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by **smoog** on Tue May 04, 2010 5:23 am

This question concerns the integral $\int_{0}^{2}\int_{0}^{\sqrt{4-y^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{8-x^2-y^2}}z dzdxdy$
(a) Sketch or describe in words the domain of integration.

(b) Rewrite the integral in cylindrical coordinates.

(c) Rewrite the integral in spherical coordinates.

(d) Which version of the integral would you choose as being the easiest to evaluate?
Evaluate the integral using the version you have chosen.

Is this correct for (b)?
$\int_{0}^{2\pi}\int_{0}^{2}\int_{r}^{\sqrt{8-r}}z dzdrd\theta$
hope it is!

Can anyone help me with (a) and (c)? If I can get (c), I should be able to do (d). Thanks!
edit: I've done some thinking and looking at the limits for x and y:
$0 \leq y \leq 2$ and $0\leq x \leq \sqrt{4-y^2}$
which means upper limit of x is $x = sqrt{4-y^2}$
solving this: $x^2 + y^2 = 4$ , which is a circle with radius 2.
Thus D is a 1/4 circle radius 2 centered around the origin. But how do I get to the next part and use z?

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by **Martingale** on Sat May 08, 2010 11:03 pm

smoog wrote:This question concerns the integral $\int_{0}^{2}\int_{0}^{\sqrt{4-y^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{8-x^2-y^2}}z dzdxdy$

Is this correct for (b)?
$\int_{0}^{2\pi}\int_{0}^{2}\int_{r}^{\sqrt{8-r}}z dzdrd\theta$
hope it is!

no,

$\int_{0}^{\pi/2}\int_{0}^{2}\int_{r}^{\sqrt{8-r^2}}z r dzdrd\theta$

If you still need help with the others let us know

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triple integral, spherical coordinates

triple integral, spherical coordinates

Re: triple integral, spherical coordinates