Converting from Cylindrical Triple Integral to Spherical Triple Integral

Question

Use Spherical Coordinates to Evaluate $$\int_{0}^{2\pi}\int_{1}^{2}\int_{0}^{2-r}\frac{r+z}{(r^2+z^2)^{\frac{3}{2}}}r\ dzdrd\theta$$ My Concern: I know how to convert from cylindrical to spherical but, this is my first time to encounter this type of integral such that its radius is starting from $r=1$ to $r=2$

Answer 1

Think about what each variable becomes when written in spherical coordinates - I will use the ones where $$(x,y,z)=(\rho\sin\varphi \cos\theta, \rho\sin\varphi\sin\theta, \rho\cos\varphi).$$

Clearly $\theta$ is the same, $z=\rho\cos\varphi$ by definition and $r=\sqrt{x^2+y^2}=\rho\sin\varphi$. Then the bounds on $\theta$ are the same, and the other ones become $$\begin{cases} 1\leq r\leq 2 \\ 0\leq z\leq 2-r \end{cases} \Rightarrow \begin{cases} 1\leq \rho\sin\varphi\leq 2 \\ 0\leq \rho\cos\varphi\leq 2-\rho\sin\varphi \end{cases} $$ The fact that $z\geq 0$ gives that $0\leq \varphi\leq \pi/2$, thus both $\sin\varphi$ and $\cos\varphi$ are non-negative. From the first we get $$\frac{1}{\sin\varphi}\leq \rho\leq \frac{2}{\sin\varphi}$$ and from the second one $$\rho \leq \frac{2}{\cos\varphi+\sin\varphi}$$ (we have already tackled $0\leq \rho\cos\varphi=z$). Since both must hold, we get that $$\frac{1}{\sin\varphi}\leq \rho\leq \frac{2}{\cos\varphi+\sin\varphi}.\qquad (\star)$$ To finish things off, we have to find the appropriate bounds for $\varphi$. Using $(\star)$ we get that $$\cos\varphi+\sin\varphi\leq 2\sin\varphi\Rightarrow \tan\varphi\geq 1$$ from which we conclude that $\pi/4\leq \varphi\leq \pi/2$.

Notice that this inequality for $\varphi$ can be immediately deduced by drawing a section in the $xz$-plane, since the region is the solid of revolution generated by the triangle with vertices $(1,0)$, $(2,0)$ and $(1,1)$.