Find the area lying outside r=4sinθ
and inside r=2+2sinθ

To find the area that lies outside \( r = 4\sin\theta \) and inside \( r = 2 + 2\sin\theta \), we can approach the problem using polar coordinates and calculus.

### Step 1: Sketch and Identify Boundaries
Both curves are given in polar form:
1. \( r = 4\sin\theta \) represents a circle with a diameter 4, centered at \( (0, 2) \).
2. \( r = 2 + 2\sin\theta \) represents a cardioid.

We need to find the area that lies inside the cardioid and outside the circle.

### Step 2: Find the Points of Intersection
To find where the curves intersect, we set the two equations equal to each other:
\[
4\sin\theta = 2 + 2\sin\theta
\]
Solving for \( \theta \):
\[
4\sin\theta - 2\sin\theta = 2
\]
\[
2\sin\theta = 2
\]
\[
\sin\theta = 1
\]
This gives \( \theta = \frac{\pi}{2} \) as the point of intersection. Therefore, the curves intersect at \( \theta = \frac{\pi}{2} \), and the corresponding \( r \) is:
\[
r = 4\sin\frac{\pi}{2} = 4
\]

### Step 3: Set up the Integral for Area
The area between two polar curves is given by the formula:
\[
A = \frac{1}{2} \int_{\theta_1}^{\theta_2} \left[ r_{\text{outer}}^2 - r_{\text{inner}}^2 \right] d\theta
\]
Here, the outer curve is \( r = 2 + 2\sin\theta \), and the inner curve is \( r = 4\sin\theta \). We will integrate from \( \theta = 0 \) to \( \theta = \pi \) because the curves are symmetric about the y-axis.

Thus, the area is:
\[
A = \frac{1}{2} \int_0^\pi \left[ (2 + 2\sin\theta)^2 - (4\sin\theta)^2 \right] d\theta
\]

### Step 4: Simplify the Integrand
We expand the squares:
\[
(2 + 2\sin\theta)^2 = 4 + 8\sin\theta + 4\sin^2\theta
\]
\[
(4\sin\theta)^2 = 16\sin^2\theta
\]
Now, the integrand becomes:
\[
A = \frac{1}{2} \int_0^\pi \left[ 4 + 8\sin\theta + 4\sin^2\theta - 16\sin^2\theta \right] d\theta
\]
\[
A = \frac{1}{2} \int_0^\pi \left[ 4 + 8\sin\theta - 12\sin^2\theta \right] d\theta
\]

### Step 5: Compute the Integral
We break the integral into three parts:
\[
A = \frac{1}{2} \left[ \int_0^\pi 4 \, d\theta + \int_0^\pi 8\sin\theta \, d\theta - \int_0^\pi 12\sin^2\theta \, d\theta \right]
\]
1. \( \int_0^\pi 4 \, d\theta = 4\pi \)
2. \( \int_0^\pi 8\sin\theta \, d\theta = -8\cos\theta \Big|_0^\pi = -8(-1 - 1) = 16 \)
3. To integrate \( \sin^2\theta \), we use the identity \( \sin^2\theta = \frac{1 - \cos(2\theta)}{2} \):
   \[
   \int_0^\pi 12\sin^2\theta \, d\theta = 12 \int_0^\pi \frac{1 - \cos(2\theta)}{2} \, d\theta = 6 \left[ \int_0^\pi 1 \, d\theta - \int_0^\pi \cos(2\theta) \, d\theta \right]
   \]
   \[
   = 6 \left[ \pi - 0 \right] = 6\pi
   \]

### Step 6: Final Calculation
Now, combining the results:
\[
A = \frac{1}{2} \left[ 4\pi + 16 - 6\pi \right] = \frac{1}{2} \left[ -2\pi + 16 \right] = -\pi + 8
\]
Thus, the area is:
\[
A = 8 - \pi
\]

### Final Answer:
The area lying outside \( r = 4\sin\theta \) and inside \( r = 2 + 2\sin\theta \) is \( \boxed{8 - \pi} \).

---

Would you like more details on any specific step or have any questions?

Here are 5 related questions to expand your understanding:
1. How do you find the area between two polar curves in general?
2. What are the properties of the polar curves \( r = 4\sin\theta \) and \( r = 2 + 2\sin\theta \)?
3. How can you apply symmetry to simplify polar area problems?
4. What are common polar coordinate curve shapes (e.g., cardioids, circles)?
5. How do you convert between polar and Cartesian coordinates?

**Tip:** When finding the area between polar curves, always check the points of intersection first to determine your limits of integration.

Find the area lying outside r=4sinθ and inside r=2+2sinθ

Explore how to calculate the area between the polar curves r=4sinθ and r=2+2sinθ using integral calculus and polar coordinates. The step-by-step solution simplifies the problem by identifying intersections and applying calculus principles. Suitable for undergraduate students.

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