Use cylindrical coordinates to evaluate the triple integral

$$\iiint_E \sqrt{x^2+y^2}dV, $$ where $E$ is the solid bounded by the circular paraboloid $z=16−4(x^2+y^2)$ and the $xy$-plane. Please Help I am confused.

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    $\begingroup$ you should edit your post to include a more descriptive title. Then, recognise that the integrand is r (the radial co-ordinate), change variables as normal and remember to include the jacobian. Then convert your ranges in cartesians into ranges for cylindrical co-ordinates, so, for instance, z will range from $0$ to $16 - r^2$. Now you should be able to just do the integral. $\endgroup$ – user27182 Mar 22 '13 at 1:51

First draw a picture:

enter image description here

You then just need to set up the integral according to the picture as follows:

$$\int_0^{16} dz \: \int_0^{\sqrt{4-z/4}} dr \: r^2 \: \int_0^{2 \pi} d\theta$$

What is going on here? The volume element $dV = r\,dr\,d\theta\,dz$. The volume is rotationally symmetric as you can see, so there's no dependence on $\theta$. Note also that I choose to integrate disks parallel to the $xy$ plane through $z$; this involves solving for $r$ as a function of $z$. Note the extra factor of $r$ comes from your specification of the integral of $r \, dV$. Finally, we integrate over $z$ from $z=0$, i.e., the $xy$ plane, through to the top of the solid at $z=16$.

We then need to evaluate the integral. I'll reduce it to a single integral for you to evaluate:

$$\frac{2 \pi}{3} \int_0^{16}\: dz (4-z/4)^{3/2}$$

You should be able to do this one out.

  • $\begingroup$ Thanks Im learning so much now. A++ $\endgroup$ – Michael Rametta Mar 22 '13 at 1:51
  • $\begingroup$ @MichaelRametta If you liked this answer, then you can accept it by clicking the green check mark underneath the score. It's considered bad manners on this site to not accept helpful answers. $\endgroup$ – A.S Mar 22 '13 at 2:24
  • $\begingroup$ @Ron Gordon It might be helpful to change the order of integration from d(theta)drdz to dzdrd(theta). In that case the resulting single integral would be more trivial to solve than the one provided by your answer. $\endgroup$ – Pratik Singhal Nov 17 '13 at 7:30
  • $\begingroup$ @user189505: thanks for your input, but really, how much more trivial does one need than an integral of $u^{3/2}$? I think the form I provided here is plenty simple enough for the OP to carry out his calculation. If you think you have a better way, then by all means, provide it in another answer. $\endgroup$ – Ron Gordon Nov 17 '13 at 7:37

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