Find the area of the shaded region.
r = 4 + 3 sin(𝜃)

To find the area of the shaded region defined by the polar equation \( r = 4 + 3 \sin(\theta) \), you can use the formula for the area \( A \) in polar coordinates:

\[
A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta
\]

First, find the points where the curve intersects itself to determine the limits of integration. For \( r = 0 \):

\[
4 + 3 \sin(\theta) = 0 \implies \sin(\theta) = -\frac{4}{3}
\]

This doesn't yield real solutions, so we typically integrate from \( 0 \) to \( \pi \) for this type of polar curve.

Now, compute \( r^2 \):

\[
r^2 = (4 + 3 \sin(\theta))^2 = 16 + 24 \sin(\theta) + 9 \sin^2(\theta)
\]

The area becomes:

\[
A = \frac{1}{2} \int_{0}^{\pi} (16 + 24 \sin(\theta) + 9 \sin^2(\theta)) \, d\theta
\]

Calculating this integral gives:

1. The integral of \( 16 \) over \( 0 \) to \( \pi \) is \( 16\pi \).
2. The integral of \( 24 \sin(\theta) \) is \( -24 \cos(\theta) \) evaluated from \( 0 \) to \( \pi \) gives \( 0 \).
3. The integral of \( 9 \sin^2(\theta) \) can be evaluated using the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \).

Thus, the final area is:

\[
A = \frac{1}{2} \left( 16\pi + 0 + \frac{9\pi}{2} \right) = \frac{1}{2} \left( \frac{32\pi}{2} + \frac{9\pi}{2} \right) = \frac{41\pi}{4}
\]

The area of the shaded region is \( \frac{41\pi}{4} \).

Would you like more details on any part of this process? Here are some related questions:

1. How do you convert polar coordinates to Cartesian coordinates?
2. What are the properties of polar curves?
3. Can you explain the significance of the integral limits?
4. How can you find the area of other types of polar curves?
5. What are common applications of polar coordinate systems?

**Tip:** Always check the symmetry of the curve to simplify calculations!

Find the area of the shaded region. r = 4 + 3 sin(𝜃)

Explore the calculation of the area of the shaded region defined by the polar equation r = 4 + 3 sin(𝜃). This detailed solution utilizes integration in polar coordinates to find that the area is 41π/4. Perfect for advanced math students.

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