Find the equation of the line that passes from (3,0) to (0,2).
We find it has equation
Now,
Find the surface area of that portion of the planez = 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????