Question

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

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

Step-by-step solution

Sketch the curve r = 8sinθ

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

Identify the bounding curves and interval of θ

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

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 \(16\pi\).