Question

Verify that Stokes' Theorem is true for the vector field

Help Entering Answers (1 point) Verify that Stokes' Theorem is true for the vector field F -yi+ zj + xkand the surface S the hemisphere x2 + y2 + z2-25, y > 0oriented in the direction of the positive y- axis To verify Stokes' Theorem we will compute the expression on each side. First compute curl F dS curl F The surface S can be parametrized by S(s, t) - (5 cos(t) sin(s), S2 curl F . ds- iidt ds where curl F dS Now compute |F-dr The boundary curve C of the surface S can be parametrized by: r(t)-5 cos(t), (Use the most natural parametrization) 2n F dr - ii dt F. dr-

Answer 1

Given that

we have to verify that stokes'Theorem is true for the vector field

F = yi + zj + xk

Your first part seems fine and looks like you are having trouble with the second part where you are supposed to calculate the line integral. For this, we need to find the parametrization of the boundary curve correctly. You have to keep the directions in check.

Since the hemisphere is oriented in the +y direction, the boundary curve 'C' of this hemisphere (which is a circle) will be in the xz-plane. Now if you curl fingers of your right hand around this circle, the thumb should point towards +y axis. This gives clockwise rotation of C. Recall that we measure the angle positive for anticlockwise rotation. Therefore for this case, we need to use -t instead of t. Thus parametric form of 'C' will be:

$\mathbf{r}(t) = <5cos(-t), 0, 5sin(t)> = <5cos(t), 0, -5sin(t)>$

0<t<2π

such that

r'(t) =<ー5sin(t), 0,-5cos(t) >

The line integral can be calculated using:

F(r(t)).r(t)dt F.dr

Here,

$\mathbf{F}=<y, z, x> \Rightarrow \mathbf{F}(\mathbf{r}(t)) = <0, -5sin(t), 5cos(t)>$

and,

r'(t) =<ー5sin(t), 0,-5cos(t) >

Therefore

F(r(t)).r'(t)dt < 0,-5sin(t), 5cos(t) > . <-5sin(t),0,-5cos(t) >

$\Rightarrow \mathbf{F}(\mathbf{r}(t)).\mathbf{r}'(t)dt = -25cos^2(t)$

Thus,

2π 25cos tdt F(r(t))r(t)dt F.dr

25(, sin(2t)〉 F.dr --25

sin(4π)-5272 ( 0 sin(o) ー_25π