Evaluate the integral $$\int\limits_{}^{}\int\limits_{}^{}R(x ^{2}-2y ^{2})dA$$ where R is the first quadrant region between the circles of radius 3 and radius 6

Question

Evaluate the integral $$\int\limits_{}^{}\int\limits_{}^{}R(x ^{2}-2y ^{2})dA$$  where R is the first quadrant region between the circles of radius 3 and radius 6

mathmate · Answer

Starting with what you started before,
∫∫R(x2−2y2)dA
=∫∫(rcos(t))^2-2(rsin(t))^2rdtdr
=∫∫r^3(cos(t)^2-2sin(t)^2)dtdr
=∫∫r^3(cos(t)^2-2sin(t)^2)dt
=[r^4/4]∫((1/2)(1+cos(2t))-(1-cos(2t))dt
=(6^4-3^4)/4 * ∫(-1/2 + cos(2t)/2 + cos(2t))dt
=(1215/4) [-t/2 + sin(2t)/4 + sin(2t)/2] from 0 to pi/2
=(1215/4)(-pi/4)
=-1215pi/16
I believe the negative value is from the -2 in -2y^2.