Calculate the area bounded by the polar curve #r=cos theta#?

Question

Isabella Ackerman · Accepted Answer

# pi/4 #

# r=costheta #

The area we seek is.

If we convert to Polar Coordinates then the region #R# is:

an angle from #theta=-pi/2# to #theta=pi/2#
a ray from #r=0# to #r=cos theta#

And as we convert to Polar coordinates we get:

# x \ \ \ = rcos theta#
# y \ \ \ = rsin theta#
# dA = dy dy = r dr d theta#

So then the bounded area is given by#

# A = int int_R \ dA #
# \ \ = int_(-pi/2)^(pi/2) \ int_0^(cos theta) \ r dr d theta #
# \ \ = int_(-pi/2)^(pi/2) \ [1/2r^2]_0^(cos theta) \ d theta #
# \ \ = int_(-pi/2)^(pi/2) \ 1/2cos^2theta \ d theta #
# \ \ = int_(-pi/2)^(pi/2) \ 1/2 (1+cos2theta)/2 \ d theta #
# \ \ = 1/4 \ int_(-pi/2)^(pi/2) \ 1+cos2theta \ d theta #
# \ \ = 1/4 \ [theta + (sin2theta)/2]_(-pi/2)^(pi/2) #
# \ \ = 1/4 \ { (pi/2-1) - (-pi/2-(-1)) } #
# \ \ = 1/4 \ ( pi/2-1+pi/2+1) #
# \ \ = 1/4(pi) #

Observations:

Note that #r=cos theta# is the polar equation of a circle of radius #1/2# centred on #(1/2,0)#, hence the area is:

# A = pi(1/2)^2 = pi/4#

Also, note that if we apply the double integration progress to a generic function #r=r(theta)# we get the polar area formula:

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

Ezekiel Alexander · Answer

undefined To calculate the area bounded by the polar curve $ r = \cos(\theta) $, you need to evaluate the definite integral of $ \frac{1}{2} r^2 $ with respect to $ \theta $ over the interval where $ r = \cos(\theta) $ is positive. This interval is typically from $ \theta = 0 $ to the first positive root of $ r = \cos(\theta) $, which is $ \theta = \frac{\pi}{2} $.

So, the integral to compute the area is:

$$ A = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} (\cos(\theta))^2 \, d\theta $$

After integrating, you'll get the value of the area $ A $.