# Calculate the double integral of (xsiny-ysinx)dA; R={(x,y); 0 <x<pi/2, 0<y<pi/3}.I'm really confused about how you find the integral in the first place, i know it's integral from 0-pi/2...

Calculate the double integral of (xsiny-ysinx)dA; R={(x,y); 0 <x<pi/2, 0<y<pi/3}.

I'm really confused about how you find the integral in the first place, i know it's integral from 0-pi/2 integral from 0-pi/3 (xsiny-ysinx) dy dx, but what do I do after that? I tried to sub in the values 0 for x and pi/2 for y but this just gives me 0, and I know that can't be right? Help is greatly appreciated! Thanks! By the way, it is 0<(or equal to)x<or equal to pi/2, and same for the range of y!

### 1 Answer | Add Yours

You just integrate as if y was constant, then substitute and then integrate with respect to y.

`int^(pi/3)_0 int^(pi/2)_0 xsiny-ysinx dx dy`

`= int^(pi/3)_0 [x^2/2 sin y+ y cos x]^(pi/2)_0 dy`

`= int^(pi/3)_0 [pi^2/8 sin y + y(cos pi/2) - 0^2/8 sin y - y cos 0] dy`

`= int^(pi/3)_0 [pi^2/8 sin y - y] dy`

`= [pi^2/8 (-cos y) - y^2/2]^(pi/3)_0`

`= pi^2/8 (-cos pi/3) - pi^2/18 - pi^2/8(-cos 0) + 0^2/2`

`= pi^2/8 (-1/2) - pi^2/18 + pi^2/8 = (-1/16 - 1/18 + 1/8)pi^2`

`= (-9/144 - 8/144 + 18/144)pi^2`

`= pi^2/144` Final answer...