jq2uz]

You explained it very well . Thank you so much sir .

[@samytanjaoui8178]

It is a very good explanation. Thanks.

[@muhammadumarsotvoldiev8768]

amazing video. thanks u professor!

[@AtliTobiasson]

Amazing stuff, thanks!

[@studiesplusdotlk978]

A great explanation sir. Really it's the best I ever heard. Today I found this channel when I search. I added my subscribe for this informative valuable channel. I am a Sri Lankan University student who follows Bachelor's degree (Special in Mathematics) It's really help me to improve my maths knowledge.Thanks again sir

[@redbritish]

When we do single variable integral, the output is an area. When we do a double integral, the output is a volume. Why when we do this triple integral, we get a volume again? Isn’t the volume the region we are integrating? Is the output volume because we just integrated 1? And if there were an actual function in the integral, would we get a 4d volume as output? Great video though!

[@dannyatherton7557]

Thanks a ton! Explained very well

[@MsLegobuilders]

wait so what equation are you integrating? or do you just integrate the number 1 using these bounds?

[@DrTrefor]

Exactly. A triple integral of 1 gives a volume.

[@hikmatullahpakhtoon3694]

Sir! Both double and triple integral gives us volume then what's the difference between these two.

[@DrTrefor]

Triple integrals only give volume if you integrate the function f(x,y,z)=1, but you can integrate any other function too

[@hewwo3743]

@@DrTrefor this changes how I see integrals completely... thank you

[@Anythiny]

math.stackexchange.com/questions/649034/finding-volumes-when-to-use-double-integrals-and-triple-integrals#:~:text=Here%20we%20have%20obtained%20the,x%2Cy)%20for%20free.&text=You%20can%20use%20both%20double%20and%20triple%20integrals%20when%20calculating%20a%20volume.&text=The%20only%20difference%20is%20that,double%20integral%20is%20a%20shortcut.

[@vaderanomaly1573]

I love your videos but can you please get a lavalier mic for $30 pleaaase, or maybe its the room that needs foam idk. Please please plase. It will genuinely improve your videos. Thanks for these explanations.

[@DrTrefor]

I’ve actually got one for newer videos!

[@vaderanomaly1573]

@@DrTrefor ahh beautiful, let me pass calc 3 first i guess lol

[@udkspi9234]

would you integrate the same way if the circle of intersection would have its centre in (0,0,0)? So the Body that is constructed has -2

[@hubenbu]

This is a great conceptualization. I cut the corners to compute the volume of the blue paraboloid cut between xy plane and the x^2 + y^2 = 3/2, and then double it. the 2 differentials I use are dA and dh. The result I got is 27pi/4, which I'm not quite certain.

[@nicholasrodriguez1234]

Why isn't the upper bound of the x integral sqrt(3/2 - y^2) and vice versa?

[@lucieneyvrard5414]

Because at each step you suppose to get rid of one of the variable . When integrating in z you get rid of the z (and you integrate like if it was a vertical line) When you integrate in respect with y you get rid of your y variable and you still integrate in line but this time imagine you are integrating in a "crossline" you can interpret that as a plan And in x you finallly get rid of your x and integrate the last plan all around the x value so minus the radius and plus the radius (because i suppose the circle got his circle at zero) I hope that help, and i hope i am right

[@UniformDelta00]

Because x is the last variable. In this order (z then y then x), it is z and y that are express in terms of x. And x really can vary from - sqrt3/2 and +sqrt 3/2. You have to give x real boundaries independant from z and y so it makes sense (or you will be left with an indeterminate result). You cant express x in terms of y and y in terms of x. And you have to choose an order, thats what he says at the end of the video

[@AbjSir]

you can do in any order that makes sense to you

[@dhruvupadhyay8126]

I would recommend you to use a different mic not the camera's mic if you are using it

[@dvirjacobross1826]

@@DrTrefor try a lav mic

[@HRISHIKESHSANDIPDESHPANDE]

i pretty muchgot it when we were finding volume using double integration. but in thriple integration, like in this example, u assumed F(x,y,z) = 1. could u give some intuiton behind this. ( we kinda used this similiar thing to find area using double integration(by assumng the height( z = 1) is constant). i am not able to relate this hright term in triple integration.

[@fatmakslal8103]

Your videos are great, and it would be much greater if your voice not sounded like you were talking in an empty room, I mean echo.

[@happytrigger3778]

Hi Sir, could you please explain why x bounds are between -sqrt(3/2) and +sqrt(3/2)?

[@happytrigger3778]

is this by projecting the intersection on the x axis thus y becomes zero and x assumes -sqrt(3/2) and +sqrt(3/2)?

[@aravindhsm1287]

The circle is centred at the origin,integrating for the entire circle,you need to take both the limits.

[@NEHAYADAV-jq2uz]

Love from India ❤️

[@andrejcernansky6520]

How is the general method of setting limits in the integral ? Thank you

[@rewardkhaled6359]

can we solve it using double integral of 3-x^2-y^2-(x^2+y^2)dydx ?

[@ayushabad1862]

sir how limit of x came?? x from root of 3/2 to -root of 3/2

[@carultch]

Set the two equations equal to each other, and you'll find a circle given by the equation x^2 + y^2 = 3/2. The general equation for a circle centered on the origin is x^2 + y^2 = R^2. Thus, the radius R = sqrt(3/2). This means the range of x-values is from -3/2 to +3/2.

[@benjaminyellin5095]

Any chance for the solution? I tried following along by solving and got 27π/16 but I'm pretty sure I messed up somewhere... Thanks!

[@mannomanno2570]

What's the difference between double and triple both give volume

[@carultch]

A triple integral has an additional advantage if it isn't volume that you are interested in, but rather mass. Suppose this solid were not uniform, and were made out of some kind of resin cast with a varying concentration of a heavy sand. You could have a density function that depends on x, y, and z, and doing a triple integral with the density function would allow you to find the total mass of the solid. This example could be done with a double integral.

[@Lexyvil]

So making equations equal to each other always gives their intersection?

[@DrTrefor]

If both written as z=Blah

[@stevenwilson5556]

Thanks for this video.

[@nuthakantirohan4685]

Well I heard that triple integrals give hyper volume in four dimensions what does that mean and what if there is a function inside triple integral what does it mean like under surface integral it means we are calculating volume under the function then what does it mean for a triple integral

[@carultch]

In 4 dimensions, yes, a function inside a triple integral would give hyper-volume. A real-life application in our 3-dimensional universe of a triple integral, would be the mass of region of space of varying density, where density would be the integrand. Another real life application, is moment of inertia of a solid, where rho*r^2 is the integrand, with rho being density, and r being radius from the axis of rotation. Moment of inertia would be a triple integral, even with uniform density, although it often can simplify to lower order integrals for situations with symmetry to use to your advantage.

[@userozancinci]

Hocam you are the freakin best!

[@BadAss_691]

Thanks doc it’s nice

[@VishwamHemangPatel]

Thank you so much bro

[@hassanali-ii8yx]

How can l solve it x^2+y^2=a^2 x^2+z^2=a^2 find the volume between tow surfaces that describe above

[@dalibormaksimovic6399]

Hi. I am interested in how one can determine boundaries of integration when there is no a explicit function for z in terms of y, or y in terms of x. For instance, calculate the volume of body bounded by following surfaces: x^2+y^2 = cz, x^4+y^4=a^2(x^2+y^2) and z=0.

[@nuthakantirohan4685]

hey one of the function is a curve not a surface as it has no z values it lies in xy plane and if you want volume under the two curves id think its hard

[@nicka4881]

Good video but was somewhat confusing at first without the axes being labelled

[@axisepsilon514]

This is honestly a better explanation than my professor who just said "triple integrals are integrating a 4d volume" . Not that my professor was wrong but this gave me a much better understanding and application of triple integrals

[@DrTrefor]

Glad it helped!

[@441harinder]

Thank you sir

[@samytanjaoui8178]

Dos it make difference if I only integrate the function (3- x^2-y^2) over x any y?

[@nikolozchaduneli3875]

Hey great vid! Could somebody explain how the boundaries of x were figured out?

[@alfieplant6927]

The boundaries of x are the positive and negative values of the radius of the circle that is projected onto the xy plane. If you imagine going from the far left side of the circle to the far right side, those are your x bounds

[@video_camera]

@@alfieplant6927 Yeah, but aren't so the y boundaries? In a circle like that, both boundaries go from -R to R. Why are the y limits x-dependant?

[@aswinbabu1213]

How do we know which curve is above and which one below

[@milan2499]

How do ypu make theee videos?

[@discoveryofphysics9303]

If this is a triple integral , then shouldn't the volume be 4th dimensional object?

[@DrTrefor]

Yup. Although that doesn't necessarily mean a fourth SPACIAL dimension. Could be accumulating density over a 3D region or something like this.

[@discoveryofphysics9303]

@@DrTreforunderstood sir... It helps. Please make video on physical interpretation of gradient, Divergence and curl. That will really help. Thank you sir.

[@harnishkaur95]

Integrate the function f (x, y, z) in the given region. A sphere of radius R centered on the origin; f = x2 + y2 + z2. please solve this question. i really need help in this question, please help me out.

[@Sathrandur]

The equation for your sphere shall be x^2 + y^2 + z^2 = r^2 You can rearrange the equation from there to put it in terms of z= ... or f(x,y)= ... Then you can integrate for the whole sphere or just the part with poisitive x, y and z values and multiply out by eight to get the complete volume of the sphere. Your limits of integration shall be either from -r to r or 0 to r depending on which method you use.

[@josephhajj1570]

Can you prove the jacobian plz in the next video

[@josephhajj1570]

@@DrTrefor no a proof that when I change variables I should multiply by the jacobian for any transformation Thankyou mister I like your attitude

[@sudharsanr4466]

தலைவா மிக்க நன்றி

[@freakshow1010]

Sounds like New Rockstar Guy.

[@michaelonskis]

But what are you integrating? There's nothing between the right-most integral sign and dz

[@DrTrefor]

Just 1.

[@HosRo4161]

Thank you!!

[@hdheuejhzbsnnaj]

Trevor, can we get Line Integrals, Vector Fields, and Green's Theorem? More juicy stuff plz. 😋

[@DrTrefor]

Yes! It’s all there in my vector calc playlist:D

[@hdheuejhzbsnnaj]

@@DrTrefor amazing, how did I not know?!? Thank you 🙏

[@saurabhsingh-ow7ue]

thank you sir............

[@Dikke_Jonko_010]

U are a legend

[@marcuswassenius1862]

imagine putting so much work into the visuals but having the microphone of a nokia

[@ΚωνσταντίνοςΛαζαρίδης-ξ9ι]

Thanks!

[@ericcartman1168]

How did you figure out the intersection point on the z axis, my calculator can do it with a system of equations but I'm not sure how to do it by hand. Great video, it helped a lot

[@chaoticgood8645]

I believe he mentioned it at the very beginning( 0:41 ): you set the equation of both shapes =z if they aren't already given that way(remember f(x,y)=z), and then you set those two equations equal to each other by the z. (Example: 5x^2 + 7y^2 = z, and 8x^2 + y^2 = z for the other graph. The intersection of these graphs is 8x^2 +y^2 = 5x^2 +7y^2. Then you simplify by combining like terms and you should get a number = some form of x^2 + y^2 equation which is the eq of the intersecting xy plane (which when graphed will look like a circle when you look down on it)

[@Salarr]

Realistically this is just a double integral

[@user-ci9te3yt6t]

Apparently it is just a double integral, but how would describe getting the volume or say the lengths or the area of the sum of the points of any the line between the curves Z= X2+Y2 and Z= 3-X2-Y2. The area of a line whose 2nd is 1unit of width then it's area is sane as its length, and a line whose area and length are same and has 3rd dimension with a length of 1 unit then its volume and length/height are same. So Z is what gives those lines a height and you can easily subtract the two function or take Z as 1 and integrate it giving you again the difference between the two function. It's just another way of looking at it, it's just beautiful! Thanks!

[@everettmcinnis5858]

This problem is much easier to do if you use polar coordinates instead of Cartesian coordinates. I got an answer that is different from all of the answers listed in the comments below. which are all different. (That says something in itself.) I got an answer of 5pi/2. Does anyone agree?

[@sapnasingla9038]

Pls add subtitles 🙃🙃

[@joy4467]

Kardashian? No man i choose Cartesian!

[@octopusknight2207]

"outttt" , there it is , you are canadian aren't you?

[@eee_bangla]

tnx

[@amir-ali8850]

Tnx

[@tanjinaaktar1146]

Great