Question

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

Answer 1

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)

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