We are given the iterated integral $\int_{-\pi/2}^{\pi/2} \int_{0}^{\theta} kr dr d\theta$. The goal is to sketch the lamina R, determine the density $\delta$, find the mass, and the center of mass.

1. Problem Description

We are given the iterated integral $\int_{-\pi/2}^{\pi/2} \int_{0}^{\theta} kr dr d\theta$. The goal is to sketch the lamina R, determine the density $\delta$, find the mass, and the center of mass.

2. Solution Steps

The iterated integral is given in polar coordinates. The density function is $\delta(r, \theta) = k$.

The limits of integration are:

$-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$

$0 \le r \le \theta$

To sketch the region R, we note that the angle $\theta$ ranges from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, and for each angle $\theta$, the radius $r$ ranges from $0$ to $\theta$. This region is a sector of a circle where the radius varies with the angle $\theta$.

The mass M of the lamina is given by the integral:

$M = \int_{-\pi/2}^{\pi/2} \int_{0}^{\theta} kr dr d\theta$

First, we integrate with respect to $r$:

$\int_{0}^{\theta} kr dr = k \left[ \frac{1}{2}r^2 \right]_0^{\theta} = \frac{1}{2} k \theta^2$

Now, we integrate with respect to $\theta$:

$M = \int_{-\pi/2}^{\pi/2} \frac{1}{2} k \theta^2 d\theta = \frac{1}{2}k \left[ \frac{1}{3}\theta^3 \right]_{-\pi/2}^{\pi/2} = \frac{1}{6}k \left[ (\frac{\pi}{2})^3 - (-\frac{\pi}{2})^3 \right] = \frac{1}{6} k \left[ \frac{\pi^3}{8} + \frac{\pi^3}{8} \right] = \frac{1}{6} k \left[ \frac{\pi^3}{4} \right] = \frac{k \pi^3}{24}$

To find the center of mass $(\bar{x}, \bar{y})$, we need to find the moments $M_x$ and $M_y$.

$M_x = \iint y \delta(r, \theta) dA = \int_{-\pi/2}^{\pi/2} \int_{0}^{\theta} (r \sin \theta) k r dr d\theta = k \int_{-\pi/2}^{\pi/2} \int_{0}^{\theta} r^2 \sin \theta dr d\theta$

$\int_{0}^{\theta} r^2 \sin \theta dr = \sin \theta \int_{0}^{\theta} r^2 dr = \sin \theta \left[ \frac{1}{3} r^3 \right]_0^{\theta} = \frac{1}{3} \theta^3 \sin \theta$

$M_x = k \int_{-\pi/2}^{\pi/2} \frac{1}{3} \theta^3 \sin \theta d\theta = \frac{k}{3} \int_{-\pi/2}^{\pi/2} \theta^3 \sin \theta d\theta$

Since $\theta^3 \sin \theta$ is an even function,

$\int_{-\pi/2}^{\pi/2} \theta^3 \sin \theta d\theta = 2 \int_{0}^{\pi/2} \theta^3 \sin \theta d\theta$

Using integration by parts:

$\int \theta^3 \sin \theta d\theta = -\theta^3 \cos \theta + 3 \int \theta^2 \cos \theta d\theta = -\theta^3 \cos \theta + 3(\theta^2 \sin \theta - 2 \int \theta \cdot \sin \theta + \theta \cdot \cos \theta d\theta) = -\theta^3 \cos \theta + 3\theta^2 \sin \theta - 6\theta \cos \theta + 6 \sin \theta$

$\int_{0}^{\pi/2} \theta^3 \sin \theta d\theta = [-\theta^3 \cos \theta + 3\theta^2 \sin \theta - 6\theta \cos \theta + 6 \sin \theta]_0^{\pi/2} = 0 + 3 (\frac{\pi}{2})^2 (1) - 0 + 6(1) - 0 = \frac{3\pi^2}{4} + 6$

So, $M_x = \frac{k}{3} (2 (\frac{3\pi^2}{4} + 6)) = \frac{k}{3}(\frac{3\pi^2}{2} + 12) = k(\frac{\pi^2}{2} + 4)$

$\bar{y} = \frac{M_x}{M} = \frac{k(\frac{\pi^2}{2} + 4)}{\frac{k\pi^3}{24}} = \frac{24(\frac{\pi^2}{2} + 4)}{\pi^3} = \frac{12\pi^2 + 96}{\pi^3}$

$M_y = \iint x \delta(r, \theta) dA = \int_{-\pi/2}^{\pi/2} \int_{0}^{\theta} (r \cos \theta) k r dr d\theta = k \int_{-\pi/2}^{\pi/2} \int_{0}^{\theta} r^2 \cos \theta dr d\theta$

$\int_{0}^{\theta} r^2 \cos \theta dr = \cos \theta \int_{0}^{\theta} r^2 dr = \cos \theta \left[ \frac{1}{3} r^3 \right]_0^{\theta} = \frac{1}{3} \theta^3 \cos \theta$

$M_y = k \int_{-\pi/2}^{\pi/2} \frac{1}{3} \theta^3 \cos \theta d\theta = \frac{k}{3} \int_{-\pi/2}^{\pi/2} \theta^3 \cos \theta d\theta$

Since $\theta^3 \cos \theta$ is an odd function, the integral from $-\pi/2$ to $\pi/2$ is

0. Therefore, $M_y = 0$, and $\bar{x} = \frac{M_y}{M} = 0$.

3. Final Answer

Density: $\delta(r, \theta) = k$

Mass: $M = \frac{k \pi^3}{24}$

Center of mass: $(\bar{x}, \bar{y}) = (0, \frac{12\pi^2 + 96}{\pi^3})$