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

1 Answer
Oct 19, 2017

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

Explanation:

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

The area we seek is shaded in grey.

Steve M using Autograph

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 #