# Stokes' Theorem, how to apply for this surface?

Gold Member

## Homework Statement

With the stokes' theorem transform the integral $\iint_\sigma \vec{\nabla}\times\vec{F}\cdot\vec{\mathrm{d}S}$ into a line integral and calculate.
$\vec{F}(x,y,z) = y\hat{i} -x^2\hat{j} +5\hat{k}$
$\sigma(u,v) = (u, v, 1-u^2)$
$v\geq0$, $u\geq0$, $u+v\leq1$

## Homework Equations

Stokes' Theorem
$$\oint_\gamma\vec{F}\cdot\vec{\mathrm{d}S} = \iint_\sigma\vec{\nabla}\times\vec{F}\cdot\vec{\mathrm{d}S}$$

## The Attempt at a Solution

The surface in this case is formed by a surface with several cuts: in the plane x+y=1, x=0 and y=0. So I have no idea on how to apply stokes' theorem here. I know this way: given a closed simple curve with not conservative vectorial field in it, the line integral of this vector field along this curve is equal to the integral of surface in which the surface is bounded by this curve. Thinking this way I can only imagine curves being bounds of a bounded surface, but here we got a surface that is part of a square and part of a curve. How to deal with this case?

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LCKurtz
Homework Helper
Gold Member
Your surface is not "part of a square and part of a curve". You are given a parametric equation of the surface $\sigma(u,v) = \langle u,v,1 - u^2\rangle$. That is $x = u,~y=v,~z = 1-x^2$. Do you recognize what that is? Plot a picture of it, then look at the part of that surface that satisfies the inequalities. The boundaries of that surface will give you the curves you need for your line integral. Come back when (if) you need more help.

Gold Member
Oh, I can see. I was looking more than just the surface, it seemed to be something like a peace of cake hehe, but I saw it's just a slice of a parabola. It's solved now, thank you !!

Orodruin
Staff Emeritus
Just to add a piece of notational advice: Do not use $d\vec S$ for the line element if you are also using it to represent the area element. You risk confusing not only others but also yourself.
Just to add a piece of notational advice: Do not use $d\vec S$ for the line element if you are also using it to represent the area element. You risk confusing not only others but also yourself.
Oh I'm sorry I meant $\vec{\mathrm{d}r}$ .