Is there a technique to identify the $r$ limits, and function which is being integrated?

Question

I am given the following example of which I have the solutions the already. I just wonder how the regions which are in this particular format: \begin{equation}\iint_Rf(r,\theta)r\ dr \ d\theta\end{equation} Using Polar regions I am given these one region: \begin{align}\text{Inside}\ x^2+y^2+z^2&=9\\ \text{Outside} \ x^2+y^2&=1\end{align} I have to find the volume of the solid enclosed by these two multi-variable functions. I have the answer for this problem is given as the following: \begin{equation}8\int_\limits{0}^\frac{\pi}{2}\int_\limits{1}^{3}r\sqrt{9-r^2}drd\theta\end{equation} This is where my confusion lies, I understand the 8 is that since its a sphere, and it transverse the eight octants. My question is why the $r$ is going from $1$ to $3$, and why is my function $\sqrt{9-r^2}$?

Answer 1

Note that “inside $x^{2}+y^{2}+z^{2}=9$” means $z^{2}\le9-x^{2}-y^{2}=9-r^{2}$, which means $-\sqrt{9-r^{2}}\le z\le\sqrt{9-r^{2}}$. Note that if $r^{2}>9$ then $9-r^{2}<0$, so $z^{2}\le9-r^{2}$ would be impossible. Therefore, this also forces $r^{2}\le9$.

“outside $x^{2}+y^{2}=1$” means $x^{2}+y^{2}\ge1$, so $r^{2}\ge1$. Since you probably want to integrate only over positive values of $r$, we can just take the $r\ge1$ part of this. Combining this with $r^{2}\le9$, we have $1\le r\le3$.

Therefore, the volume should be $$\int_{0}^{2\pi}\int_{1}^{3}\left(\sqrt{9-r^{2}}-\left(-\sqrt{9-r^{2}}\right)\right)r\,\mathrm{d}r\mathrm{d}\theta$$

$$=\int_{0}^{2\pi}\int_{1}^{3}\left(2\sqrt{9-r^{2}}\right)r\,\mathrm{d}r\mathrm{d}\theta=2\int_{0}^{2\pi}\int_{1}^{3}r\sqrt{9-r^{2}}\,\mathrm{d}r\mathrm{d}\theta$$

$$=2\int_{0}^{2\pi}\left(\int_{1}^{3}r\sqrt{9-r^{2}}\,\mathrm{d}r\right)1\,\mathrm{d}\theta=\cdots$$

If we had cut the xy-plane into four quadrants and noted the volume would be the same for each quadrant, we would have instead started with $$4\int_{0}^{\pi/2}\int_{1}^{3}\left(\sqrt{9-r^{2}}-\left(-\sqrt{9-r^{2}}\right)\right)r\,\mathrm{d}r\mathrm{d}\theta=8\int_{0}^{\pi/2}\int_{1}^{3}r\sqrt{9-r^{2}}\,\mathrm{d}r\mathrm{d}\theta\text{.}$$