Question

evaluate the surface integral

double integral (y dS)

z= 2/3 (x^3/2 + y^3/2)

0 ≤ x ≤ 5, 0 ≤ y ≤ 2

Answer 1

∫∫_S f(x,y,z) dS
we have f(x,y,z) = y
z = 2/3 (x^3/2 + y^3/2)
∂z/∂x = x^1/2
∂z/∂y = y^1/2
so the integral becomes

∫[0,5]∫[0,2] y √(x+y+1) dy dx

Looks pretty straightforward now, right?

Answer 2

I am having trouble taking the first integral. I get y^2/2 (x+y+1)^3/2 (1) is this correct? or do I take the integral of the inside instead of the derivative.