Evaluate this double Integral in polar coordinates. $$ \iint y^2(a^2-x^2)^{0.5}dxdy $$ over $x^2+y^2\le a^2$
I have evaluated this in $x-y$ plane and got $32a^5/45$. Please help in evaluating the same in polar coordinates.
Ive tried putting $x=r\cos\theta$ and $y=r\sin\theta$ and After considering the Jacobian I am getting this expression
$$ \int\limits_0^a\int\limits_0^{2π}r^3\sin^2θ(a^2-r^2cos^2θ)^{0.5}\,drdθ $$
Please help me evaluate this.
 A: Let $D$ be the disk of radius $a$ centered at the origin. We can parameterize the region in polar coordinates by letting
$$\begin{cases}x = a r \cos(\theta) \\ y = a r \sin(\theta) \end{cases}$$
with $(r,\theta) \in [0,1] \times [0,2\pi]$. The Jacobian in this case is almost identical to the typical one you get when converting to standard polar coordinates (sans $a$):
$$J = \begin{bmatrix}a \cos(\theta) & -ar \sin(\theta) \\ a\sin(\theta) & ar\cos(\theta)\end{bmatrix} \implies |\det J| = a^2r$$
Now the integral is
$$\begin{align*}
I &= \iint_D y^2 \sqrt{a^2 - x^2} \, dx \, dy \\[1ex]
&= \int_0^{2\pi} \int_0^1 a^2r^2\sin^2(\theta) \sqrt{a^2 - a^2 r^2 \cos^2(\theta)} \cdot a^2r \, dr \, d\theta \\[1ex]
&= 2a^5 \int_0^\pi \int_0^1 r^3 \sin^2(\theta) \sqrt{1 - r^2 \cos^2(\theta)} \, dr \, d\theta \tag{1} \\[1ex]
&= -\frac{2a^5}3 \int_0^\pi \left[\frac{\sin^5(\theta)}{\cos^2(\theta)} - 2 \tan^2(\theta) \int_0^1 r \left(1-r^2\cos^2(\theta)\right)^{3/2} \, dr\right] \, d\theta \tag{2} \\[1ex]
&= -\frac{2a^5}3 \int_0^\pi \left[\frac{\sin(\theta)(1-\cos^2(\theta))^2}{\cos^2(\theta)} - \frac{\sin^2(\theta)}{\cos^4(\theta)} \int_{\sin^2(\theta)}^1 s^{3/2} \, ds \right] \, d\theta \tag{3} \\[1ex]
&= -\frac{2a^5}3 \cdot -\frac{16}{15} \tag{4} \\[1ex]
&= \frac{32a^5}{45}
\end{align*}$$


*

*$(1)$ : exploit symmetry about $\theta=\pi$; that is, replace $\theta\mapsto\pi-\theta$ to get an integral of an even function (with respect to $\theta$) over $[-\pi,\pi]$. We end up with an integral over $[0,\pi]$, and on this interval we have $\sin(\theta)\ge0$ so we can pick positive square roots whenever they show up.

*$(2)$ : integrate by parts with respect to $r$, letting $u=r^2$ :
$$\begin{align*}
\int_0^1 r^3 \sqrt{1 - r^2 \cos^2(\theta)} \, dr &= uv\big|_0^1 - \int_0^1 v\,du \\[1ex]
&= -\frac{\sin^3(\theta)}{3\cos^2(\theta)} + \frac2{3\cos^2(\theta)} \int_0^1 r \left(1-r^2\cos^2(\theta)\right)^{3/2} \, dr
\end{align*}$$

*$(3)$ : substitute $s = 1 - r^2 \cos^2(\theta)$

*$(4)$ : partial fractions and evaluate; in particular, we rewrite to the more readily integrable form,
$$\sec^2(\theta) - \frac{3\sin(\theta)}{\cos^2(\theta)} - \sec^2(\theta)(1+\tan^2(\theta)) + \frac{\sin(\theta)}{\cos^4(\theta)} + 3\sin(\theta) - \sin(\theta)\cos^2(\theta)$$
A: Since the integrand and the region $x^2 + y^2 \le a^2$ is symmetric, it suffices to calculate the quarter.
$$
\begin{align}
\iint_{x^2 + y^2 \le a^2} y^2\sqrt{a^2 - x^2} dx dy &= 4 \iint_{0 \le x,y \land x^2+y^2 \le a^2} y^2\sqrt{a^2 - x^2} dx dy \\
&=4 \int_{0}^{a} \sqrt{a^2 - x^2} dx \int_0^{\sqrt{a^2 - x^2}} y^2 dy \\
&= \frac43 \int_0^a (a^2 - x^2)^2 dx \\
&= \frac43 \int_0^a (x^4 - 2a^2 x^2 + a^4) dx \\
&= \frac 43 (\frac15 a^5 - \frac23 a^5 + a^5) \\
&= \frac{32}{45}a^5
\end{align}
$$
