# Stoke's Theorem

Exus

## Homework Statement

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##.

## Homework Equations

##\iint_S (\nabla × \vec F) ⋅ \, d\vec S = \int_C \vec F ⋅\,d\vec r ##

## 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.

## Answers and Replies

Staff Emeritus
Homework Helper
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.

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 ##

Staff Emeritus