<s>[INST] How to calculate $\iint_D{} (x^2 + y^2) \, dxdy$ with $D = \{(\frac{x}{2})^2 + (\frac{y}{3})^2 \leq 1\}$? 
I'm asked to calculate 
  $$\iint_D{} (x^2 + y^2)\, dxdy,\quad D = \left\{\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 \leq 1\right\}.$$

My attempt:
Set 
$$\frac{x}{2} = r\cos{\theta},\quad\frac{y}{3} = r\sin{\theta}$$
which nets the functional matrix i $6r$.
$$\iint_D{(4r^2\cos^2{\theta} + 9r^2\sin^2{\theta})6r \, drd\theta}$$
$$\iint_D{24r^3 + 30r^3\sin^2{\theta} \, drd\theta}$$
$$\int^{2\pi}_0\int^1_0{24r^3 + 30r^3\sin^2{\theta} \,drd\theta}$$
$$\int^{2\pi}_0{6 + 10\sin^2{\theta} \, d\theta}$$
$$\left[6\theta + \frac{10}{2}(\theta - \sin{\theta}\cos{\theta})\right]^{2\pi}_0$$
which equals to $22\pi$. 
However, the answer is supposed to be $\frac{39\pi}{2}$. What am I doing wrong / missing?
 [/INST] In going from the third-to-last displayed expression to the second-to-last, you have $\int_0^1 30r^3 \: dr = 10$.  This integral is $30/4$, so the second-to-last displayed expression should be
$$ \int_0^{2\pi} 6 + {30 \over 4} \sin^2 \theta \: d\theta $$
which after integration will give you $39\pi/2$.
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