Find the area of the region enclosed by one loop of the curve: $r = 4 + 8 \sin(\theta)$ (inner loop)

Neetesh Kumar | January 1, 2025

Calculus Homework Help

This is the solution to Math 1c
Assignment: 10.4 Question Number 6
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Step-by-step solution:

To calculate the area enclosed by one loop of the given polar curve, we use the formula for the area in polar coordinates:

$A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta$

Here, $r = 4 + 8\sin(\theta)$.

Since we are focusing on the inner loop, we need to determine the limits of integration $\theta_1$ and $\theta_2$ that correspond to the loop.

Step 1: Identify the limits of integration for the inner loop:

For an inner loop in a polar curve, the loop corresponds to the portion where $r = 0$.

Setting $r = 0$:

$4 + 8\sin(\theta) = 0$

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

The solutions for $\sin(\theta) = -\frac{1}{2}$ in the interval $[0, 2\pi]$ are:

$\theta = \frac{7\pi}{6}$ and $\theta = \frac{11\pi}{6}$.

Thus, the inner loop occurs as $\theta$ varies from $\frac{7\pi}{6}$ to $\frac{11\pi}{6}$.

Step 2: Set up the integral for the area:

The area of the inner loop is:

$A = \frac{1}{2} \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} \left(4 + 8\sin(\theta)\right)^2 \, d\theta$

Step 3: Expand $r^2$:

Expand $\left(4 + 8\sin(\theta)\right)^2$:

$\left(4 + 8\sin(\theta)\right)^2 = 16 + 64\sin(\theta) + 64\sin^2(\theta)$

So, the integral becomes:

$A = \frac{1}{2} \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} \left(16 + 64\sin(\theta) + 64\sin^2(\theta)\right) \, d\theta$

Step 4: Simplify and compute the integral:

Split the integral into three parts:

$A = \frac{1}{2} \left[ \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 16 \, d\theta + \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 64\sin(\theta) \, d\theta + \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 64\sin^2(\theta) \, d\theta \right]$

First Term:

$\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 16 \, d\theta = 16 \cdot \left(\frac{11\pi}{6} - \frac{7\pi}{6}\right) = 16 \cdot \frac{4\pi}{6} = \frac{64\pi}{6} = \frac{32\pi}{3}$

Second Term:

$\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 64\sin(\theta) \, d\theta$

The integral of $64\sin(\theta)$ over the interval $[\frac{7\pi}{6}, \frac{11\pi}{6}]$ is nonzero. Compute:

$\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 64\sin(\theta) \, d\theta = 64 \cdot \left[-\cos(\theta)\right]_{\frac{7\pi}{6}}^{\frac{11\pi}{6}}$

Evaluate $-\cos(\theta)$ at the limits:

$-\cos\left(\frac{11\pi}{6}\right) = -\left(\frac{\sqrt{3}}{2}\right), \quad -\cos\left(\frac{7\pi}{6}\right) = -\left(-\frac{\sqrt{3}}{2}\right)$

So:

$\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 64\sin(\theta) \, d\theta = 64 \cdot \left(-\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\right) = -64\sqrt{3}$

Third Term:

Use the identity $\sin^2(\theta) = \frac{1}{2}(1 - \cos(2\theta))$:

$\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 64\sin^2(\theta) \, d\theta = 64 \cdot \frac{1}{2} \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} (1 - \cos(2\theta)) \, d\theta$

$= 32 \left[ \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 1 \, d\theta - \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} \cos(2\theta) \, d\theta \right]$

For the first part:

$\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 1 \, d\theta = \frac{11\pi}{6} - \frac{7\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$

For the second part:

The integral of $\cos(2\theta)$ over $[\frac{7\pi}{6}, \frac{11\pi}{6}]$ is zero because $\cos(2\theta)$ is symmetric over this interval:

$\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} \cos(2\theta) \, d\theta = 0$

Thus:

$\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 64\sin^2(\theta) \, d\theta = 32 \cdot \frac{2\pi}{3} = \frac{64\pi}{3}$

Step 5: Combine the results:

Now, combine all the terms:

$A = \frac{1}{2} \left( \frac{32\pi}{3} - 64\sqrt{3} + \frac{64\pi}{3} \right)$

$A = \frac{1}{2} \cdot \left( \frac{96\pi}{3} - 64\sqrt{3} \right)$

$A = \frac{48\pi}{3} - 32\sqrt{3} = 16\pi - 32\sqrt{3}$

Final Answer:

The area enclosed by one loop of the curve is $\boxed{16\pi - 32\sqrt{3}}$

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