Question

Evaluate the surface integral \(\int\int_{\sigma}\sqrt{2e^{2(x+y)}+1}\text{ dS}\) given that S is the surface \(z=e^{x}e^y\) above the triangular region in the xy-plane enclosed by \(y=1-x\) and the coordinate axes.

Answer 1

Here's our region Dso our bounds will be\[0\le y\le1-x\]\[0\le x\le 1\]now we need dS, and since this surface is given by z=g(x,y) our formula for that will be\[dS=\sqrt{(\frac{dz}{dx})^2+(\frac{dz}{dy})^2+1}dA=\sqrt{(e^xe^y)^2+(e^xe^y)^2+1}dA\]\[dS=\sqrt{2e^{2(x+y)}+1}dA\]which is extremely convenient, because now our integral will be much easier\[\int\int_\sigma\sqrt{2e^{2(x+y)}+1}dS=\int_{0}^{1}\int_{0}^{1-x}2e^{2(x+y)}+1dydx\]you should be able to integrate this yourself without too much trouble.