# Stoke's Theorem

1. Aug 26, 2016

### Exus

1. The problem statement, all variables and given/known data
Let S be the portion of the paraboloid $z = 4 - x^2 - y^2$ that lies above the plane $z = 0$ and let $\vec F = < z-y, x+z, -e^{ xyz }cos y >$. Use Stoke's Theorem to find the surface integral $\iint_S (\nabla × \vec F) ⋅ \vec n \,dS$.
2. Relevant equations
$\iint_S (\nabla × \vec F) ⋅ \, d\vec S = \int_C \vec F ⋅\,d\vec r$

3. The attempt at a solution
Parameterize C as $\vec r(t) = 2<\cos t, \sin t, 0> , 0≤t≤2π$

$\int_C \vec F ⋅\,d\vec r = \int_0^{2π} \vec F (\vec r(t))⋅\vec r'(t) \,dt$

$\vec r' (t) = 2<\sin t, -\cos t, 0>$

$\vec F (\vec r (t)) = <-\sin t , \cos t, -\cos (\sin t)>$

$4\int_0^{2π} <-\sin t , \cos t, -\cos (\sin t)>⋅<\sin t, -\cos t, 0> \, dt = 4\int_0^{2π} (-\sin ^2 t -\cos ^2 t) \, dt = -4 \int_0^{2π} \, dt = -8π$

Just looking for someone to verify this was done correctly, it's been almost a year since I last took calculus and I don't remember if there's any way to check my work.

2. Aug 26, 2016

### vela

Staff Emeritus
Close. You made a sign error when calculating $\vec{r}'$, and your expression for $\vec{F}(\vec{r}(t))$ is missing factors of 2 here and there.

3. Aug 26, 2016

### Exus

that's embarrassing, thank you.
$\vec r' (t) = <-2\sin t, 2\cos t, 0>$

$\vec F (\vec r (t)) = <-2\sin t , 2\cos t, -\cos (2\sin t)>$

$\int_0^{2π} <-2\sin t , 2\cos t, -\cos (2\sin t)>⋅<-2\sin t, 2\cos t, 0> \, dt = \int_0^{2π} (4\sin ^2 t +4\cos ^2 t) \, dt = 4 \int_0^{2π} \, dt = 8π$

4. Aug 26, 2016

### vela

Staff Emeritus
Another way you can check your result is to do the surface integral over a simpler surface with the same boundary. In this case, try integrating over the circle of radius 2 with its center at the origin and lies in the xy plane. Since the normal points in the z direction, you only need to calculate the z-component of the curl.

5. Aug 26, 2016

### Exus

$\iint_S (∇×\vec F)⋅\vec n \,dS = \iint_S <P, Q,\frac ∂ {∂x} (x+z) - \frac ∂ {∂y} (z-y) >⋅<0, 0, 1> \, dS = \iint_S 2\,dS = 8π$

That makes a lot of sense, thanks for the tip.