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Chapter 10: Problem 24

Make a sketch of the region and its bounding curves. Find the area of the region. The region inside the cardioid \(r=4+4 \sin \theta\)

Short Answer

Expert verified
Answer: The area of the region bounded by the cardioid is \(24\pi\) square units.

Step by step solution

01

Sketching the cardioid

To sketch the cardioid, we can plot a few points and observe their behavior: \(\theta = 0, r = 4\) \(\theta = \frac{\pi}{2}, r = 8\) \(\theta = \pi, r = 4\) \(\theta = \frac{3\pi}{2}, r = 0\) We can see that the cardioid starts at \((4,0)\), expands to \((0,8)\), contracts back to \((-4,0)\), and then returns to \((4,0)\). Connecting these points will give us a rough idea of the cardioid's shape.
02

Determining the limits of integration

In order to find the area enclosed by the cardioid, we need to determine the limits of integration for our polar integral. Since the cardioid starts and ends at \((4,0)\), the limits of integration for the angle \(\theta\) will be from \(0\) to \(2\pi\).
03

Setting up the polar integral

To set up the integral for the area of the region bounded by the cardioid, we can use the formula for the area of a polar curve: \(A = \frac{1}{2} \int_\alpha^\beta r^2 d\theta\) where \(A\) is the area of the region, \(r\) is the polar coordinate distance function, and \(\alpha\) and \(\beta\) are the limits of integration for the angle \(\theta\). In our case, \(\alpha = 0\) and \(\beta = 2\pi\), and \(r(\theta) = 4+4\sin\theta\). So our integral becomes \(A = \frac{1}{2} \int_0^{2\pi} (4+4\sin\theta)^2 d\theta\)
04

Evaluating the integral

Now, we need to evaluate the integral to find the area of the region bounded by the cardioid. Expand the integrand: \(A = \frac{1}{2} \int_0^{2\pi} (16 + 32\sin\theta + 16\sin^2\theta) d\theta\) For the second term, we have \( \int_0^{2\pi} 32\sin\theta d\theta = 0\) For the third term, we will use the trigonometric identity: \(\sin^2\theta=\frac{1-\cos2\theta}{2}\) \(A = \frac{1}{2} \int_0^{2\pi} (16 + 8 - 8\cos2\theta) d\theta\) Now our integral looks like: \(A = \frac{1}{2} \int_0^{2\pi} (24 - 8\cos2\theta) d\theta\) Evaluate the integral: \(A = \frac{1}{2} \left[24\theta - 4\sin2\theta \right]_0^{2\pi}\) \(A = \frac{1}{2} (48\pi - 0) = 24\pi\)
05

Final Solution

So the area of the region bounded by the cardioid \(r=4+4\sin\theta\) is \(24\pi\) square units.

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