Find the area of the region that lies inside both curves: $r = \sqrt{3}\cos(\theta), \quad r = \sin(\theta)$

Neetesh Kumar | January 2, 2025

Calculus Homework Help

This is the solution to Math 1c
Assignment: 10.4 Question Number 16
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Step-by-step solution:

To find the area of the region that lies inside both curves, we need to determine the bounds of integration where the two curves intersect and use the formula for area in polar coordinates:

$A = \frac{1}{2} \int_{\theta_1}^{\theta_2} \min\{r_1^2, r_2^2\} \, d\theta$

1. Find the Points of Intersection:

The two curves are given as:

1. $r_1 = \sqrt{3}\cos(\theta)$
2. $r_2 = \sin(\theta)$

At the points of intersection, $r_1 = r_2$.

Thus:

$\sqrt{3}\cos(\theta) = \sin(\theta)$

Divide through by $\cos(\theta)$ (for $\cos(\theta) \neq 0$):

$\sqrt{3} = \tan(\theta)$

Solve for $\theta$:

$\theta = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$

The other angle of intersection is $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ (from symmetry).

Thus, the bounds of integration are:

$\theta_1 = \frac{\pi}{3}, \quad \theta_2 = \frac{2\pi}{3}.$

2. Set Up the Integral:

The area of the region common to both curves is determined by integrating the smaller of $r_1^2$ and $r_2^2$.

Within $\theta \in \left[\frac{\pi}{3}, \frac{2\pi}{3}\right]$, $\sqrt{3}\cos(\theta)$ is the smaller value.

Thus:

$A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} (\sqrt{3}\cos(\theta))^2 \, d\theta$

Simplify $r_1^2$:

$r_1^2 = (\sqrt{3}\cos(\theta))^2 = 3\cos^2(\theta).$

The area becomes:

$A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} 3\cos^2(\theta) \, d\theta$

3. Simplify Using a Trigonometric Identity:

Use the identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$:

$3\cos^2(\theta) = \frac{3}{2} (1 + \cos(2\theta))$

Substitute this into the integral:

$A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{3}{2} (1 + \cos(2\theta)) \, d\theta$

Factor out $\frac{3}{2}$:

$A = \frac{3}{4} \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} (1 + \cos(2\theta)) \, d\theta$

Split the integral:

$A = \frac{3}{4} \left[ \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} 1 \, d\theta + \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \cos(2\theta) \, d\theta \right].$

4. Evaluate Each Integral:

1. First Integral:

$\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} 1 \, d\theta = \left[ \theta \right]_{\frac{\pi}{3}}^{\frac{2\pi}{3}} = \frac{2\pi}{3} - \frac{\pi}{3} = \frac{\pi}{3}.$

2. Second Integral:

$\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \cos(2\theta) \, d\theta = \frac{1}{2} \sin(2\theta) \bigg|_{\frac{\pi}{3}}^{\frac{2\pi}{3}}.$

Evaluate $\sin(2\theta)$ at the bounds:

• At $\theta = \frac{2\pi}{3}$:

$\sin(2\theta) = \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}.$

• At $\theta = \frac{\pi}{3}$:

$\sin(2\theta) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}.$

Thus:

$\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \cos(2\theta) \, d\theta = \frac{1}{2} \left(-\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right) = \frac{1}{2} \cdot \left(-\sqrt{3}\right) = -\frac{\sqrt{3}}{2}.$

5. Combine Results:

Now substitute the results into the area formula:

$A = \frac{3}{4} \left[ \frac{\pi}{3} + \left(-\frac{\sqrt{3}}{2}\right) \right].$

Simplify:

$A = \frac{3}{4} \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right).$

Distribute $\frac{3}{4}$:

$A = \frac{\pi}{4} - \frac{3\sqrt{3}}{8}.$

Final Answer:

The area of the region that lies inside both curves is:

$A = \boxed{\frac{\pi}{4} - \frac{3\sqrt{3}}{8}}$

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