<s>[INST] Evaluate $\iint_{R}(x+y)^2dxdy$ in $0\leq r\leq 1 \,\, ,\frac{\pi}{3}\leq \theta\leq\frac{2\pi}{3}$ 
$$\iint_{R}(x+y)^2dxdy$$
$$0\leq r\leq 1 \,\, ,\frac{\pi}{3}\leq \theta\leq\frac{2\pi}{3}$$

My attempt number 1:
$$=\iint_{R}(x^2+2xy+y^2)dxdy$$
$$x:=r\cos \theta \,\,\,,y:=r\cos \theta$$
$$\sqrt{x^2+y^2}=r$$
$$\int_{\pi/3}^{2\pi/3}\bigg[\int_{0}^{1}\bigg(r^2+(r\cos \theta)(r\sin \theta)\bigg)dr\bigg]d\theta$$
$$=\int_{\pi/3}^{2\pi/3}\bigg(\frac{1}{2}+\cos \theta \sin \theta\bigg)d \theta$$

Is it correct so far?

Attempt number 2:
$$\int_{\pi/3}^{2\pi/3}\bigg[\int_{0}^{1}r\bigg(r^2+\color{red}2(r\cos \theta)(r\sin \theta)\bigg)dr\bigg]d\theta$$
$$=\int_{\pi/3}^{2\pi/3}\bigg(\frac{1}{4}+ \frac{1}{4}\sin(2 \theta) \bigg)d \theta$$
$$=\boxed{\pi/12}$$
 [/INST] So you are trying to determine the surface area of the function $(x+y)^2$ in a part of a circle with radius 1. Let's define $R=\left( (r,\theta)\in \mathbb{R}^2 : 0 \leq r \leq 1 , \frac{\pi}{3} \leq \theta \leq \frac{2 \pi}{3}\right)$.
We have the co-ordinate transformation $X(x,y)=(r \cos \theta , r \sin \theta)$.
The Jacobian is:
$$J(r,\theta) = \left[ \begin{array}{cc} \frac{∂x}{∂r} &\frac{∂x}{∂\theta} \\ \frac{∂y}{∂r} & \frac{∂y}{∂\theta} \end{array} \right] = \left[ \begin{array}{cc} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{array} \right] $$ To determine the integral we have to calculate $|J(r,\theta)|=|r \cos^2\theta + r \sin^2 \theta| = r$.
$$ \int\int_R (x+y)^2 d(x,y) = \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \int_{0}^{1} (r \cos \theta + r \sin \theta)^2 \cdot r \ dr d\theta =$$
$$\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \int_{0}^{1} r(r^2+2 r^2 \sin\theta \cos \theta)\ dr d\theta =$$
$$\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \int_{0}^{1} r^3+2 r^3 \sin\theta \cos \theta\ dr d\theta =$$
$$\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \left[\frac{r^4}{4}+\frac{r^4}{2}\sin\theta \cos\theta \right]^{1}_{r=0} d\theta=\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{1}{4}+\frac{1}{2}\sin\theta \cos\theta d\theta=$$
$$\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{1}{4} d\theta +\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{1}{4}\sin(2\theta) d\theta= \frac{1}{4}\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} 1 d\theta +\frac{1}{4}\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \sin(2\theta) d\theta=$$
$$\frac{1}{4}(\frac{2\pi}{3}-\frac{\pi}{3}) + \frac{1}{4} \left[-\frac{1}{2}\cos(2\theta)  \right]_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} =$$
$$\frac{\pi}{12} -\frac{1}{8}(\frac{-1}{2}-\frac{-1}{2})= \frac{\pi}{12}$$
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