# Thread: Surface Area (2 Parts)

1. ## Surface Area (2 Parts)

Find the surface area of that portion of the plane
z = 1+2 x-3 y that lies above the triangular region in the xy -plane with vertices (0, 0 ), (3, 0) , and (0, 2) .

I dont understand how to set up the integral or evaluate it....

Using the same plane, z=1+2x-3y, find the surface area that lies above the xy-plane

I am lost...can any one help????

2. Find the equation of the line that passes from (3,0) to (0,2).

We find it has equation $y=\frac{-2}{3}x+2$

Now, $z_{x}=2, \;\ z_{y}=-3, \;\ z_{x}^{2}+z_{y}^{2}+1=\sqrt{14}$

$\int_{0}^{3}\int_{0}^{\frac{-2x}{3}+2}\sqrt{14}dydx$

3. ok i get that...i dont understand the xy-plane part???

4. Originally Posted by galactus
Find the equation of the line that passes from (3,0) to (0,2).

We find it has equation $y=\frac{-2}{3}x+2$

Now, $z_{x}=2, \;\ z_{y}=-3, \;\ {\color{red}\sqrt{z_{x}^{2}+z_{y}^{2}+1}} =\sqrt{14}$

$\int_{0}^{3}\int_{0}^{\frac{-2x}{3}+2}\sqrt{14}dydx$
Small typo (corrected in red) that has no affect on the final double integral given by the big G.

Originally Posted by jme44
Is the answer 3 times the square root of 14 for the first question?
Yes.