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kap26 [50]
3 years ago
5

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = xzey i − xzey j + z k S is the part of the plane x + y + z = 3 in the first octant and has upward orientation Incorrect: Your answer is incorrect. 9/2
Mathematics
1 answer:
I am Lyosha [343]3 years ago
4 0

I'm pretty sure your answer of 9/2 is correct...

Parameterize S by

\vec s(u,v)=3v(1-u)\,\vec\imath+3uv\,\vec\jmath+3(1-v)\,\vec k

with 0\le u\le1 and 0\le v\le1.

In case youre' not sure where this came from: recall the standard parameterization for a line segment connecting two points \vec p_1 and \vec p_2,

\vec r(u)=(1-u)\vec p_1+u\vec p_2

with 0\le u\le1. Then treating \vec r(u) as a "point", we can parameterize a plane containing \vec r(u) and another point \vec p_3 by using

\vec s(u,v)=(1-v)\vec p_3+u\vec r(u)

with 0\le v\le 1.

Now, take the normal vector to S to be

\vec r_v\times\vec r_u=9v\,\vec\imath+9v\,\vec\jmath+9v\,\vec k

Then the flux of \vec F across S is

\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^1\int_0^19v(3-3v)\,\mathrm du\,\mathrm dv

(since the first two components of \vec F cancel in the dot product)

=\displaystyle27\int_0^1(v-v^2)\,\mathrm dv=\boxed{\frac92}

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