Question

Evaluate the surface integral of the function g over the surface S. g(x,y,z) = x^3 + y^3 + z^3; S is the surface of the cube formed from the coordinate planes and the planes x = 1, y = 1, z = 1

A) 10 B) 6 C) 3/2 D) 9/2

Answer 1

would you like the answer, or process, or hint, or what?

Answer 2

the surface integral taking D to be in the xy-plane is\[\int \int\limits_Sg(x,y,z)dS=\int \int\limits_D g(x,y,f(x,y))\sqrt{(f_x)^2+(f_y)^2+1}dA\]taking the region D to be the 1x1 square in the xy-plane, we can parameterize the top of the cube with z=1 and the bottom with z=0

Answer 3

this gives\[g(x,y,z)=x^3+y^3+1\]for the top of the cuve and\[g(x,y,z)=x^3+y^3\]for the bottom
and since \(f(x,y)=0\) or \(1\) depending on whether we look at the top or bottom of the cube, we also have that\[dS=\sqrt{0^2+0^2+1}dA=dA=dxdy\]for both sides

finally, to determine our bounds we look at the region in the xy-plane Dthe bounds of each integral are clearly just the square \[0\le x\le1;0\le y\le1\]