Find the area bounded by the polar curves? #r=4+4cos theta# and #r=6#

Question

Christopher Agresta · Accepted Answer

Bounded Area = # 18sqrt(3) - 4pi #

# r=4+4costheta \ \ \ \ \ \ (color(red)(red)) #
# r=6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(color(blue)(blue)) #

The area we seek is shaded in grey.

Let us first find the points of intersection:

# 4+4costheta=6 #
# :. 4cos theta = 2 #
# :. cos theta = 1/2 #
# :. theta = +-pi/3#

We calculate area in polar coordinates using :

# A = 1/2 \ int_alpha^beta \ r^2 \ d theta #

The area bounded by #r=4+4costheta# and #r=+-pi/3# is:

# A_1 = 1/2 \ int_(-pi/3)^(pi/3) \ (4+4costheta)^2 \ d theta #
# \ \ \ \ = 1/2 \ int_(-pi/3)^(pi/3) \ 16(1+costheta)^2 \ d theta #
# \ \ \ \ = 8 \ int_(-pi/3)^(pi/3) \ 1+2costheta+cos^2theta \ d theta #
# \ \ \ \ = 8 \ int_(-pi/3)^(pi/3) \ 1+2costheta+(1+cos2theta)/2 \ d theta#
# \ \ \ \ = 8 \ int_(-pi/3)^(pi/3) \ 1+2costheta+1/2+(cos2theta)/2 \ d theta#
# \ \ \ \ = 8 \ int_(-pi/3)^(pi/3) \ 3/2+2costheta+(cos2theta)/2 \ d theta#
# \ \ \ \ = 8 [3/2theta+2sintheta+(sin2theta)/4]_(-pi/3)^(pi/3) #
# \ \ \ \ = 8 { (3/2(pi/3) + 2sin(pi/3)+1/4sin((2pi)/3)) - (3/2(-pi/3) + 2sin(-pi/3)+1/4sin((-2pi)/3))} #
# \ \ \ \ = 8 { ( pi/2 + 2sqrt(3)/2+1/4sqrt(3)/2) - (-pi/2 - 2sqrt(3)/2-1/4sqrt(3)/2)} #
# \ \ \ \ = 8 { pi/2 + 2sqrt(3)/2+1/4sqrt(3)/2 +pi/2 + 2sqrt(3)/2+1/4sqrt(3)/2} #
# \ \ \ \ = 8 { pi + 2sqrt(3)+1/4sqrt(3)} #
# \ \ \ \ = 8pi + 18sqrt(3) #

The area bounded by #r=6# and #r=+-pi/3# is a sector of angle #(2pi)/3# so we can use the sector area #1/2r^2theta#

# A_2 = 1/2(6^2)(2pi)/3 #
# \ \ \ \ = 1/2 xx 36 xx (2pi)/3 #
# \ \ \ \ = 12pi #

Then, the difference is:

# A = A_1 - A_2 #
# \ \ \ = (8pi + 18sqrt(3)) - (12pi) #
# \ \ \ = 18sqrt(3) - 4pi #

Note:

We could also construct a single integral to get the solution using:

# A = 1/2 \ int_(-pi/3)^(pi/3) \ (4+4costheta)^2 - (6)^2 \ d theta #

Isaiah Allen · Answer

undefined To find the area bounded by the polar curves $r = 4 + 4\cos(\theta)$ and $r = 6$, integrate the difference of the curves' areas over the interval where the curves intersect.

1. Find the intersection points by setting the two equations equal to each other and solving for θ.
$$4 + 4\cos(\theta) = 6$$
$$4\cos(\theta) = 2$$
$$\cos(\theta) = \frac{1}{2}$$
$$\theta = \frac{\pi}{3}, \frac{5\pi}{3}$$

2. Integrate the area formula from θ = π/3 to θ = 5π/3.
$$A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (6^2 - (4 + 4\cos(\theta))^2) \, d\theta$$
$$A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (36 - (16 + 32\cos(\theta) + 16\cos^2(\theta))) \, d\theta$$

3. Simplify the integral and evaluate it.
$$A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (20 - 32\cos(\theta) - 16\cos^2(\theta)) \, d\theta$$

4. Use trigonometric identities to simplify further.

5. Evaluate the integral.

6. The result will be the area bounded by the given polar curves.