# Thread: using stokes' theorem to evaluate

1. ## using stokes' theorem to evaluate

I have to evaluate ∬ curlF.dS where F (x,y,z)= (xyz, xy, yzx^2)
s
S consists of the top and the four sides( but not the bottom) of the cube with vertices(±1, ±1, ±1) oriented onward.

I calculated curl F but I wasnt sure how to go from there.

2. If you have computed $\displaystyle \mathop{curl}(\vec{F})$, then the surface integral on, for instance, the face of the cube with $\displaystyle x=1$ is: $\displaystyle \int_{-1}^1\int_{-1}^1 \mathop{curl}(\vec{F})(1,y,z)\cdot \vec{e}_x dy\,dz$, where $\displaystyle \vec{e}_x$ is a unit vector in direction $\displaystyle x$ (hence orthogonal to the face), meaning that you integrate the $\displaystyle x$-component of $\displaystyle \mathop{curl}(\vec{F})$ over the face. And you have to do this for each of the 5 faces. Good luck!

Or you remember the title "using Stokes' thm to evaluate..." and write your surface integral (over $\displaystyle \Sigma$: the cube minus the bottom face) as the integral $\displaystyle \int_{\partial\Sigma}\vec{F}\cdot d\vec{r}$ over the boundary of $\displaystyle \Sigma$, which is a square. This reduces the problem to the computation of four line integrals (and you don't need the curl). For instance, one of them is: $\displaystyle \int_{-1}^1 \vec{F}(1,y,-1)\cdot \vec{e}_y dy.$

You also have to care about the orientation of the boundary: it is defined by the orientation of the surface.

Laurent.

3. would you have:

$\displaystyle curl F = (x^2z) \ \vec{i} -(2xyz-xy) \ \vec{j} + (y-xz) \ \vec{k}$

where $\displaystyle P=(x^2z), \ Q=-(2xyz-xy), \ R=(y-xz)$

so since we want all sides but the bottom then would it be:

top: $\displaystyle z=g(x,y)=1$
left side: $\displaystyle y=g(x,z)=-1$, right side: $\displaystyle y=g(x,z)=1$
front: $\displaystyle x=g(y,z)=1$, back: $\displaystyle x=g(y,z)=-1$

now since $\displaystyle \iint_S F \cdot \ dS = \iint_D \left(-P \frac{\partial g}{\partial x} -Q \frac{\partial g}{\partial y} +R \right) \ dA$

so for the top I would have:

$\displaystyle \int_{-1}^1 \int_{-1}^1 (y-xz) \ dx \ dy = \int_{-1}^1 \int_{-1}^1 (y-x) \ dx \ dy$

now for the left side:

$\displaystyle \int_{-1}^1 \int_{-1}^1 -(-(2xyz-xy)) \ dx \ dz = \int_{-1}^1 \int_{-1}^1 (2x(-1)z+x)) \ dx \ dz$

and for the front:

$\displaystyle \int_{-1}^1 \int_{-1}^1 -(x^2z) \ dz \ dy = \int_{-1}^1 \int_{-1}^1 -(z) \ dz \ dy$

where the right and back would simply be replaced by 1 and -1.

is this correct?

4. Originally Posted by lllll
is this correct?
Sorry for the delay... This seems correct.

Besides, I've just noticed this can be shortened: if you integrate $\displaystyle \mathop{curl}(\vec{F})$ over the surface of the whole cube (bottom included), this gives 0. This is a consequence of the divergence theorem and $\displaystyle \mathop{div}(\mathop{curl})=0$ (or a consequence of Stokes' theorem, with an empty boundary). Therefore, it suffices to find the integral of $\displaystyle \mathop{curl}(\vec{F})$ of the bottom, and take its opposite. This finally gives $\displaystyle -\int_{-1}^1\int_{-1}^1 -(y-x(-1)) dx dy = 0$, disappointedly...