Problem 1
1. Calculate \(\displaystyle\int_0^\pi\!\int_{\sin(x)}^{3\sin(x)} x(1+y)\, dy\, dx\) and sketch the domain of integration. (5 points)
Solution. We begin by computing the inner integral, after distributing the \(x\). This gives
\[ \int_0^\pi\!\int_{\sin(x)}^{3\sin(x)} x+xy\, dy\, dx = \int_0^\pi \left(xy+\frac{xy^2}{2}\right)\Bigg|_{\sin(x)}^{3\sin(x)} dx = \int_0^\pi \left(3x\sin(x)+\frac{9x\sin^2(x)}{2} - x\sin(x) - \frac{x\sin^2(x)}{2}\right) dx. \]
This simplifies down to \(\displaystyle\int_0^\pi 2x\sin(x) + 4x\sin^2(x)\, dx\). We compute each piece by parts. First, note that \[ \int_0^\pi x\sin(x)\, dx = -x\cos(x)\Big|_0^\pi - \int_0^\pi(-\cos(x))\, dx = \left(-x\cos(x)+\sin(x)\right)\Big|_0^\pi = \pi. \]
Thus, the first term, \(2\displaystyle\int_0^\pi x\sin(x)\, dx = 2\pi\). As for the second term, note that \(\sin^2(x) = 1/2 - (1/2)\cos(2x)\), and so we have \[ \int_0^\pi x\sin^2(x)\, dx = \frac{1}{2}\int_0^\pi x - x\cos(2x)\, dx. \]
We then proceed with parts on the \(x\cos(2x)\) term, giving us \[ \int_0^\pi x\cos(2x)\, dx = \frac{1}{2}x\sin(2x)\Big|_0^\pi - \int_0^\pi \frac{1}{2}\sin(2x)\, dx = \frac{1}{2}\left[x\sin(2x)+\frac{1}{2}\cos(2x)\right]_0^\pi = 0. \]
We finally see that \(\dfrac{1}{2}\displaystyle\int_0^\pi x\, dx = \dfrac{x^2}{4}\Big|_0^\pi = \dfrac{\pi^2}{4}\), so \(4\displaystyle\int_0^\pi x\sin^2(x)\, dx = \pi^2\). Combining the two terms, we get \[ \int_0^\pi\!\int_{\sin(x)}^{3\sin(x)} x(1+y)\, dy\, dx = \pi^2+2\pi. \]
Problem 2
2. Calculate \(\displaystyle\int_0^2\!\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} (x^2+y^2)^2\, dx\, dy\). (5 points)
Solution. The bounds suggest that we are integrating over a disc. Note that \(-\sqrt{4-y^2}\le x\le \sqrt{4-y^2}\) is equivalent to \(x^2\le 4-y^2\) or \(x^2+y^2\le 4\). That is, we are on a disc of radius 2. We also note that since \(y\ge 0\), we are only on the top half of the disc. We make the standard polar transformation to obtain
\[ \int_0^\pi\!\int_0^2 r^4\cdot r\, dr\, d\theta = \int_0^\pi \left(\frac{r^6}{6}\right)\Bigg|_0^2 d\theta = \int_0^\pi \frac{64}{6}\, d\theta = \frac{64\pi}{6}. \]