Double Integration - Surface Area

pious&peevish
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Homework Statement



Find the surface area of the surface defined by:

z = (2/3)(x^(3/2) + y^(3/2)), 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

Homework Equations



SA = ∫∫ √(fx)^2 + (fy)^2 + 1 dA

The Attempt at a Solution



Solved for the partial derivatives and plugged them into the formula, but got a really messy answer in the end. I think I was supposed to do a change of variables somewhere but I'm not sure how...
 
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pious&peevish said:

Homework Statement



Find the surface area of the surface defined by:

z = (2/3)(x^(3/2) + y^(3/2)), 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

Homework Equations



SA = ∫∫ √(fx)^2 + (fy)^2 + 1 dA

The Attempt at a Solution



Solved for the partial derivatives and plugged them into the formula, but got a really messy answer in the end. I think I was supposed to do a change of variables somewhere but I'm not sure how...
Use adequate parentheses.

Do you mean SA = ∫∫ √((fx)^2 + (fy)^2 + 1) dA , or perhaps something else.

Added in Edit:
What did you get for ∂f/∂x ?

The integrand turns out to be fairly simple.
 
Yes, sorry - I meant ∫∫ √((fx)^2 + (fy)^2 + 1) dA .

I had √(x) for ∂f/∂x and √(y) for ∂f/∂y.
 
pious&peevish said:
Yes, sorry - I meant ∫∫ √((fx)^2 + (fy)^2 + 1) dA .

I had √(x) for ∂f/∂x and √(y) for ∂f/∂y.

It doesn't matter what order of integration you go in this case. If you plug in your ##f_x## and ##f_y## into your integral it becomes very easy with a regular substitution.

i.e Let's say you integrate x first. So u = x + y + 1 implies du = dx.
 
Ah, now it makes perfect sense! Thanks!
 
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