What are the bounds for the double integral in this curved surface area problem?

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SUMMARY

The discussion focuses on determining the bounds for the double integral in the context of calculating the surface area of the surface defined by the equation z² = 2xy. The specific region of interest is above the xy-plane and bounded by the planes x=0, x=2, and y=0, y=1. The correct bounds for the double integral are established as 0 ≤ x ≤ 2 and 0 ≤ y ≤ 1, allowing for the integration of the surface area formula derived from the partial derivatives of z.

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harrietstowe
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Homework Statement



Find the area of the part of the surface z^2 = 2*x*y that lies above the xy plane and is bounded by the planes x=0, x=2 and y=0, y=1.

Homework Equations





The Attempt at a Solution


z = Sqrt[2*x*y]
Sqrt[(partial z/partial x)^2 + (partial z/partial y)^2 +1] =
(Sqrt[2/(x*y)] * (x+y))/2
So Integrate over that but I can't figure out the bounds for this double integral.
Thank You
 
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The boundary is a nice, neat box, so you would just integrate z over 0 <= x <= 2, 0 <= y <= 1.
 
Ok I see. Thank You
 

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