Calculate the flux of the vector field f⃗ (x, y,z)=(exy+9z+4)i⃗ +(exy+4z+9)j⃗ +(9z+exy)k⃗ through the square of side length 3 with one vertex at the origin, one edge along the positive y-axis, one edge in the xz-plane with x≥0 and z≥0, oriented downward with normal n⃗ =i⃗ −k

The square (call it ) has one vertex at the origin (0, 0, 0) and one edge on the y-axis, which tells us another vertex is (0, 3, 0). The normal vector to the plane is , which is enough information to figure out the equation of the plane containing :

We can parameterize this surface by

for and . Then the flux of , assumed to be

,

is

answer: the answer is 127 degrees