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

since the airplane is moving with constant speed then the distance travelled can be calculated by x=u*t

in km/h would be:

x=640*3.5=2.240 km

in m/s would be:

u=640 km/h= 640.000 m/h =640.000/ 3600 m/s =177.8 m/s

t= 3.5 h= 3.5 * 3600 s=12.600 s

then x=u*t= 177.8*12.600= 2.240.280 m

do you have the measurements?