Math need help. How to solve this

Question

If F (x,y,z)=tan z i +(x + z) j+ xsiny k and σ is the portion of the cylinder z=9-x²-y² above the xy-plane oriented so that the unit normal vector n points to the outside of the cylinder, use Stokes' theorem to evaluate

 ∫∫_σ(∇ x F) • n dS.

Answer 1

Recall the statement of Stokes' Theorem:

∫∫_S curl F • dS = ∫_C F • dr

We are asked to find the left-hand side, where S is the surface of the paraboloid z = 9 - x² - y² above the xy-plane. It's quite difficult to integrate over this region, so we can use Stokes' Theorem and calculate the right-hand side instead, which is much easier to do.

Let C be the curve that bounds our paraboloid. Because our surface is bounded from below by the plane z = 0 (the xy-plane), C is the circle x² + y² = 9 with z = 0. Next, we parameterize this curve. Recall that our curve must have positive orientation, meaning it must travel in an overall counterclockwise direction.

Then r(t) = 3cos(t) i + 3 sin(t) j + 0 k, where t ranges from 0 to 2π.

Recall F • dr = F(r(t)) • dr/dt * dt = F(r(t)) • r'(t) dt. Let's find these components.

F(r(t)) = tan(0) i + (3 cos(t) + 0) j + 3 cos(t) sin(3 sin(t)) k = 0 i + 3 cos(t) j + 3 cos(t) sin(3 sin(t)) k.

r'(t) = -3 sin(t) i + 3 cos(t) j + 0 k.

Then F(r(t)) • r'(t) = 9 cos²(t).

Finally, we integrate to find

∫_C 9 cos²(t) dt = 9π.