Problem 1
1. Calculate the improper integral \(\displaystyle\int\!\!\int_D \ln\sqrt{x^2+y^2}\, dA\) for \(D = \{0\le x^2+y^2\le 1\}\). (5 points)
Solution. We begin by making a polar substitution. Noting that \(x^2+y^2=r^2\), we get
\[ \int\!\!\int_D \ln\sqrt{x^2+y^2}\, dA = \int_0^{2\pi}\!\int_0^1 \ln r\cdot r\, dr\, d\theta. \]
We note this is an improper integral as \(\ln(0)\) is undefined. Therefore, we instead look at \(\displaystyle\lim_{a\to 0}\int_0^{2\pi}\!\int_a^1 r\ln r\, dr\, d\theta\). To evaluate this integral, we perform integration by parts. Letting \(u=\ln r\) and \(dv=r\), one has \[ \begin{align*} \lim_{a\to 0}\int_0^{2\pi}\!\int_a^1 r\ln r\, dr\, d\theta &= \lim_{a\to 0}\int_0^{2\pi} \left(\frac{r^2}{2}\ln r - \int_a^1 \frac{r^2}{2}\cdot\frac{1}{r}\, dr\right)\Bigg|_a^1 d\theta \\ &= \lim_{a\to 0}\int_0^{2\pi} \left(\frac{r^2}{2}\ln r - \frac{r^2}{4}\right)\Bigg|_a^1 d\theta \\ &= \lim_{a\to 0}\int_0^{2\pi} \left(\frac{1}{2}\cdot 0 - \frac{1}{4}\right) - \left(\frac{a^2}{2}\ln a - \frac{a^2}{4}\right) d\theta \\ &= \lim_{a\to 0}\int_0^{2\pi} \frac{a^2}{4} - \frac{1}{4} - \frac{a^2}{2}\ln a\, d\theta \\ &= \lim_{a\to 0}\ \frac{\pi}{2}\left(a^2 - 1 - 2a^2\ln a\right) \\ &= -\frac{\pi}{2}. \end{align*} \]
This last limit is known but can also be evaluated with L'Hospital's rule.Problem 2
2. Calculate \(\displaystyle\int\!\!\int\!\!\int_B z\, dV\) where \(B\) is the region bounded by the planes \(x=0\), \(y=0\), \(z=0\), \(z=1\), and the cylinder \(x^2+y^2=1\) with \(x\ge 0\) and \(y\ge 0\). (5 points)
Solution. We note that this region is \(z\)-simple. (Also \(x\)- and \(y\)-simple.) Then, the resulting bounds of integration are
\[ \int\!\!\int\!\!\int_B z\, dV = \int_0^1\!\int_0^{\sqrt{1-y^2}}\!\int_0^1 z\, dz\, dx\, dy = \int_0^1\!\int_0^{\sqrt{1-y^2}} \frac{z^2}{2}\Bigg|_0^1 dx\, dy = \int_0^1\!\int_0^{\sqrt{1-y^2}} \frac{1}{2}\, dx\, dy = \frac{1}{2}\int_0^1 \sqrt{1-y^2}\, dy. \]
This last integral we evaluate using a trig substitution. Let \(y=\sin(\theta)\), so that \(dy=\cos(\theta)\, d\theta\). This gives \[ \frac{1}{2}\int_0^{\pi/2} \sqrt{1-\sin^2(\theta)}\cos(\theta)\, d\theta = \frac{1}{2}\int_0^{\pi/2} \cos^2(\theta)\, d\theta = \frac{1}{4}\int_0^{\pi/2} 1+\cos(2\theta)\, d\theta = \frac{1}{4}\left[\theta+\frac{1}{2}\sin(2\theta)\right]_0^{\pi/2} = \frac{\pi}{8}. \]