area of 3-dimensional surface, z = 1 + 3x + 2y^2, above

[mathstresser]

Find the area of the surface.
The part of the surface z = 1 + 3x + 2y^2 that lies above the triangle with vertices (0, 0), (0, 1), and (2, 1).

I get this
\(\displaystyle \L\\\int_{0}^{2}\int_{0}^{1}1+3x+2y^2dydx\)

But I don’t know what else to do, or what to do differently. Please help!

6

 

[galactus]

Remember your line equation from (0,0) to (2,1) is y=x/2.