Surface area Double Integral Problem

[CoryB]

I am having issues solving this problem, it is getting way too big on me so think I am making a mistake somewhere.

Here Is the problem as it is written:
Express the area of the given surface as an iterated double integral, and then find the surface area. The portion of the surface z=2x+y^2 that is above the triangular region with vertices (0,0), (0,1) and (1,1)

To set it up I ended up with int(sqrt(4+4y^2+1)dydx) My outer limits were from 0 to 1 while my inner limits were from x to 1

Any help would be greatly appreciated

 

[CoryB]

Never mind, I figured it out. For some reason I did not think to switch the integrals, if you do that and integrate with respect to x first instead of y it simplifies down very nicely and you do not end up with the huge problem I was facing