Question

Find the area of the region enclosed by the polar curve r=sin(2theta) for 0 is less than or equal to theta is less than or equal to pi/2. (Answer: pi/8)

Answer 1

@phi

Answer 2

OK, here is what I got. In polar coordinates with angle x and radius r, do a double integral to find the area of the figure.

See the attached figure (¼ of a "rose")
the radius ranges from 0 to sin(2x)
the angle x ranges from 0 to pi/2

the double integral is
\[ \int_0^{\frac{\pi}{2}} \int_0^{\sin(2x)} r\ dr\ dx\]

you will get a \(\sin^2(u) \) that you will have to integrate... you can use
\[ \sin^2(u) = \frac{1}{2} \left( 1- \cos(2u)\right) \]