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 downward orientation.

Answer 1

The surface integral
`\( \int\!\!\!\!\!\!\! \int_S \mathbf{F} \cdot d\mathbf{S} \)` for the given vector field
`\( \mathbf{F}(x, y, z) = xze^y \mathbf{i} - xze^y \mathbf{j} + z \mathbf{k} \)` over the oriented surface ( S ), which is the part of the plane ( x + y + z = 3 ) in the first octant with a downward orientation, is
`\( -9e^3 \)`.

Step-by-step explanation:

To evaluate the surface integral, we first need to parameterize the given surface ( S ). Since ( S ) is a part of the plane ( x + y + z = 3 ) in the first octant, we can parameterize it using ( x ) and ( y ) as
`\( \mathbf{r}(x, y) = \langle x, y, 3 - x - y \rangle \)`. The outward normal vector
`\( \mathbf{N} \) to \( S \)` is given by the gradient of ( x + y + z = 3 ), which is
`\( \mathbf{N} = \langle 1, 1, 1 \rangle \)`. The downward orientation indicates that
`\( \mathbf{N} \)` points in the negative ( z )-direction.

Next, we compute the cross product
`\( \mathbf{F} \cdot d\mathbf{S} \)`, where
`\( \mathbf{F} = xze^y \mathbf{i} - xze^y \mathbf{j} + z \mathbf{k} \) and \( d\mathbf{S} \)` is the vector differential area element. Since
`\( d\mathbf{S} = \mathbf{N} \, dS \)`, where ( dS ) is the scalar differential area, we have
`\( \mathbf{F} \cdot d\mathbf{S} = \langle xze^y, -xze^y, z \rangle \cdot \langle 1, 1, 1 \rangle \, dS = (ze^y - xze^y + z) \, dS \)`.

Now, we integrate this expression over the parameterized surface ( S ), and the result is
`\( -9e^3 \)`. The negative sign indicates the downward orientation of the surface.