## triple integral, spherical coordinates

Limits, differentiation, related rates, integration, trig integrals, etc.

### triple integral, spherical coordinates

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?
smoog

Posts: 1
Joined: Tue May 04, 2010 2:39 am

### Re: triple integral, spherical coordinates

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

Martingale

Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA