Last Question of the Semester (I hope)

[hank]

And it has to do with Stokes theorem.

I need to use Stokes theorem to evaluate the following:

F(x,y,z) = (x-y)i +(y-z)j + (z-x)k; over the portion of the plane x + y + z = 1 in the first octant.

Ok.

curl F: i + j + k

n = i + j + k

So,

SS (i + j + k) * (i + j + k) dS (* = dot product)

= SS 3 dydx with the limits of integration 0 <= x <= 1, 0 <= y <= 1.

= 3.

My problem is the book says the answer is 3/2. Indeed, if I evaluate the line integral, I get 3/2.

Can someone show me what I'm missing?

 

[galactus]

If \(\displaystyle \sigma\)is oriented with upward normals then C has three parts parametrized as:

\(\displaystyle C_{1}:r(t)=(1-t)i+tj, \;\ 0\leq{t}\leq{1}\)

\(\displaystyle C_{2}:r(t)=(1-t)j+tk, \;\ 0\leq{t}\leq{1}\)

\(\displaystyle C_{3}:r(t)=ti+(1-t)k, \;\ 0\leq{t}\leq{1}\)

\(\displaystyle \int_{C_{1}}Fdr=\int_{C_{2}}Fdr=\int_{C_{3}}Fdr=\int_{0}^{1}(3t-1)dt=\frac{1}{2}\)

So, \(\displaystyle \oint_{C}Fdr=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\)

\(\displaystyle \text{Curl}F=i+j+k, \;\ z=1-x-y\)

R is the triangular region in the xy plane enclosed by x+y=1, x=0, y=0

\(\displaystyle \int\int_{\sigma}\cdot(curlF)ndS=3\int\int_{R}dA=3(\text{area of R})=3\left(\frac{1}{2}(1)(1)\right)=\frac{3}{2}\)