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Chapter 11: Problem 23

Make a sketch of the region and its bounding curves. Find the area of the region. The region inside the circle \(r=8 \sin \theta\)

Short Answer

Expert verified
Answer: The area of the region enclosed by the curve \(r = 8\sin\theta\) in polar coordinates is \(16\pi\).

Step by step solution

01

Sketch the curve r = 8sin胃

To make a sketch of the curve, we should first notice that when 胃 = 0, r = 0. The coordinate (r,胃) (0,0) in the polar system has the Cartesian coordinate (0,0). For \(0<胃<\frac{\pi}{2}\), \(\sin \theta\) is positive and so is r. When 胃 = 蟺/2, r = 8 since it is maximum over that interval of 胃. At each value of 胃 in the interval (0, 蟺/2], we get a corresponding value of r that determines a point in the curve. Notice that for 胃 in the range (蟺/2, 蟺), r is still positive as.sin胃 is positive while r is negative over (0,-蟺/2). Thus, the curve is symmetric about the y-axis. Now, you can sketch the curve by plotting some points on the interval (0, 蟺/2] and their mirrored counterparts along the y-axis.
02

Identify the bounding curves and interval of 胃

The given region is inside the curve r = 8sin胃, which means that the curve itself is one of the bounding curves. Since the curve starts at 胃 = 0 and ends at 胃 = 蟺, it covers the entire region of the polar plane when 胃 is in the range of [0, 蟺]. Thus, the bounded region is determined by the interval 胃 = [0, 蟺].
03

Compute the area of the region

To find the area of the region inside the curve \(r = 8 \sin \theta\), we will use the polar area formula: Area = \(\frac{1}{2}\int_{\alpha}^{\beta} r^2 d\theta = \frac{1}{2} \int_0^{\pi} (8\sin\theta)^2 d\theta\) Now we compute the integral: Area = \(\frac{1}{2} \int_0^{\pi} 64\sin^2\theta d\theta = 32 \int_0^{\pi} \frac{1-\cos(2\theta)}{2} d\theta\) Area = \(32 \left[\frac{1}{2}\theta - \frac{1}{4}\sin(2\theta)\right]_0^{\pi}\) Area = \(32 \left[\frac{\pi}{2} - 0\right] = 16\pi\) So, the area of the region inside the curve \(r=8\sin\theta\) is \(16\pi\).

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