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How to evaluate Surface Integral?
How do I evaluate the double integral over the region S of yz dS, where S is the part of the plane x + y + z = 2 that lies in the first octant.
1 Answer
- John SLv 51 decade agoFavorite Answer
x + y + z = 2 and also x, y, z > 0 as the surface lies in the first octant
So,
z = 2 - x - y, 0 < x < 2, 0 < y < 2 - x,
and @z/@x = -1 and @z/@y = -1. (@ = partial differential)
So, the integral
= Int{x = 0 to 2} Int{y = 0 to 2 - x} y(2 - x - y) sqrt((-1)^2 + (-1)^2 + 1) dydx
= Evaluate