Answer:
4i.
Explanation:
To find the flux through the square, we use the divergence theorem for the flux. So Flux of F(x,y) = ∫∫divF(x,y).dA
F(x,y) = hxy,x - yi
div(F(x,y)) = dF(x,y)/dx + dF(x,y)dy = dhxy/dx + d(x - yi)/dy = hy - i
So, ∫∫divF(x,y).dA = ∫∫(hy - i).dA
= ∫∫(hy - i).dxdy
= ∫∫hydxdy - ∫∫idxdy
Since we are integrating along the boundary of the square given by −1 ≤ x ≤ 1, −1 ≤ y ≤ 1, then
∫∫divF(x,y).dA = ∫₋₁¹∫₋₁¹hydxdy - ∫₋₁¹∫₋₁¹idxdy
= h∫₋₁¹{y²/2}¹₋₁dx - i∫₋₁¹[y]₋₁¹dx
= h∫₋₁¹{1²/2 - (-1)/2²}dx - i∫₋₁¹[1 - (-1)]dx
= h∫₋₁¹{1/2 - 1)/2}dx - i∫₋₁¹[1 + 1)]dx
= 0 - i∫₋₁¹2dx
= - 2i[x]₋₁¹
= 2i[1 - (-1)]
= 2i[1 + 1]
= 2i(2)
= 4i