Surface area problem in 3-d calculus

[dukefanfromch]

Homework Statement

Find the area of the surface.
The surface z = (2/3)(x^(3/2) + y^(3/2)), 0 </= x </= 1, 0 </= y </= 1

Homework Equations

Double integral over S of the magnitude of dr/du cross dr/dv dS, which equals the double integral over D of the magnitude of dr/du cross dr/dv dA.
(SSs |dz/dx X dz/dy|dS = SS D |dz/dx X dz/dy|dA)

The Attempt at a Solution

_ 1 1
I found the integral to be SS D (1+x+y)^(1/2) dA = SS (1+x+y)^(1/2) dxdy
_ 0 0
but my answer keeps coming out wrong. I might be making a mistake with algebra because I get a lot of different answers when i do it different ways. This is from an NC State Calculus 3 homework assignment, if anyone may have seen this problem before and remember how to do it. Help?

 

[HallsofIvy]

Homework Statement

Find the area of the surface.
The surface z = (2/3)(x^(3/2) + y^(3/2)), 0 </= x </= 1, 0 </= y </= 1

Homework Equations

Double integral over S of the magnitude of dr/du cross dr/dv dS, which equals the double integral over D of the magnitude of dr/du cross dr/dv dA.
(SSs |dz/dx X dz/dy|dS = SS D |dz/dx X dz/dy|dA)

The Attempt at a Solution

_ 1 1
I found the integral to be SS D (1+x+y)^(1/2) dA = SS (1+x+y)^(1/2) dxdy
_ 0 0
but my answer keeps coming out wrong. I might be making a mistake with algebra because I get a lot of different answers when i do it different ways. This is from an NC State Calculus 3 homework assignment, if anyone may have seen this problem before and remember how to do it. Help?

What you have shown is completely correct. Now, how did you do that integration and what answer did you get?

 

[dukefanfromch]

I got the right answer finally, 1.4066. It was an algebraic error when I was evaluating the integral. Thanks for telling me I was doing it right though, that helped me know it wasn't a mistake in my set-up.