Evaluate integral using Stokes Theorem

**Alpina540** · July 13th 2010, 11:14 AM

Use Stokes Theorem to evaluate the integral of F*dx , where F(x,y,z)= e^-7x i+e^-4y j+e^5z and C is the boundary of the part of the plane 8x+y+8z=8 in the first octant.

Anyone know how to go about this problem I am beyond lost.

Please help

Thanks!

Possible answers include-
0
23
49
16
69

**HallsofIvy** · July 13th 2010, 12:51 PM

Do you know what "Stoke's Theorem" is? What does it say?

**Alpina540** · July 13th 2010, 12:56 PM

Stokes theorem is defined by the below function-

link- http://mathworld.wolfram.com/StokesTheorem.html

**HallsofIvy** · July 13th 2010, 02:13 PM

Okay, good! Since you are asked to use Stoke's theorem to integrate around a path, $\int_{\partial S} \vec{f}\cdot d\vec{s}$ , this problem must intend you to integrate $\int\int_S \nabla\times \vec{f} dA$ over the area inside the path.

You are told that $\vec{f}= e^{-7x} \vec{i}+e^{-4y} \vec{j}+e^{5z}\vec{k}$ so you need to find the "curl" of that.

Also, you are told that C (the $\partial S$ in the formula) is the boundary of the portion of the plane 8x+y+8z=8 in the first octant. lThat plane cuts off a triangle in the first quadrant. When x and y are both 0, that is 8z= 8 so z= 1 and (0, 0, 1) is one corner of that triangle. When y and z are both 0, 8x= 8 so x= 1 and (1, 0, 0) is another corner. When x and z are both 0, y= 8 so (0, 8, 0) is the third corner.

What I recommend you do is this: from the equation, 8x+ y+ 8z= 8, we can solve for y= 8- 8x- 8z= 8(1- x- z). That means that we can write that plane a as a vector equation: $\vec{r}(x, z)= x\vec{i}+ 8(1- x- z)\vec{j}+ z\vec{k}$ . The two derivative vectors, $\vec{r}_x= \vec{i}- 8\vec{j}$ and $\vec{r}_z= -8\vec{j}+ \vec{k}$ are in the tangent plane to the surface (here, in the plane itself) and their cross product,
$\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 1 & -8 & 0 \\ 0 & -8 & 1\end{array}\right|= -8\vec{i}- \vec{j}- 8\vec{k}$ . Its length, times "dxdz", $\sqrt{129}dxdz$ gives the "differential of area". (All that is very general. For a plane it reduces to the square root of the sum of squares of the coefficients in the equation- you probably have that formula in your textbook.)

The plane cuts the xz-plane where y= 0: 8x+ 8z= 8 which is the same as x+ z= 1 or z= 1- x. x ranges from 0 up to 1 and, for each x, z ranges from from 0 up to 1- x. Your integral should look like
$\int_{x=0}^1\int_{z= 0}^{1- x} \nabla\times \vec{f} \sqrt{129}dz dx$ .

**Alpina540** · July 14th 2010, 05:30 AM

Originally Posted by HallsofIvy

$\int_{x=0}^1\int_{z= 0}^{1- x} \nabla\times \vec{f} \sqrt{129}dz dx$ .

Thanks

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