Find the area of the surface z = 2/3(x^(3/2) + y^(3/2))

0 <= x <=1 , 0 <= y <=1

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- Nov 11th 2008, 06:16 PMjlt1209Find the area of the surface ...
Find the area of the surface z = 2/3(x^(3/2) + y^(3/2))

0 <= x <=1 , 0 <= y <=1 - Nov 11th 2008, 07:10 PMTKHunny
What is your plan? Can you relate the task to the arc length in 2D?

- Nov 11th 2008, 07:19 PMjlt1209
Well i know dz/dx = x^1/2 and dz/dy = y^1/2.

Then putting it into the formula sqrt(1 + (dz/dx)^2 + (dz/dy)^2) dA

would be double integral of sqrt(1 + x + y) dA

Then I think it would be double integrals both from 0 to 1 sqrt(1 + cos(theta) + sin(theta)) r dr dtheta

Is this correct so far? I think I did something wrong since that gets really hard to integrate.