Evaluate using polar coordinates:
f(x,y)=4, and the region R is given by r=sin(2theta) in the first quadrant.

Question

Evaluate using polar coordinates: 
f(x,y)=4, and the region R is given by r=sin(2theta) in the first quadrant.

hacer · Answer

according to first quadrant we have  $$0\le \theta \le \Pi/2$$ then for the integral  $$\int\limits_{0}^{\Pi/2}\int\limits_{0}^{\sin(2\theta)}4rdrd \theta $$ now taking 4 out of the integral firs I'll  solve according to r's integral such that $$\int\limits_{0}^{\sin(2\theta)}rdr=r^2/2 $$ then putting the equation in order to upper and lower bound .I have $$\sin^2(2\theta)/2-0=\sin^2(2\theta)/2$$ I know $$\sin^2(2\theta)/2=1-\cos(4\theta)/2\times1/2=1-\cos(4\theta)/4$$ Then I have new integral such that  $$4\int\limits_{0}^{\Pi/2}(1-\cos(4\theta))/4  d \theta  $$    taking 1/4 out of the integral I get $$\int\limits_{0}^{\Pi/2}(1-\cos(4\theta))d \theta    $$ $$=\theta-\sin(4\theta)/4$$ putting upper and lower bound in order to $$\theta  $$ the I get $$[\Pi /2-\sin(4\times \Pi/2)/4]-[0-\sin(0)/4]=\Pi/2$$ so the result is $$\Pi/2$$