# Another Surface Area question

• April 17th 2008, 07:12 PM
apw235@nyu.edu
Another Surface Area question
we have $z=1+3x+2y^2$ and I am to find the part of the surface that lies above a triangle with vertices at (0,0),(0,1),(2,1).
So far:
$dS=\sqrt{1+(\frac{dz}{dx})^2+(\frac{dz}{dy})^2}dA= \sqrt{1+(3)^2+(4y)^2}dA=\sqrt{16y^2+10}dA$, so the integral looks ugly...
• April 18th 2008, 05:47 AM
apw235@nyu.edu
bump
• April 18th 2008, 06:04 AM
mr fantastic
Quote:

Originally Posted by apw235@nyu.edu
we have $z=1+3x+2y^2$ and I am to find the part of the surface that lies above a triangle with vertices at (0,0),(0,1),(2,1).
So far:
$dS=\sqrt{1+(\frac{dz}{dx})^2+(\frac{dz}{dy})^2}dA= \sqrt{1+(3)^2+(4y)^2}dA=\sqrt{16y^2+10}dA$, so the integral looks ugly...

Have you sketched the region of integration in the xy-plane?

Your integral is $S = \int_{x = 0}^{x = 2} \int_{y = x/2}^{y = 1} \sqrt{16y^2+10} \, dy \, dx$.

Life is much easier if you reverse the order of integration:

$S = \int_{y = 0}^{y = 1} \int_{x = 0}^{x = 2y} \sqrt{16y^2+10} \, dx \, dy$.

This should cause little trouble.