Difficulty in the evaluation of double integral of a continuous function. Evaluate the double integral
$$\iint_D \frac{xy}{\sqrt{1-y^2}} dxdy,$$ 
where $D$ is the region of the first quadrant bounded by standard circle with unit radius $1$: that is $x^2+y^2=1$. 
I have difficulty with its evaluation. 
First of all I want an answer to verify. I tried this using polar coordinate. I have also tried by taking $x$ changing from $0$ to $1$ and $y$ changing from $0$ to $\sqrt{1-y^2}$. 
The answer comes to be 1/6.
Need help. As I am not getting the same answer always when I try to use new method.
 A: The computation you show us correct. Here us another way:
$$
\begin{align}
\int_0^1\int_0^{\sqrt {1-y^2}}\frac {xy}{\sqrt {1-y^2}}\,dx\,dy
&=\int_0^1\frac {y ( {1-y^2})}{\sqrt {1-y^2}}\,dy
=\int_0^1 y\sqrt {1-y^2}\,dy\\ \ \\
&=\frac12\int_0^1v^{1/2}\,dv=\frac12\,\frac23=\frac16.
\end{align}
$$
It can also be done with polar coordinates: take $u=1-r^2\sin^2 t $. Then $du=-2r^2\sin t\cos t\,dt $, and then
\begin{align}
\int_0^1\int_0^{\pi/2}\frac {r^2\cos t\sin t}{\sqrt {1-r^2\sin ^2t}}\,r\,dt\,dr
&=-\frac12\int_0^1\int_1^{1-r^2}u^{-1/2}{r}\,du\,dr\\ \ \\
&=\int_0^1r\int_{1-r^2}^1\left.u^{1/2}\right|_{1-r^2}^1\,dr\\ \ \\
&=\int_0^1r-r(1-r^2)^{1/2}\,dr\\ \ \\
&=\frac12-\frac12\int_0^1v^{1/2}\,dv\\ \ \\
&=\frac12-\frac12\,\frac23=\frac16.
\end{align}
A: The fact that $D$ is a disc tells you that you should use polar coordinates. Therefore, let $x = r \cos t$ and $y = r \sin t$ with $r \in [0,1]$ and $t \in [0, \frac \pi 2]$, in order to transform your integral into
$$\iint \limits _{[0,1] \times [0, \frac \pi 2]} \frac {r \cos t r \sin t } {\sqrt {1 - r^2 \sin^2 t}} r \Bbb d r \Bbb d t .$$
Notice that the derivative with respect to $t$ of the expression under the square root is $-r^2 2 \sin t \cos t$, which is almost what you already have in the numerator (save for a factor of $-2$). Threrefore, the computation can be continued as
$$\int \limits _0 ^1 -r \left( \int \limits _0 ^{\frac \pi 2} \frac {\frac {\partial} {\partial t} (1 - r^2 \sin^2 t)} {\sqrt {1 - r^2 \sin^2 t}} \Bbb d t \right) \Bbb d r = \int \limits _0 ^1 -r \left( \sqrt {1 - r^2 \sin^2 t} \ \Bigg| _0 ^{\frac \pi 2} \right) \Bbb d r = \int \limits _0 ^1 -r \left( \sqrt {1 - r^2} - 1\right) \Bbb d r = \\
\int \limits _0 ^1 r \left( 1 - \sqrt {1 - r^2} \right) \Bbb d r = \frac {r^2} 2 \Bigg| _0 ^1 + \frac 1 2 \int \limits _0 ^1 (1-r^2)' \sqrt {1-r^2} \ \Bbb d r = \frac 1 2 + \frac 1 2 \frac 2 3 \sqrt{1 - r^2} \Big| _0 ^1 = \frac 1 2 - \frac 1 3 = \frac 1 6.$$

The fact that the result should be $\frac 1 6$ can be also seen by using Fubini's thorem: your integral is
$$\int \limits _0 ^1 x \left( \int \limits _0 ^{\sqrt {1 - x^2}} \frac y {\sqrt {1-y^2}} \ \Bbb d y \right) \ \Bbb d x = \int \limits _0 ^1 x \left( - \sqrt{1 - y^2} \Bigg| _0 ^{\sqrt{1-x^2}} \right) = \int \limits _0 ^1 x ( -x + 1) \ \Bbb d x = - \frac 1 3 + \frac 1 2 = \frac 1 6.$$
A: In polar coordinates,
$$\int_{r=0}^1\int_{\theta=0}^{\pi/2}\frac{r^2\cos\theta\sin\theta}{\sqrt{1-r^2\sin^2\theta}}r\,dr\,d\theta
=-\int_{r=0}^1\left.r\sqrt{1-r^2\sin^2\theta}\right|_{\theta=0}^{\pi/2}dr=\int_{r=0}^1\left(r-r\sqrt{1-r^2}\right)dr\\
=\frac12\left.r^2+\frac13(1-r^2)^{3/2}\right|_{r=0}^1=\frac16.$$

In Cartesian coordinates,
$$\int_{x=0}^1\int_{y=0}^{\sqrt{1-x^2}}\frac{xy}{\sqrt{1-y^2}}dy\,dx=
-\int_{x=0}^1\left.x\sqrt{1-y^2}\right|_{y=0}^{\sqrt{1-x^2}}dx=
\int_{x=0}^1\left(x-x\sqrt{1-x^2}\right)dx$$
is the same integral as above !

With the changes of variable $u=x^2,v=y^2$,
$$\int_{u=0}^1\int_{v=0}^{1-u}\frac{\sqrt{uv}}{\sqrt{1-v}}\frac{du\,dv}{4\sqrt u\sqrt v}=
\frac12\int_{u=0}^1\left.\sqrt{1-v}\right|_{v=0}^{1-u}du=
\frac12\int_{u=0}^1(\sqrt u-1)du\\
=\left.\frac u2-\frac{u^{2/3}}3\right|_{u=0}^1=\frac16.
$$
