area of a surface: part of y = 4x + z^2 between planes

[mathstresser]

Find the area of the surface.
The part of the surface y=4x+z^2 that lies between the planes x = 0, x = 1, z = 0, and z = 1

Is this right so far?

\(\displaystyle \L \\ \int_{0}^{1}\,\int_{0}^{1}\, \left( 4x\,+\,z^2 \right) \,dx\,dz\)

Do I need to do anything differently?

 

[galactus]

Surface area formula:

\(\displaystyle \L\\S=\int\int_{R}\sqrt{\left(\frac{\partial{y}}{\partial{x}}\right)^{2}+\left(\frac{\partial{y}}{\partial{z}}\right)^{2}+1}dA\)

\(\displaystyle \L\\y=4x+z^{2}\)

\(\displaystyle \L\\y_{x}=4\)

\(\displaystyle \L\\y_{z}=2z\)

 

[mathstresser]

thank you so much