Calculate the following integral: double integral over D
Calculate the following integral: $$ \iint_D (x^2+y^2)\,dx\,dy, $$ where $D$ is the disk $x^2+y^2\leqslant2ay$ (radius $a$, centered at $(0,a)$, tangent to the $x$-axis at the origin).
Explanation of Method
In polar coordinates the boundary circle $x^2+y^2=2ay$ becomes the simple polar curve $r=2a\sin\theta$, with $\theta$ ranging over $[0,\pi]$ to trace the whole disk. Since $x^2+y^2=r^2$, the integrand becomes $r^2$, and the extra factor of $r$ from the area element makes the radial integral a pure power of $\sin\theta$.
- Convert the boundary circle to polar form: $r=2a\sin\theta$.
- Integrate with respect to $r$ from $0$ to $2a\sin\theta$.
- Integrate the resulting $\sin^4\theta$ term over $[0,\pi]$ using the Wallis formula.
Step 1 · Convert to polar coordinates
With $x=r\cos\theta$, $y=r\sin\theta$, the boundary $x^2+y^2=2ay$ becomes $r^2=2ar\sin\theta$, i.e. $r=2a\sin\theta$. This traces the full disk as $\theta$ runs from $0$ to $\pi$ (the circle lies entirely in $y\geqslant0$).
Step 2 · Set up the integral
$$ I = \int_0^{\pi}\int_0^{2a\sin\theta} r^2\cdot r\,dr\,d\theta. $$
Step 3 · Integrate with respect to $r$
$$ \int_0^{2a\sin\theta} r^3\,dr = \left[\frac{r^4}{4}\right]_0^{2a\sin\theta} = \frac{16a^4\sin^4\theta}{4} = 4a^4\sin^4\theta. $$
Step 4 · Integrate with respect to $\theta$
$$ I = 4a^4\int_0^{\pi}\sin^4\theta\,d\theta. $$
By the Wallis formula, $\displaystyle\int_0^{\pi/2}\sin^4\theta\,d\theta = \frac34\cdot\frac12\cdot\frac{\pi}{2} = \frac{3\pi}{16}$, and by symmetry about $\theta=\frac\pi2$, $\displaystyle\int_0^{\pi}\sin^4\theta\,d\theta = 2\cdot\frac{3\pi}{16} = \frac{3\pi}{8}$.
Step 5 · Combine the results
$$ I = 4a^4\cdot\frac{3\pi}{8} = \frac{3\pi a^4}{2}. $$
Additional Notes
- As a consistency check, this exceeds the corresponding integral over a disk of the same radius $a$ centered at the origin, $\frac{\pi a^4}{2}$, by a factor of $3$ — reflecting the extra distance-from-origin contributed by the off-center placement.
Result
The solution to the integral is: