Math need help. thank you

Question

Given that σ is the part of the surface z=2-x in the first octant from y=0 to y=1. If 𝜓(x,y,z) =x²yz, F(x,y,z)=yi+zj+xk and n is the unit normal vector to the surface pointing outwards, evaluate the following surface intergals:

i) ∫∫_σ𝜓(x,y,z) dS.

ii) ∫∫_σ F•n dS.

Answer 1

For (i), our surface is the plane f(x,y) = 2 - x over the region 0 <= x <= 2 and 0 <= y <= 1. (The bounds for the x variable come from the constraint that this plane has to be in the first octant.) Thus, our integral becomes

∫∫σ𝜓(x,y,z) dS = ∫₀¹∫₀²𝜓(x,y,f(x,y)) (f_x(x,y)²+f_y(x,y)² + 1)^1/2 dxdy.

Here, we used the formula dS = (f_x(x,y)²+f_y(x,y)² + 1)^1/2 dxdy when our surface is given by a function z = f(x,y). The subscripts f_x and f_y denote partial derivatives of f with respect to x and y.

Once we have our integral set up, we can pretty quickly evaluate it

∫₀¹∫₀²𝜓(x,y,f(x,y)) (f_x(x,y)+f_y(x,y) + 1)^1/2 dxdy = ∫₀¹∫₀²x²y(2-x)√2 dxdy

= √2 ∫₀¹ydy∫₀²x²(2-x)dx

= 2√2/3.

For (ii), our surface again is the plane f(x,y) = 2 - x over the region 0 <= x <= 2 and 0 <= y <= 1. Thus, our integral becomes

∫∫σ F•n dS = ∫₀¹∫₀² F(x,y,f(x,y))•n(f_x(x,y)²+f_y(x,y)² + 1)^1/2 dxdy.

Again, we're using the formula dS = (f_x(x,y)²+f_y(x,y)² + 1)^1/2 dxdy. Since our surface is a plane, we know the normal vector of the surface is N = <1,0,1>. This vector points out and up because of the positive x and z components, which is what we want. However, N is not a unit vector. We need to divide by its magnitude to the correct unit vector of the surface n = <1/√2,0,1/√2>. Given our unit vector, we're readying to plug everything into our integral to get

∫₀¹∫₀² F(x,y,f(x,y))•n(f_x(x,y)²+f_y(x,y)² + 1)^1/2 dxdy = ∫₀¹∫₀² <y,2-x,x>•<1/√2,0,1/√2>√2 dxdy

= ∫₀¹∫₀² (y+x) dxdy

= 3.