Questions: Find the area of the region that lies inside the first curve and outside the second curve. (23.) r=4 sin theta, r=2

Solution Steps

To find the area of the region that lies inside the first curve \( r = 4 \sin \theta \) and outside the second curve \( r = 2 \), we need to:

1. Determine the points of intersection of the two curves by setting \( 4 \sin \theta = 2 \) and solving for \( \theta \).
2. Calculate the area of the region enclosed by the curve \( r = 4 \sin \theta \) from the intersection points.
3. Subtract the area of the circle \( r = 2 \) from the area calculated in step 2 over the same range of \( \theta \).

To find the points of intersection of the curves \( r = 4 \sin \theta \) and \( r = 2 \), we set them equal to each other:

\[ 4 \sin \theta = 2 \]

Solving for \( \theta \), we get:

\[ \sin \theta = \frac{1}{2} \implies \theta = \frac{\pi}{6} \]

The area \( A_1 \) enclosed by the curve \( r = 4 \sin \theta \) from \( \theta = 0 \) to \( \theta = \frac{\pi}{6} \) is given by the integral:

\[ A_1 = \frac{1}{2} \int_0^{\frac{\pi}{6}} (4 \sin \theta)^2 \, d\theta \]

Calculating this integral, we find:

\[ A_1 = \frac{1}{2} \int_0^{\frac{\pi}{6}} 16 \sin^2 \theta \, d\theta = 8 \int_0^{\frac{\pi}{6}} \sin^2 \theta \, d\theta \]

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

\[ A_1 = 8 \cdot \frac{1}{2} \int_0^{\frac{\pi}{6}} (1 - \cos(2\theta)) \, d\theta = 4 \left[ \theta - \frac{1}{2} \sin(2\theta) \right]_0^{\frac{\pi}{6}} \]

Evaluating this gives:

\[ A_1 = 4 \left( \frac{\pi}{6} - 0 \right) = \frac{2\pi}{3} \]

The area \( A_2 \) enclosed by the circle \( r = 2 \) from \( \theta = 0 \) to \( \theta = \frac{\pi}{6} \) is given by:

\[ A_2 = \frac{1}{2} \int_0^{\frac{\pi}{6}} (2)^2 \, d\theta = \frac{1}{2} \int_0^{\frac{\pi}{6}} 4 \, d\theta = 2 \left[ \theta \right]_0^{\frac{\pi}{6}} = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} \]

The area of the region that lies inside the first curve and outside the second curve is:

\[ A = A_1 - A_2 = \frac{2\pi}{3} - \frac{\pi}{3} = \frac{2\pi}{3} - \frac{1\pi}{3} = \frac{\pi}{3} \]

Final Answer