We are asked to evaluate six iterated integrals.

1. Problem Description

We are asked to evaluate six iterated integrals.

2. Solution Steps

1. $\int_{0}^{\pi/2}\int_{0}^{cos\theta} r^2 \sin\theta \, dr \, d\theta$

First integrate with respect to $r$:

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

Now integrate with respect to $\theta$:

$\int_{0}^{\pi/2} \frac{1}{3} \cos^3\theta \sin\theta \, d\theta$

Let $u = \cos\theta$, $du = -\sin\theta \, d\theta$.

When $\theta = 0$, $u = 1$. When $\theta = \pi/2$, $u = 0$.

$\frac{1}{3} \int_{1}^{0} u^3 (-du) = -\frac{1}{3} \int_{1}^{0} u^3 \, du = \frac{1}{3} \int_{0}^{1} u^3 \, du = \frac{1}{3} \left[ \frac{u^4}{4} \right]_0^1 = \frac{1}{3} \left( \frac{1}{4} - 0 \right) = \frac{1}{12}$

2. $\int_{0}^{\pi/2} \int_{0}^{sin\theta} r \, dr \, d\theta$

First integrate with respect to $r$:

$\int_{0}^{sin\theta} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{sin\theta} = \frac{sin^2\theta}{2}$

Now integrate with respect to $\theta$:

$\int_{0}^{\pi/2} \frac{sin^2\theta}{2} \, d\theta = \frac{1}{2} \int_{0}^{\pi/2} sin^2\theta \, d\theta$

We use the identity $sin^2\theta = \frac{1 - cos(2\theta)}{2}$.

$\frac{1}{2} \int_{0}^{\pi/2} \frac{1 - cos(2\theta)}{2} \, d\theta = \frac{1}{4} \int_{0}^{\pi/2} (1 - cos(2\theta)) \, d\theta = \frac{1}{4} \left[ \theta - \frac{1}{2}sin(2\theta) \right]_0^{\pi/2} = \frac{1}{4} \left[ \frac{\pi}{2} - \frac{1}{2}sin(\pi) - (0 - \frac{1}{2}sin(0)) \right] = \frac{1}{4} \left[ \frac{\pi}{2} - 0 - 0 \right] = \frac{\pi}{8}$

3. $\int_{0}^{\pi} \int_{0}^{sin\theta} r^2 \, dr \, d\theta$

First integrate with respect to $r$:

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

Now integrate with respect to $\theta$:

$\int_{0}^{\pi} \frac{sin^3\theta}{3} \, d\theta = \frac{1}{3} \int_{0}^{\pi} sin^3\theta \, d\theta$

We know that $sin^3\theta = sin\theta (sin^2\theta) = sin\theta (1 - cos^2\theta) = sin\theta - sin\theta cos^2\theta$

$\frac{1}{3} \int_{0}^{\pi} (sin\theta - sin\theta cos^2\theta) \, d\theta = \frac{1}{3} \left[ -cos\theta + \frac{cos^3\theta}{3} \right]_0^{\pi} = \frac{1}{3} \left[ -cos(\pi) + \frac{cos^3(\pi)}{3} - (-cos(0) + \frac{cos^3(0)}{3}) \right] = \frac{1}{3} \left[ -(-1) + \frac{(-1)^3}{3} - (-1 + \frac{1}{3}) \right] = \frac{1}{3} \left[ 1 - \frac{1}{3} - (-\frac{2}{3}) \right] = \frac{1}{3} \left[ 1 - \frac{1}{3} + \frac{2}{3} \right] = \frac{1}{3} \left[ \frac{3 - 1 + 2}{3} \right] = \frac{1}{3} \left[ \frac{4}{3} \right] = \frac{4}{9}$

4. $\int_{0}^{\pi} \int_{0}^{1-cos\theta} r \sin\theta \, dr \, d\theta$

First integrate with respect to $r$:

$\int_{0}^{1-cos\theta} r \sin\theta \, dr = \sin\theta \int_{0}^{1-cos\theta} r \, dr = \sin\theta \left[ \frac{r^2}{2} \right]_{0}^{1-cos\theta} = \sin\theta \frac{(1-cos\theta)^2}{2} = \frac{1}{2} \sin\theta (1 - 2cos\theta + cos^2\theta)$

Now integrate with respect to $\theta$:

$\frac{1}{2} \int_{0}^{\pi} (\sin\theta - 2cos\theta \sin\theta + cos^2\theta \sin\theta) \, d\theta = \frac{1}{2} \left[ -cos\theta + cos^2\theta - \frac{cos^3\theta}{3} \right]_0^{\pi} = \frac{1}{2} \left[ (-cos(\pi) + cos^2(\pi) - \frac{cos^3(\pi)}{3}) - (-cos(0) + cos^2(0) - \frac{cos^3(0)}{3}) \right] = \frac{1}{2} \left[ (-(-1) + (-1)^2 - \frac{(-1)^3}{3}) - (-1 + 1 - \frac{1}{3}) \right] = \frac{1}{2} \left[ (1 + 1 + \frac{1}{3}) - (0 - \frac{1}{3}) \right] = \frac{1}{2} \left[ 2 + \frac{1}{3} + \frac{1}{3} \right] = \frac{1}{2} \left[ 2 + \frac{2}{3} \right] = \frac{1}{2} \left[ \frac{8}{3} \right] = \frac{4}{3}$

5. $\int_{0}^{\pi} \int_{0}^{2} r cos(\frac{\theta}{4}) \, dr \, d\theta$

First integrate with respect to $r$:

$\int_{0}^{2} r cos(\frac{\theta}{4}) \, dr = cos(\frac{\theta}{4}) \int_{0}^{2} r \, dr = cos(\frac{\theta}{4}) \left[ \frac{r^2}{2} \right]_0^2 = cos(\frac{\theta}{4}) (\frac{4}{2}) = 2cos(\frac{\theta}{4})$

Now integrate with respect to $\theta$:

$\int_{0}^{\pi} 2cos(\frac{\theta}{4}) \, d\theta = 2 \int_{0}^{\pi} cos(\frac{\theta}{4}) \, d\theta = 2 \left[ 4sin(\frac{\theta}{4}) \right]_{0}^{\pi} = 8 \left[ sin(\frac{\pi}{4}) - sin(0) \right] = 8 \left[ \frac{\sqrt{2}}{2} - 0 \right] = 8 \cdot \frac{\sqrt{2}}{2} = 4\sqrt{2}$

6. $\int_{0}^{2\pi} \int_{0}^{\theta} r \, dr \, d\theta$

First integrate with respect to $r$:

$\int_{0}^{\theta} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{\theta} = \frac{\theta^2}{2}$

Now integrate with respect to $\theta$:

$\int_{0}^{2\pi} \frac{\theta^2}{2} \, d\theta = \frac{1}{2} \int_{0}^{2\pi} \theta^2 \, d\theta = \frac{1}{2} \left[ \frac{\theta^3}{3} \right]_{0}^{2\pi} = \frac{1}{2} \left[ \frac{(2\pi)^3}{3} - 0 \right] = \frac{1}{2} \frac{8\pi^3}{3} = \frac{4\pi^3}{3}$

3. Final Answer

1. 1/12

2. $\pi/8$

3. 4/9

4. 4/3

5. $4\sqrt{2}$

6. $\frac{4\pi^3}{3}$