Compute the flux of the vector field f with rightarrow xzjw

Compute the flux of the vector field f with rightarrow = xzjwith rightarrow - xyk with rightarrow through the open-top, outward-oriented cylinder S. x^2 + y^2 = 5, 0 lessthanorequalto z lessthanorequalto 3.

Solution

Solution: Given F = xzj - xyk, x² + y² = 5, 0\\leq z \\leq 5

Here outward normal n = k, so F.n = -xy

So flux = \\int\\int_{S}F.nds

= \\int_{-\\sqrt{5}}^{\\sqrt{5}} \\int_{-\\sqrt{5-x²}}^{\\sqrt{5-x²}(-xy)dydx

=2x2 \\int_{0}^{\\sqrt{5}}\\int_{0}^{\\sqrt{5-x²}(-xy)dydx

=4 \\int_{0}^{\\sqrt{5}}[(-xy²/2)]₀^\\sqrt{5-x2}dx

=-2 \\int_{0}^{\\sqrt{5}}x(5-x²)dx

= 2 \\int_{0}^{\\sqrt{5}}(x³ -5x)dx

=2(x⁴ /4 - 5x²/2)_{0}^{\\sqrt{5}}

= 2(25/4 - 25/2)

= - 25/2