Double integrals

Solution

1. Write it as an iterated integral and integrate with respect to $y$ first, treating $x$ as a constant.

   $$\begin{aligned}\int_{1}^{2} \left(\int_{0}^{3} (-x + y + 3)\,dy\right)\,dx &= \int_{1}^{2} \left[-xy + \frac{1}{2}y^{2} + 3y\right]_{0}^{3}\,dx \\ &= \int_{1}^{2} \left(-3x + \frac{27}{2}\right)\,dx \\ &= \left[-\frac{3}{2}x^{2} + \frac{27}{2}x\right]_{1}^{2} \\ &= 9\end{aligned}$$

Solution

1. An antiderivative of $\frac{\sin y}{y}$ with respect to $y$ is not elementary, so change the order of integration. The region is $\{0 \le x \le \pi,\ x \le y \le \pi\} = \{0 \le y \le \pi,\ 0 \le x \le y\}$.

2. Hence

   $$\begin{aligned}\int_{0}^{\pi} \left(\int_{x}^{\pi} \frac{\sin y}{y}\,dy\right)\,dx &= \int_{0}^{\pi} \left(\int_{0}^{y} \frac{\sin y}{y}\,dx\right)\,dy \\ &= \int_{0}^{\pi} \sin y\,dy \\ &= \left[-\cos y\right]_{0}^{\pi} \\ &= 2\end{aligned}$$

Solution

1. Switching to polar coordinates $x = r\cos\theta$ and $y = r\sin\theta$ gives $dx\,dy = r\,dr\,d\theta$, and the region becomes a rectangle in $r$ and $\theta$.

2. $$\begin{aligned}\iint_D e^{-(x^{2} + y^{2})}\,dx\,dy &= \int_{0}^{2\pi} \left(\int_{0}^{3} re^{-r^{2}}\,dr\right)\,d\theta \\ &= \left(\int_{0}^{2\pi} d\theta\right)\left(\int_{0}^{3} re^{-r^{2}}\,dr\right) \\ &= 2\pi\left[-\frac{1}{2}e^{-r^{2}}\right]_{0}^{3} \\ &= \pi\left(1 - \frac{1}{e^{9}}\right)\end{aligned}$$

Solution

1. The integrand is a product of a function of $x$ and a function of $y$, and the region is a rectangle, so the integral splits into a product.

   $$\begin{aligned}\int_{0}^{\frac{\pi}{2}} \left(\int_{0}^{\frac{\pi}{2}} \sin x\cos y\,dy\right)\,dx &= \left(\int_{0}^{\frac{\pi}{2}} \sin x\,dx\right)\left(\int_{0}^{\frac{\pi}{2}} \cos y\,dy\right) \\ &= \left[-\cos x\right]_{0}^{\frac{\pi}{2}}\left[\sin y\right]_{0}^{\frac{\pi}{2}} \\ &= 1 \\ &= 1\end{aligned}$$

Solution

1. An antiderivative of $e^{y^2}$ with respect to $y$ is not elementary, so change the order of integration. The region is $\left\{0 \le x \le 2,\ \dfrac{x}{2} \le y \le 1\right\} = \{0 \le y \le 1,\ 0 \le x \le 2y\}$.

2. Hence

   $$\begin{aligned}\int_{0}^{2} \left(\int_{\frac{x}{2}}^{1} e^{y^{2}}\,dy\right)\,dx &= \int_{0}^{1} \left(\int_{0}^{2y} e^{y^{2}}\,dx\right)\,dy \\ &= \int_{0}^{1} 2ye^{y^{2}}\,dy \\ &= \left[e^{y^{2}}\right]_{0}^{1} \\ &= e - 1\end{aligned}$$

Solution

1. Switching to polar coordinates $x = r\cos\theta$ and $y = r\sin\theta$ gives $dx\,dy = r\,dr\,d\theta$, and the region becomes a rectangle in $r$ and $\theta$.

2. $$\begin{aligned}\iint_D \sqrt{9 - x^{2} - y^{2}}\,dx\,dy &= \int_{0}^{2\pi} \left(\int_{0}^{3} \sqrt{9 - r^{2}}r\,dr\right)\,d\theta \\ &= \left(\int_{0}^{2\pi} d\theta\right)\left(\int_{0}^{3} \sqrt{9 - r^{2}}r\,dr\right) \\ &= 2\pi\left[-\frac{1}{3}(9 - r^{2})^{\frac{3}{2}}\right]_{0}^{3} \\ &= 18\pi\end{aligned}$$

This matches the volume $\frac{2}{3}\pi a^3$ of a hemisphere of radius $a$.