Hi, I need some help on these problems. I'm not sure what to do. 1 Find the area of the plane with vector equation r(u, v) =< 1+v, u-2v, 3-5u+v> that is given by 0<u<1, 0<v<1. So far, I took the partial derivatives with respect to u and v. I don't know if I was supposed to or not and I'm really stuck there. Since I'm trying to find surface area a relevant equation may be S = (double integral) SQRT(1 + (partial derivative of u)^2 + (partial derivative of v)^2) dA 2. The part of the surface z = 1 + 3x +2y^2 that lies above the triangle with vertices (0,0), (0,1), (2,1) Relevant equation should be same as above. I think I got the answer but I'm not sure and would really appreciate some help if I got it wrong. Here's how I did it: partial x: 3 Partial y: 4y The boundary of the triangle is x=2y so it took the integral from 0 to 1 and from 0 to 2y SQRT(10 + 16y^2)dx dy and I got the integral from 0 to 1 times 2y SQRT(10 +16y^2) dy then I took that integral and got (2/3)*(2/32)*SQRT(10 + 16Y^2)^(3/2) from 0 to 1 My final answer is (1/24)(26)^(3/2) - (1/24)(10)^(3/2) Is this correct?