# Find the area of the region inside these curves?

##
curves:

#r=1+\cos\theta# (red)
#r=1-\cos\theta# (blue)

curves:

#r=1+\cos\theta# (red)#r=1-\cos\theta# (blue)

##### 1 Answer

# (3pi)/2 - 4 #

#### Explanation:

The question is ambiguous, asking us to find the area inside the polar curves:

# r = 1 + cos theta # ..... [A]

# r = 1 - cos theta # ..... [B]

Assuming that we seek the area bounded by both curves:

First we test for symmetry (which would appear obvious) about

# r = 1 + cos theta |-> r=1 + cos -theta = 1 - cos theta#

which is the equation of [B]. Similarly we test for symmetry about

# r = 1 + cos theta |-> r=1 + cos (pi-theta)#

# => r = 1 + cos pi cos theta + sin pi sin theta = 1 - cos theta #

which, again, is the equation of [B].

These test confirm that we have symmetry about

We calculate area in polar coordinates using :

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

Thus, the enclosed area of a single half petal is:

# A_1 = 1/2 \ int_(0)^(pi/2) \ (1 - cos theta)^2 \ d theta #

# \ \ \ \ = 1/2 \ int_(0)^(pi/2) \ 1 - 2cos theta + cos^2 theta \ d theta #

# \ \ \ \ = 1/2 \ int_(0)^(pi/2) \ 1 - 2cos theta + (1+cos 2theta)/2 \ d theta #

# \ \ \ \ = 1/2 \ int_(0)^(pi/2) \ 3/2 - 2cos theta + 1/2cos 2theta \ d theta #

# \ \ \ \ = 1/2 \ [ \ (3 theta)/2 - 2sin theta + 1/4sin 2theta \ ]_(0)^(pi/2) #

# \ \ \ \ = 1/2 \ {(3/2*pi/2 - 2sin(pi/2)+1/4sin(pi))-(0-2sin0+1/4sin0) } #

# \ \ \ \ = 1/2 \ {(3pi)/4 - 2+0)0(0)} #

So that the total area of all 4 half-petals, is:

# A = 4A_1 #

# \ \ = 4 * 1/2 \ ((3pi)/4 - 2) #

# \ \ = (3pi)/2 - 4 #