Video Transcript
Use the divergence theorem to find
the outward flux of 𝐅 equal to 𝑦𝐢 plus 𝑥𝐣 minus 𝑧𝐤 across the boundary of the
region 𝐷: the solid cylinder 𝑥 squared plus 𝑦 squared less than or equal to four
between the plane 𝑧 equals zero and the paraboloid 𝑧 equals 𝑥 squared plus 𝑦
squared.
This question is a fantastic
demonstration of how dramatically the divergence theorem can simplify otherwise
extremely complicated integrals.
Let’s make a diagram of the
scenario. Let’s start with the cylinder, 𝑥
squared plus 𝑦 squared less than or equal to four. This is just a circle of radius two
stretched into three dimensions along the 𝑧-axis. We need only draw the cylinder down
to the 𝑥𝑦-plane, or 𝑧 equals zero, since the region concerned is also bound by
the plane 𝑧 equals zero.
Now, for the paraboloid 𝑧 equals
𝑥 squared plus 𝑦 squared, one way of thinking of this is that the height 𝑧
increases with the square of the distance from the 𝑧-axis, 𝑥 squared plus 𝑦
squared. The paraboloid therefore looks
something like this, a standard 2D parabola rotated 360 degrees around the
𝑧-axis. At the point where the two shapes
intersect, the distance from the 𝑧-axis or the square root of 𝑥 squared plus 𝑦
squared is the same. Therefore, the value of 𝑥 squared
plus 𝑦 squared is the same. Therefore, the intersection is at a
height of 𝑧 equals four. The region 𝐷 is therefore this
region underneath the paraboloid but still inside the cylinder and above the plane
𝑧 equals zero.
The outward flux of a vector field
𝐅 coming from this region 𝐷 is given by the double integral over the surface 𝑆,
which bounds the region 𝐷, of 𝐅 dot d𝐒, where d𝐒 is the unit normal times the
area element of the surface.
Evaluating this integral directly
would be extremely difficult. First of all, we have three
separate surfaces. We have the exterior surface of the
paraboloid, the curved surface of the cylinder, and the circular bottom of the
cylinder. Not only that, but evaluating the
dot product of 𝐅 and the unit normal times the area element on each of these
surfaces will be extremely hard. But instead we can use the
divergence theorem, which states that the surface integral over a closed surface 𝑆
of the scalar product of the vector field 𝐅 and the unit normal to the surface d𝐒
is given by the volume integral over the volume 𝑉 bound by 𝑆 of the divergence of
𝐅 or ∇ dot 𝐅.
The divergence theorem is extremely
useful for scenarios in which the divergence of 𝐅 is a simpler function than the
outward flux of 𝐅 or if the volume 𝑉 is more straightforward to integrate over
than the surface 𝑆. In this case, both are true. Whenever we have a linear function
for a vector field, the divergence will always be a constant, since we’re taking the
derivative of a linear function. Therefore, when integrating the
divergence of a linear function over a volume 𝑉, the result is just the constant
value of the divergence times the magnitude of the volume 𝑉. Let’s clear a little space before
we continue.
So taking the divergence of 𝐅, we
have 𝜕 by 𝜕𝑥 𝐢 plus 𝜕 by 𝜕𝑦 𝐣 plus 𝜕 by 𝜕𝑧 𝐤 dot 𝑦𝐢 plus 𝑥𝐣 minus
𝑧𝐤. This gives us 𝜕 by 𝜕𝑥 of 𝑦 plus
𝜕 by 𝜕𝑦 of 𝑥 plus 𝜕 by 𝜕𝑧 of negative 𝑧. This evaluates to zero plus zero
minus one, so this comes to negative one. So the triple integral over 𝐷 of
the divergence of 𝐅 comes to simply the triple integral over 𝐷 of negative one
d𝑉. And we can take this negative one
outside the integration to give us negative the integral over 𝐷 of d𝑉.
There are a few ways we can go
about this triple integration. Probably, the most natural way is
to use cylindrical coordinates. So for cylindrical coordinates, we
have 𝜌, the distance from the 𝑧-axis; 𝜙, the argument with the 𝑧-axis; and 𝑧,
the height along the 𝑧-axis. In Cartesian coordinates, 𝜌 is
given by the square root of 𝑥 squared plus 𝑦 squared. This will make defining the
equations of the surfaces much easier.
We now need to find the limits for
our triple integration. And again there are several ways of
going about this. One way is to consider a vertical
column of the region 𝐷 and integrate this over the base area, which is the circle
intersection of the cylinder and the plane 𝑧 equals zero. This way, the limits of the radius
𝜌 and the argument 𝜙 will remain constant and only the limits of the height 𝑧
will change. Since we’re integrating over a
cross section of a circle of radius two centered on the origin, the lower and upper
limits of 𝜌 will just be zero and two, respectively.
Likewise, the lower and upper
limits of 𝜙 will just be the minimum and maximum arguments zero and two 𝜋. The lower limit of 𝑧, 𝑧 one, will
just be zero, since this is where the cylinder intersects the plane 𝑧 equals
zero. The upper limit of 𝑧, 𝑧 two, will
just be the height of this column, which is where it intersects with the
paraboloid. 𝑧 two is therefore given by the
value of 𝑧 defined by the equation of the paraboloid, 𝑥 squared plus 𝑦
squared. But in cylindrical coordinates, 𝑥
squared plus 𝑦 squared is just equal to 𝜌 squared.
Finally, our volume element d𝑉 in
cylindrical coordinates is given by 𝜌 d𝜌 d𝜙 d𝑧. So our triple integral is the
negative the integral between 𝜌 equals zero and two, between 𝜙 equals zero and two
𝜋, and between 𝑧 equals zero and 𝜌 squared of 𝜌 d𝜌 d𝜙 d𝑧. We can go about this integration in
any order we please. But for now, let’s keep it the same
and we will place all of the elements inside their respective integrals.
So integrating first with respect
to 𝑧, we are integrating one with respect to 𝑧. So this comes to 𝑧 evaluated
between zero and 𝜌 squared. So this becomes 𝜌 squared minus
zero, which is just 𝜌 squared. Since the outer integral is with
respect to 𝜙 and 𝜌 does not vary with 𝜙, we can take this 𝜌 squared outside the
integration. This gives negative the integral
between zero and two of 𝜌 cubed d𝜌, between zero and two 𝜋 of d𝜙. This outer integral with respect to
𝜙 evaluates to simply two 𝜋. Taking the constant of two 𝜋
outside the integration and integrating with respect to 𝜌, we get negative two 𝜋
times one-quarter times 𝜌 to the fourth evaluated between zero and two. This evaluates to negative two 𝜋
times four minus zero, which comes to negative eight 𝜋. So this outward flux is in fact an
inward flux of magnitude eight 𝜋.
