Final Answer:
The surface integral
for the given vector field
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
.
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
. The outward normal vector
is given by the gradient of ( x + y + z = 3 ), which is
. The downward orientation indicates that
points in the negative ( z )-direction.
Next, we compute the cross product
, where
is the vector differential area element. Since
, where ( dS ) is the scalar differential area, we have
.
Now, we integrate this expression over the parameterized surface ( S ), and the result is
. The negative sign indicates the downward orientation of the surface.