Double and triple integral in spherical polar coordinates, $\iint\sqrt{4-x^2-y^2}\mathrm{d}A$

Question

Consider the double integral $$ I=\iint\limits_{\mathcal D} \sqrt{4-x^2-y^2}\mathrm{d}A $$ where $\mathcal D=\{(x,Y): x^2 + y^2 \leq 4\}$ is the disc on the $xy$ plane (source)

A.) Use polar coordinates to evaluate the double integral $I$

B.) Give an interpretation of the integral $I$ as the volume of a region $\mathcal R$ in ($x$, $y$, $z$)-space. Hence write $I$ as a triple integral over $\mathcal R$ in cylindrical polar coordinates.

For A, I changed the integral to $$ \int_0^{2\pi}\int_0^2 r\sqrt{4-r^2}\mathrm{d}r\mathrm{d}\theta $$ This eventually led me to get $\displaystyle{\int_0^{2\pi}\frac83 \mathrm{d}\theta}=\frac{16\pi}{3}$. I'm not completely sure about the integral limits I used.

For B, I rewrote the equation as $x^2 + y^2 + z^2 = a^2 = 4$ ($a=2$), and then solved normally using spherical polar coordinates. I got an answer of $\displaystyle{\frac{32\pi}{3}}$. But since the original equation only included the positive area of the sphere, should the volume be $\displaystyle{\frac{16\pi}{3}}$ instead?

Answer 1

a) You got \begin{align*} \iint_{\mathcal D} \sqrt{4-x^2-y^2}\ dA = \int_0^{2\pi}\int_0^{2}r\sqrt{4-r^2}\ drd\theta=\int_0^{2\pi}\frac{8}{3}\ d\theta=\frac{16}{3}\pi. \end{align*} and this is completely correct.

b) Now let $\mathcal R=\{(x,y,z) : x^2+y^2+z^2\leq 4, z\geq 0\}$, then \begin{align*} \iint_{\mathcal D} \sqrt{4-x^2-y^2}\ dA =\iint_{\mathcal R}\ dV=\operatorname{vol}(\mathcal R)=\frac 12\operatorname{vol}(B_2(\mathbf 0))= \frac{16}{3}\pi. \end{align*} Maybe you lost the condition $z\geq0$.

Answer 2

Technically the question asked you to do (b) using cylindrical coordinates, not spherical coordinates. This is basically the same as doing it with polar coordinates and a height function, since you just have $\int_0^{2\pi} \int_0^2 \left ( \int_0^{\sqrt{4-r^2}} 1 dz \right ) r dr d\theta$, where I use parentheses just for emphasis.

But yes, since this is a hemisphere, whichever coordinate system you use, you will end up with $2 \pi 2^3/3$ instead of $4 \pi 2^3/3$.