Find ∬R x√(y) dA where R is the region in the first quadrant bounded above by the curve y = 9−x^2

Find ∬R x√(y) dA where R is the region in the first quadrant bounded above by the curve y = 9−x^2 ??

stuck.. help!!

Note that y = 9 - x^2 crosses the x-axis when x = 3 (we ignore x = -3 since we want x > 0).

So, ∫∫R xy^(1/2) dA
= ∫(x = 0 to 3) ∫(y = 0 to 9 - x^2) xy^(1/2) dy dx
= ∫(x = 0 to 3) x * (2/3)y^(3/2) {for y = 0 to 9 - x^2} dx
= (-1/3) ∫(x = 0 to 3) -2x(9 - x^2)^(3/2) dx
= (-1/3) * (2/5)(9 - x^2)^(5/2) {for x = 0 to 3}
= (2/15) * 3^5
= 162/5.

I hope this helps!