# Thread: Finding the Surface Area above a Triangle

1. ## Finding the Surface Area above a Triangle

Hey guys, I'm trying to find the surface area z = y^2, above the triangle with vertices (0,0), (0,1),(1,1).
I'm pretty lost on how to bring the triangle into this, all help is appreciated!

2. ## Re: Finding the Surface Area above a Triangle

Given a function z=f(x,y,z) you can find the surface area in 3-dimensions using:

$\displaystyle \int \int \sqrt{(\frac{\partial f}{\partial x})^2 + (\frac{\partial f}{\partial y})^2 + 1} dx dy$ from wikipedia Surface integral - Wikipedia, the free encyclopedia where the integral is over the desired region. Since it is a triangle you're working with, the limits of x will depend on y.

3. ## Re: Finding the Surface Area above a Triangle

A more general method is this- We can write the surface $\displaystyle z= y^2$ as the vector equation $\displaystyle \vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{k}Z+ y^2\vec{k}$ using x and y as parameters. Then $\displaystyle \vec{r}_x= \vec{i}$ and $\displaystyle \vec{r}_y= \vec{j}+ 2y\vec{k}$. The cross product of those two vectors, $\displaystyle 2y\vec{j}+ \vec{k}$, is perpendicular to the surface and its length gives the "differential of surface area", [tex]\sqrt{4y^2+ 1}dxdy[tex] which is the same as the integrand Krahl gives.

The regionof integration, in the xy plane, is the triangle with vertices (0, 0), (0, 1), and (1, 1). The edges of that triangle are x= 0, y= 1, and y= x. If you choose to integrate with respect to y first and then x, x goes from 0 to 1 so the "outer integral" has lower limits x=0 and upper limit x= 1. For each x, y goes from y= x to 1 so the "inner integral" has lower limit y= x and upper limit y= 1. If you choose to integrate with respect to x first and then y, y goes from 0 to 1 so the "outer integral" has lpwer limit y= 0 and upper limit y= 1. For each y, x goes from 0 to y= x so the "integral" has lower limit x= 0 and upper limit x= y.