How do you find the area of the region bounded by the polar curves #r=sqrt(3)cos(theta)# and #r=sin(theta)# ?

Question

18286 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-09-25T22:49:41

The area of the enclosed region is #5/24 pi-sqrt{3}/4#.

It looks like this:

enter image source here

#A=int_0^{pi/3}int_0^{sin theta}rdrd theta +int_{pi/3}^{pi/2}int_0^{sqrt{3}cos theta}rdrd theta#

#=int_0^{pi/3} [r^2/2]_0^{sin theta} d theta + int_{pi/3}^{pi/2}[r^2/2]_0^{sqrt{3}cos theta} d theta#

#=int_0^{pi/3} {sin^2theta}/2 d theta + int_{pi/3}^{pi/2}{3cos^2theta}/2 d theta#

#=1/4int_0^{pi/3} (1-sin2 theta) d theta + 3/4int_{pi/3}^{pi/2}(1+cos2 theta) d theta#

#=1/4[theta-{sin2theta}/2]_0^{pi/3} +3/4[theta+{sin2theta}/2]_{pi/3}^{pi/2}#

#=1/4(pi/3-sqrt{3}/4)+3/4(pi/6-sqrt{3}/4)#

#=5/24 pi-sqrt{3}/4#