Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Hi we have the following problem:

$\iiint x\,dx\,dy\,dz$ limited by the paraboloid of equation $x=4 y^2+4z^2$ and for the plane $x=4$.

We are having difficulty on finding the limits of each integral and how to turn to polar coordinates.

Could you offer any tips? I can provide more details on the comments on what we tried. Thank you.

share|improve this question

2 Answers 2

Normally, polar coordinates are given as $x=r\cos \theta $ and $y=r\sin \theta $, but that does not necessarily have to be. In this case, I think it is wise to use $y=r\cos \theta $ and $z=r\sin \theta $. Then, your limits of integration would be from the paraboloid to the plane (in $dx$), then from zero to the radius of the bounding circle in the $y$-$z$ plane, then from zero to $2\pi$ in $\theta$ to complete the full revolution around the circle.

share|improve this answer
    
Thank you we are trying this right now! –  Edoardo Sep 7 '12 at 3:03
    
If you use y=rcos(θ) and z=rsin(θ), what do you use for x since the integral is ∫∫∫xdxdydz? –  Edoardo Sep 7 '12 at 3:11
    
You don't need to substitute anything for x... the change in variables result in a differential of the form: $r dxdrd\theta$. –  Paul Sep 7 '12 at 13:42

An idea: make a substitution change $\,x\leftrightarrow z\,$, so that you have the paraboloid $\,z=4x^2+4y^2\,$ and the plane$\,z=4\,$, and now use cylindrical coordinates and the symmetry of the paraboloid around the $\,z-\,$ axis:

$$\iiint z\,dx\,dy\,dz=4\int_0^1dr\int_0^{\pi/2}d\theta\int_{4r^2}^4 4r^3\,dz$$

Please do note that $\,4r^3=4r^2\cdot r=z|J|$, where $|J|$ is the Jacobian of the transformation into cylindrical coordinates.

Added: We can also do the following:

$$\begin{align} \iiint z\,dx\,dy\,dz &= 4\int_0^1dr\int_0^{\pi/2}d\theta\int_{4r^2}^4 zr\,dz\\ & = 2\pi\int_0^1r\left[\frac{1}{2}z^2\right]\Bigg|_{4r^2}^4\,dr \\ & = \pi\int_0^1(16r-16r^5)\,dr \\ & = 16\pi\left(\frac{1}{2}-\frac{1}{6}\right) \\ & = \frac{16\pi}{3} \end{align}$$

which is the right answer according to the book. I still am not sure what went wrong with the first method which I leave here for others to check and comment.

share|improve this answer
    
Sorry the answer doesn't matches with the calculator :/ –  Edoardo Sep 7 '12 at 3:15
    
What's *the answer" and by whom ( the book's?)? And what did you get above? Just like that it's hard to say what could have happened... –  DonAntonio Sep 7 '12 at 3:16
    
The answer according to my text book is 16pi/3 and what I got from above was 64pi/15 –  Edoardo Sep 7 '12 at 3:18
    
Well, the result of the integral in my answer is $\,\frac{8\pi}{3}\,$ , so it is off by a simple factor of 2 which I can't see right now from where it comes. Perhaps later I'll add something to my answer. Anyways, your calculation seesm to be wrong –  DonAntonio Sep 7 '12 at 3:28
    
@Edoardo , what book is that? –  DonAntonio Sep 7 '12 at 3:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.