Hi there,

I'm having great difficulty with the following problem:

Below is what I believe I have established so far...This question concerns the integral $\displaystyle \int_{0}^{2}\int_{0}^{\sqrt{4-y^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{8-x^2-y^2}}\!z\ \mathrm{d}z\ \mathrm{d}x\ \mathrm{d}y$. Sketch or describe in words the domain of integration. Rewrite the integral in both cylindrical and spherical coordinates. Which is easier to evaluate?

The projection of this integral's domain onto the $\displaystyle xy$-plane is the portion of the circle $\displaystyle x^2+y^2=4$ on $\displaystyle 0\le x\le2,\ y\ge0$.

The bounds on $\displaystyle z$ correspond to

These bounds intersect at$\displaystyle z^2=x^2+y^2$ (cone) and $\displaystyle x^2+y^2+z^2=8$ (sphere).

Below $\displaystyle z=2$ (where the bounds on $\displaystyle z$ intersect), I believe that the cone and cylinder, $\displaystyle x^2+y^2=4$, are completely inside the sphere.$\displaystyle x^2+y^2=4$.

Would it hence be correct to say that the region of integration is the solid lying between the cone and the cylinder, on $\displaystyle x\ge0$, $\displaystyle y\ge0$ and $\displaystyle 0\le z\le2$? I'm struggling to visualize this problem.

When I attempt to move on, and evaluate the integral in cylindrical/spherical coordinates, my solutions differ by a factor of 2.

That is, I evaluated this integral as,

And,$\displaystyle \int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{0}^{ \sqrt{8-r^2}}\!z\ r\ \mathrm{d}z\ \mathrm{d}r\ \mathrm{d}\theta=2\pi$

Can you please help me to identify where I am going wrong?$\displaystyle \int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\sqrt{2}}\!\rho\ \cos\phi\ \rho^2 \sin \phi\ \mathrm{d}\rho\ \mathrm{d}\theta\ \mathrm{d}\phi=4\pi$

Thank you very much.