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

(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)?

$\displaystyle \int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{r^2}^{\s qrt{8-r^2}}z dzdrd\theta$

hope it is!

Can anyone help me with (a) and (c)? If I can get (c), I should be able to do (d).

This is as far as I've got with part (a):

looking at the limits for x and y:

$\displaystyle 0 \leq y \leq 2$ and $\displaystyle 0\leq x \leq \sqrt{4-y^2}$

which means upper limit of x is $\displaystyle x = \sqrt{4-y^2}$

solving this: $\displaystyle x^2 + y^2 = 4$, 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 incorporate z?

For part (c), since it's a 1/4 circle am I right in thinking that $\displaystyle 0 \leq \theta \leq \frac{\pi}{2}$ and $\displaystyle 0 \leq \phi \leq \frac{\pi}{2}$? This is just vague intuition, so if true how do I show this mathematically?

As always, thanks for any and all help