129k views
0 votes
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.

User Bazo
by
7.8k points

1 Answer

2 votes

Final Answer:

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.

User Llogan
by
7.3k points