Calculate the iterated integral: $\int_0^6 \int_0^\pi 8r\sin^2(\theta) \, d\theta \, dr$

Neetesh Kumar | November 30, 2024

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This is the solution to Math 1D
Assignment: 15.1 Question Number 12
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Step-by-step solution:

To evaluate the iterated integral: $\int_0^6 \int_0^\pi 8r\sin^2(\theta) \, d\theta \, dr$ we will first evaluate the inner integral with respect to $\theta$, and then the outer integral with respect to $r$

Step 1: Evaluate the Inner Integral

The inner integral is:

$\int_0^\pi 8r \sin^2(\theta) \, d\theta$

Since $r$ is treated as a constant with respect to $\theta$, we can factor out $8r$:

$8r \int_0^\pi \sin^2(\theta) \, d\theta$

Now, we need to evaluate the integral $\int_0^\pi \sin^2(\theta) \, d\theta$. We can use the identity for $\sin^2(\theta)$:

$\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$

Thus, the integral becomes:

$\int_0^\pi \sin^2(\theta) \, d\theta = \int_0^\pi \frac{1 - \cos(2\theta)}{2} \, d\theta$

This can be separated into two integrals:

$\frac{1}{2} \int_0^\pi 1 \, d\theta - \frac{1}{2} \int_0^\pi \cos(2\theta) \, d\theta$

First Integral: $\int_0^\pi 1 \, d\theta$

This is straightforward:

$\int_0^\pi 1 \, d\theta = \pi$

Second Integral: $\int_0^\pi \cos(2\theta) \, d\theta$

The integral of $\cos(2\theta)$ is $\frac{\sin(2\theta)}{2}$, and evaluating this from $\theta = 0$ to $\theta = \pi$ gives:

$\left[ \frac{\sin(2\theta)}{2} \right]_0^\pi = \frac{\sin(2\pi)}{2} - \frac{\sin(0)}{2} = 0$

Thus, the second integral is zero.

Putting these results together:

$\int_0^\pi \sin^2(\theta) \, d\theta = \frac{1}{2} \left( \pi - 0 \right) = \frac{\pi}{2}$

Now, the inner integral becomes:

$8r \cdot \frac{\pi}{2} = 4\pi r$

Step 2: Evaluate the Outer Integral

Next, we evaluate the outer integral:

$\int_0^6 4\pi r \, dr$

This is straightforward:

$4\pi \int_0^6 r \, dr = 4\pi \left[ \frac{r^2}{2} \right]_0^6$

Evaluating at the bounds:

$4\pi \cdot \frac{6^2}{2} = 4\pi \cdot \frac{36}{2} = 4\pi \cdot 18 = 72\pi$

Final Answer:

The value of the iterated integral is:

$\boxed{72\pi}$

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