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?

>

Switching to polar coordinates, the Jacobian is given by $ |J|$ where $$ J = \dfrac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \dfrac{\partial x}{\partial r} & \dfrac{\partial y}{\partial r} \\ \dfrac{\partial x}{\partial \theta} & \dfrac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{vmatrix} = r$$ Therefore, your double integral is given by $$ \begin{aligned} \iint_{R} \left( x^2 + y^2 \right) \text{ d}x \text{ d}y & = \int_{\pi/4}^{3\pi/4} \int_{0}^{2} \left( (r\cos\theta)^2 + (r\sin\theta)^2 \right) |J| \text{ d}r \text{ d}\theta \\ & = \int_{\pi/4}^{3\pi/4} \int_{0}^{2} r^2 |r| \text{ d}r \text{ d}\theta \end{aligned}$$ and since $r \in \left[0,2\right]$, $|r| = +r$ so the integrand is $r^3$. I leave the rest to you.
