What is the area inside the polar curve #r=1#, but outside the polar curve #r=2costheta#?

Question

Charles Anctil · Accepted Answer

# A = pi/3 + sqrt(3)/2 ~~ 1.9132 #

Here is the graph of the two curves. The shaded area, #A#, is the area of interest:

This is a symmetrical problems so we only need find the shaded area, #B# and subtract twice this from that of a unit circle (#r=1#).

We can find the polar coordinate of the point of intersection in Q1 by simultaneously solving the polar equations:

# r=2cos theta #
# r=1 #

From which we get:

# 2cos theta = 1 =>cos theta = 1/2#
# :. theta = pi/3 #

So we can easily calculate the area, #B#, which is that of the a circle sector #C# and that bounded by the curve #r=2costheta# where #theta in (pi/3,pi/2)#

The area, #C# is simply #1/2r^2theta#:

# A_C = 1/2 * 1 * pi/3 #
# \ \ \ \ \ = pi/6 #

And we calculate the area of the segment #B#, via Calculus using:

# A = int \ 1/2r^2 \ d theta#

Thus:

# A_D = int_(pi/3)^(pi/2) \ 1/2(2costheta)^2 \ d theta#
# \ \ \ \ \ = int_(pi/3)^(pi/2) \ 1/2 \ 4cos^2 theta \ d theta#
# \ \ \ \ \ = 2int_(pi/3)^(pi/2) \ cos^2 theta \ d theta#
# \ \ \ \ \ = 2int_(pi/3)^(pi/2) \ 1/2(cos2theta+1) \ d theta#
# \ \ \ \ \ = int_(pi/3)^(pi/2) \ cos2theta+1 \ d theta#
# \ \ \ \ \ = [1/2sin2theta + theta]_(pi/3)^(pi/2)#
# \ \ \ \ \ = (1/2sin(pi)+pi/2) - (1/2sin((2pi)/3)+pi/3)#
# \ \ \ \ \ = (0+pi/2) - (1/2 sqrt(3)/2+pi/3)#
# \ \ \ \ \ = pi/2 - sqrt(3)/4-pi/3#
# \ \ \ \ \ = pi/6 - sqrt(3)/4#

So then the total area we require, is given by:

# A_A = pi(1)^2 - 2(A_C+A_D) #
# \ \ \ \ \ = pi - 2(pi/6 + pi/6 - sqrt(3)/4) #
# \ \ \ \ \ = pi - 2(pi/3 - sqrt(3)/4) #
# \ \ \ \ \ = pi - (2pi)/3 + (2sqrt(3))/4 #
# \ \ \ \ \ = pi/3 + sqrt(3)/2 #

Charles Arocha · Answer

undefined To find the area inside the polar curve $r = 1$ but outside the polar curve $r = 2\cos(\theta)$, you need to compute the definite integral of the difference between the two curves from the starting angle to the ending angle.

The area $A$ is given by:

$$ A = \frac{1}{2} \int_{\alpha}^{\beta} [(2\cos(\theta))^2 - (1)^2] \, d\theta $$

where $\alpha$ and $\beta$ are the angles where the curves intersect.

To find the intersection points, set the two equations equal to each other and solve for $\theta$:

$$ 1 = 2\cos(\theta) $$

$$ \theta = \arccos\left(\frac{1}{2}\right) $$

$$ \theta = \frac{\pi}{3}, \frac{5\pi}{3} $$

Now, integrate the difference between the curves from $\frac{\pi}{3}$ to $\frac{5\pi}{3}$:

$$ A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} [(2\cos(\theta))^2 - (1)^2] \, d\theta $$

$$ A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} [4\cos^2(\theta) - 1] \, d\theta $$

$$ A = \frac{1}{2} \left[ 4\int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \cos^2(\theta) \, d\theta - \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} 1 \, d\theta \right] $$

$$ A = \frac{1}{2} \left[ 4\int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \frac{1 + \cos(2\theta)}{2} \, d\theta - \left[\theta\right]_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \right] $$

$$ A = \frac{1}{2} \left[ 2\left[\frac{\theta}{2} + \frac{\sin(2\theta)}{4}\right]_{\frac{\pi}{3}}^{\frac{5\pi}{3}} - \left[\theta\right]_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \right] $$

$$ A = \frac{1}{2} \left[ 2\left(\frac{5\pi}{6} + \frac{\sqrt{3}}{4}\right) - \left(\frac{5\pi}{3} - \frac{\pi}{3}\right) \right] $$

$$ A = \frac{1}{2} \left[ \frac{5\pi}{3} + \frac{\sqrt{3}}{2} - \frac{4\pi}{3} \right] $$

$$ A = \frac{1}{2} \left[ \frac{\pi}{3} + \frac{\sqrt{3}}{2} \right] $$

$$ A = \frac{\pi}{6} + \frac{\sqrt{3}}{4} $$

So, the area inside the polar curve $r = 1$ but outside the polar curve $r = 2\cos(\theta)$ is $\frac{\pi}{6} + \frac{\sqrt{3}}{4}$.