Evaluate $\iint_{R}(x^2+y^2)dxdy$ $$\iint_{R}(x^2+y^2)dxdy$$ 

$$0\leq r\leq 2 \,\, ,\frac{\pi}{4}\leq \theta\leq\frac{3\pi}{4}$$

My attempt :
Jacobian=r
$$=\iint_{R}(x^2+y^2)dxdy$$
$$x:=r\cos \theta \,\,\,,y:=r\cos \theta$$
$$\sqrt{x^2+y^2}=r$$
$$\int_{\theta=\pi/4}^{\theta=3\pi/4}\bigg[\int_{r=0}^{r=2}\bigg(r^2\bigg)dr\bigg]d\theta$$
$$.....=\boxed{\frac{4\pi}{3}}$$

Is it correct?

>

while converting in polar co-ordinate you have directly taken dxdx=drd(theta),
you have to use polar transformation Jacobin to do so, that is dxdx=rdr*d(theta), then integrate normally 
answer will be (ip/2). 
