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Chapter 10: 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 inside the circle with the polar equation \(r=8\sin\theta\) is \(32\pi\).

Step by step solution

01

Sketch the Polar Curve

To sketch the polar curve \(r=8\sin\theta\), we can analyze the function to determine key features. Since the sine function has a maximum value of 1, we see that the radius r is largest when \(\sin \theta = 1\), giving us \(r = 8\). The sine function also has a period of \(2\pi\), meaning that the polar curve will repeat every \(2\pi\) radians. The sine function is positive in the first and second quadrants and negative in the third and fourth quadrants, so the circle will be centered in the first and second quadrants and approach a radius of 8.
02

Determine the Limits of Integration

Since the circle only spans the first and second quadrants, we will integrate from \(0\) to \(\pi\). This will go through all the angles through which the polar curve exists.
03

Use the Polar Coordinate Area Formula

The polar coordinate area formula is given by: $$A = \frac{1}{2} \int_{\alpha}^{\beta} [r(\theta)]^2 d\theta$$ Where \(A\) is the area, \(r(\theta)\) is the polar function, in this case, \(8\sin\theta\), and \(\alpha\) and \(\beta\) are the limits of integration. We'll integrate from \(\alpha=0\) to \(\beta=\pi\).
04

Calculate the Area

Now, let's calculate the area of the region inside the circle \(r=8\sin\theta\), using the above formula: $$A = \frac{1}{2} \int_{0}^{\pi} (8\sin\theta)^2 d\theta$$ Begin by simplifying the expression: $$A = \frac{1}{2} \int_{0}^{\pi} 64\sin^2\theta d\theta$$ Now, use the double angle formula to rewrite the expression: $$A = \frac{1}{2} \int_{0}^{\pi} 64\frac{1-\cos(2\theta)}{2} d\theta$$ Simplify the expression: $$A = 32 \int_{0}^{\pi} (1-\cos(2\theta)) d\theta$$ Integrate each term: $$A = 32 [\theta - \frac{1}{2}\sin(2\theta)]_{0}^{\pi}$$ Evaluate the definite integral: $$A = 32 [\pi - 0 - (\pi - 0)] = 32\pi$$ So the area of the region inside the circle \(r=8\sin\theta\) is \(32\pi\).

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