In this example, we're asked to set up
a triple integral to determine the volume of the solid
bounded by the paraboloid given
by F of X comma Y,
and the plane Z equals zero.
The plane Z equals zero would be the X, Y plane.
So looking at this graphically first,
here we see the graph of the paraboloid,
and in yellow we have the graph of the plane Z equals zero,
which is the X, Y plane.
So we're going to set up a triple integral
in rectangular coordinates that will give us
the volume of this solid here.
So going back to our work,
the volume is equal to the triple integral
over the region D,
and notice how the integrand function here is just one,
and the region D is a solid region
bounded by the paraboloid in the plane Z equals zero.
And differential V is equal to DZ, DY, DX,
or any of these six possible orders of integration.
But for this example,
we are going to use DZ, DY, DX.
So the volume V is going to be equal
to the triple integral,
and let's go ahead and put the integrand function
here as one,
and then we have DZ, DY, DX.
So now we need to determine the limits of integration
for Z, Y, and X.
Remember the solid region
is bounded below by the plane Z equals zero
and bounded above by Z equals nine
minus X squared minus Y squared,
given by our function,
F of X comma Y.
So we know the limits of integration for Z
will be from zero to nine
minus X squared
minus Y squared.
So now we need to find the limits of integration
with respect to Y and Z,
so we'll need the X, Y trace
to help us determine these limits of integration.
To find the X, Y trace,
we set Z equal to zero
for the equation of the paraboloid.
So the equation of the paraboloid
is Z equals nine minus X squared
minus Y squared,
so again, to find the X, Y trace,
we set Z equal to zero.
Move the variable terms to the left,
so we have X squared plus Y squared
equals nine,
which is this circle here
centered at the origin
with a radius of three.
If we go back to our graph just for a moment,
we can view the X, Y trace
by looking down on the X, Y plane,
which we see here.
So now to find the limits of integration for Y,
we do need to solve this equation for Y
to determine the equation that bounds this area
in the X, Y plane below and above.
So solving this equation here for Y,
we'd have Y squared
equals nine minus X squared,
and now if we take the square root
of both sides of the equation,
we're going to have Y equals
plus or minus the square root
of nine minus X squared.
So looking at the X, Y trace,
this area in the X, Y plane is bounded below
by the equation Y equals negative square root
of the quantity nine minus X squared,
and the area is bounded above
by Y equals positive square root
nine minus X squared,
and these are limits of integration
with respect to Y.
So we have negative square root of nine
minus X squared
to positive square root,
nine minus X squared
for the limits of integration with respect to Y.
And now with respect to X,
the limits of integration will be
from X equals negative three
all the way to X equals positive three.
This is the triple integral
in rectangular coordinates
that would give us the volume
of the solid bounded by the paraboloid
given by F of X comma Y
and Z equals zero,
which would be this solid here,
again, bounded by that paraboloid
and this yellow plane.
So in this video,
we just set up a triple integral
in rectangular coordinates.
We'll actually evaluate this
using cylindrical coordinates in a future video.
I hope you found this helpful.