Question #75f31

1 Answer
Dec 19, 2017

Answer:

#96 pi#

Explanation:

The graph of the polar curve would be as shown below:
enter image source here

The area enclosed by the curve would be twice the area enclosed by the curve lying over x-axis. The formula for the area enccloseed by a polar curve is given by the formula #int_alpha^beta 1/2 r^2 d theta #.
In the present case #alpha =0# which comes by solving #r=16=8(1+ cos theta)# and #beta =pi #, which is obtained by solving #r=0=8(1+ cos theta)#

Accordingly, the desired area would be #2int_0^pi 1/2 r^2 d theta#

=#int_0^pi 64(1+cos theta)^2 d theta#

=#64int_0^pi (1 +2 cos theta + cos^2 theta) d theta#

=#64int_0^pi (1+2 cos theta +1/2 (1 +cos 2theta)d theta#

=#64[3/2 theta + 2sin theta +1/4 sin 2theta]_0^pi #

=#96pi#