using stokes' theorem to evaluate

**panda** · September 4th 2008, 05:59 AM

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.

please help!

**Laurent** · September 4th 2008, 07:30 AM

If you have computed $\mathop{curl}(\vec{F})$ , then the surface integral on, for instance, the face of the cube with $x=1$ is: $\int_{-1}^1\int_{-1}^1 \mathop{curl}(\vec{F})(1,y,z)\cdot \vec{e}_x dy\,dz$ , where $\vec{e}_x$ is a unit vector in direction $x$ (hence orthogonal to the face), meaning that you integrate the $x$ -component of $\mathop{curl}(\vec{F})$ 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 $\Sigma$ : the cube minus the bottom face) as the integral $\int_{\partial\Sigma}\vec{F}\cdot d\vec{r}$ over the boundary of $\Sigma$ , 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: $\int_{-1}^1 \vec{F}(1,y,-1)\cdot \vec{e}_y dy.$

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

Laurent.

**lllll** · February 24th 2009, 04:26 PM

would you have:

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

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

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

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

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

so for the top I would have:

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

now for the left side:

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

and for the front:

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

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

is this correct?

**Laurent** · March 10th 2009, 10:50 AM

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

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