We want to calculate the flux integral,
which is a double integral over the surface s of F
dot differential S, where the vector field F is given here,
and the surface S is the surface of the box bounded by
these three planes, and the coordinate planes.
So let's first look at this graphically.
The surface S is bounded by the six light blue planes,
and the vector field F is graft in gray.
So the value of the flux integral
will give us the net flow across these six surfaces
of the bounded box.
And we'll evaluate the flux integral
using the divergence theorem.
So before we set this up for review,
the divergence theorem states that total divergence
of a vector field in a solid region V,
equals the total flow across the boundary surface, S.
So the region V must be a solid region bounded by S,
oriented by a unit normal vector directed outward,
and the partial derivatives must be continuous
over the region V.
So under these conditions, the flux integral
is equal to the triple integral over the region V
of the divergence of the vector field F, differential V,
which can also be written in this form here.
So this shows the relationship between a triple integral
over a solid region, and a surface integral over a surface.
This is a 3D version of the flux form of Green's Theorem.
So going back to our example, looking at our formula below,
and the vector field F, P equals e times sin y,
Q equals e to the x cosine y, and R
equals y z squared.
So let's go ahead and find our partial derivatives.
So the partial derivative of P, with respect to x,
is equal to the derivative of e to the x sin y
with respect to x, which should be e to the x sin y,
and the partial derivative of Q with respect to y
is equal to the derivative of e to the x cosine y
with respect to y, which would be e to the x
times negative sin y, or negative e to the x sin y,
and the partial of R, with respect to z,
is equal to the derivative of y z squared with respect to z,
which would be two y z.
So the flux integral, or the given surface integral
is equal to the triple integral
over the solid region V of the sum
of these partial derivatives, which would be
e to the x sin y plus negative e to the x sin y,
or minus e to the x sin y,
plus two y z, and then for differential v,
let's go ahead and use the order of integration
given by d x, d y, d z.
The limits of integration for x would be from zero to four,
the limits of integration for y will be from zero to five,
and limits of integration for z will be from zero to one.
Let's go ahead and evaluate this on the next slide.
Notice how difference will be zero,
so this simplifies to the triple integral,
of just two y z, again, we have d x, d y, d z,
so you first integrate with respect to x,
treating y and z as constants.
So the antiderivative would be just be two y z times x,
or two x y z.
Performing substitution for x,
we would just have two times four,
times y z, and then when x is zero, it's just zero.
So you got the double integral,
of eight y z,
d y d z.
And now we'll integrate with respect to y,
treating z as a constant.
So we'd have eight z times y squared divided by two,
which would be four y squared z.
Now, substituting for y, when y is five,
we have four times five squared, times z,
minus when y is zero, we have zero.
So we have the integral from zero to one,
this is going to be 25 times four, that's 100 z,
and now we integrate this back to z,
so we have an antiderivative of 100 times z squared
divided by two, which would be 50 z squared,
evaluating here, we'd just have 50
times the quantity one squared minus zero squared,
which is just equal to 50 times 150.
So the value of the given surface integral,
or the flux integral, is equal to 50.
And this value represents the total flow
across the boundary surface S of the region,
as well as the total divergence of the vector field
in the solid region V.
I hope you found this helpful.