Find the area inside the lemniscate r squared equals 30 cosine 2 theta and outside the circle r equals StartRoot 15 EndRoot.


The problem asks for the area enclosed between the lemniscate \( r^2 = 30 \cos(2\theta) \) and the circle \( r = \sqrt{15} \). Let's break this problem down into the necessary steps:

### Step 1: Lemniscate equation and interpretation
The equation of the lemniscate is given as:

\[
r^2 = 30 \cos(2\theta)
\]

This equation represents a symmetric figure about the origin, with two lobes. We rewrite it as:

\[
r = \sqrt{30 \cos(2\theta)}
\]

### Step 2: Circle equation
The circle is represented by the equation \( r = \sqrt{15} \).

### Step 3: Setting up bounds
We need to find the region where the lemniscate is outside the circle. To do this, find the angles where the two curves intersect by equating the two equations:

\[
\sqrt{30 \cos(2\theta)} = \sqrt{15}
\]

Squaring both sides:

\[
30 \cos(2\theta) = 15
\]

\[
\cos(2\theta) = \frac{1}{2}
\]

The solutions to this equation are:

\[
2\theta = \pm \frac{\pi}{3} \quad \text{so} \quad \theta = \pm \frac{\pi}{6}
\]

Thus, the limits of integration are from \( -\frac{\pi}{6} \) to \( \frac{\pi}{6} \).

### Step 4: Area between the curves
The formula for the area in polar coordinates is:

\[
A = \frac{1}{2} \int_{\alpha}^{\beta} \left( r_{\text{outer}}^2 - r_{\text{inner}}^2 \right) d\theta
\]

Here, \( r_{\text{outer}} \) is the radius of the lemniscate \( r = \sqrt{30 \cos(2\theta)} \) and \( r_{\text{inner}} \) is the radius of the circle \( r = \sqrt{15} \). Therefore, the area is:

\[
A = \frac{1}{2} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \left( 30 \cos(2\theta) - 15 \right) d\theta
\]

### Step 5: Solving the integral

Let's compute the integral:

\[
A = \frac{1}{2} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \left( 30 \cos(2\theta) - 15 \right) d\theta
\]

This can be split into two integrals:

\[
A = \frac{1}{2} \left( 30 \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \cos(2\theta) d\theta - 15 \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} d\theta \right)
\]

1. The integral of \( \cos(2\theta) \):

\[
\int \cos(2\theta) d\theta = \frac{1}{2} \sin(2\theta)
\]

Evaluating at the limits:

\[
\left[ \frac{1}{2} \sin(2\theta) \right]_{-\frac{\pi}{6}}^{\frac{\pi}{6}} = \frac{1}{2} \left( \sin\left(\frac{\pi}{3}\right) - \sin\left(-\frac{\pi}{3}\right) \right) = \frac{1}{2} \left( \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{2}
\]

So the first integral becomes:

\[
30 \times \frac{\sqrt{3}}{2} = 15\sqrt{3}
\]

2. The second integral is straightforward:

\[
\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} d\theta = \frac{\pi}{3}
\]

### Step 6: Final area calculation

Now, substitute both results back into the expression for \( A \):

\[
A = \frac{1}{2} \left( 15\sqrt{3} - 15 \times \frac{\pi}{3} \right)
\]

\[
A = \frac{1}{2} \left( 15\sqrt{3} - 5\pi \right)
\]

\[
A = \frac{15\sqrt{3}}{2} - \frac{5\pi}{2}
\]

Thus, the area inside the lemniscate and outside the circle is:

\[
A = \frac{15\sqrt{3}}{2} - \frac{5\pi}{2}
\]

### Conclusion:
The area is \( \frac{15\sqrt{3}}{2} - \frac{5\pi}{2} \).

---

Do you want details on any of the steps or have any further questions? Here are some related questions you might find interesting:

1. How do you find the area enclosed by a polar curve?
2. What is the geometric interpretation of a lemniscate?
3. How do you determine the points of intersection between polar curves?
4. What are common techniques for integrating trigonometric functions?
5. Can you solve similar problems with other polar curves, like roses?

**Tip:** In polar coordinates, always set up the limits carefully, as they play a critical role in defining the correct region.

Find the area inside the lemniscate r squared equals 30 cosine 2 theta and outside the circle r equals StartRoot 15 EndRoot.

Learn how to calculate the area between a lemniscate r^2 = 30 cos(2θ) and a circle r = √15 using polar coordinates. This problem covers key mathematical concepts like trigonometric integration, finding intersection points, and using the area formula for polar curves. Suitable for Calculus II students.

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