Find the volume of the solid that lies above the cone $z^2=x^2+y^2, z\geq0$, and below the sphere $x^2+y^2+z^2=4z$.
My attempt:
$z=\sqrt{x^2+y^2}$ and $z=\sqrt{4-y^2-x^2}+2$, then
$$\int\int_D(\sqrt{4-y^2-x^2}+2)-(\sqrt{x^2+y^2})dA$$
Switching both surfaces to polar coordinates we have the cone given by $z=r$ and the sphere by $z=\sqrt{4-r^2}+2$
Solving we obtain $r=2$
So the region $D$ is a circle of radius $r=2$. Hence we can describe the region as
$$D=((r,\theta)|0\leq\theta\leq2\pi ,0\leq r\leq2)$$
So we compute:
$$Volume=\int\int_D(\sqrt{4-y^2-x^2}+2)-(\sqrt{x^2+y^2})dA$$
$$=\int\int_D(\sqrt{4-r^2}+2-r)dA$$
$$=\int_0^{2\pi}\int_0^2 (r\sqrt{4-r^2}+2r-r^2)drd\theta$$
$$=\int_0^{2\pi}4d\theta$$
$$=8\pi$$
Best Answer
Your solution is correct. Here I'll show how to visualize this as a body of revolution. It's always helpful to plot what you are looking for. The figure below shows a vertical cross-section on the $x-z$ plane for rotation of the shaded area about the $z$-axis.
First, the simple geometric solution,
$$V=V_{cone}+V_{hemisphere}=\frac{1}{3}\pi r^2h~+~\frac{1}{2}\frac{4}{3}\pi r^3=8\pi$$
since $r=h=2$.
Now, the calculus method. Using any of the standard methods for bodies of revolution (I prefer Pappus's theorem) we have in general for rotation about the $z$-axis that
$$V=\pi\int x^2~dz$$
which in this case amounts to
$$ \begin{align} V &=\pi\left[ \int_0^2 z^2~dz~+~\int_2^4(4z-z^2)~dz\right]\\ &=\pi\left[ \frac{z^3}{3}\biggr|_0^2~+~\left(2z^2- \frac{z^3}{3}\right)\biggr|_2^4\right]\\ &=8\pi \end{align} $$