We want to use a triple integral
to find the volume of the solid
bounded by all of these equations.
We have x equals zero, x equals one,
y equals zero, y equals one, z equals zero and z equals xy.
Let's first take a look at this graphically.
z equals xy is a graph of the blue surface.
The yellow planes are given by the remaining equations.
So notice how the bounded solid region
is going to be this region here that is
below the blue surface and above the xy plane
over this square region in the xy plane.
We can determine the volume of a solid
using a triple integral as defined here.
The volume is equal to the triple integral
over the region D, where the region D is
the solid region we're trying to find the volume of.
And then differential V, using rectangular coordinates,
is equal to dz dy dx,
or any of the six possible orders of integration
shown here at the bottom.
So our volume is equal to the triple integral
again over the region D.
And then notice how the integrand function
would just be one.
And then we'd have differential V.
And for our example, let's use the order of integration
given by dz dy dx as shown here.
So the volume is going to be equal to the triple integral.
The integrand function is one.
Then we have dz dy dx.
Now let's work on determining
the limits of integration for x, y, and z.
Well we know that the solid is bounded below
by the xy plane given by z equals zero.
And it's bounded above by the surface given by z equals xy.
So these will be the limits of integration for z.
Integrate from zero to xy with respect to z.
And now for the limits of integration for y and x,
it's helpful to find the xy trace,
which in this case is pretty straightforward.
The xy trace is the area bounded
by x equals zero, x equals one,
which should be this vertical and this vertical line,
as well as y equals zero and y equals one,
which would be this horizontal line
and this horizontal line here.
And therefore this square is the xy trace.
And this will help us determine
the limits of integration for x and y.
And because we have a square, a rectangular region,
it's very straightforward.
The limits of integration for y will be from zero to one.
The limits of integration for x
will also be from zero to one.
So if we evaluate this triple integral,
it'll give us the volume of the solid
bounded by all of these equations.
So let's evaluate this on the next slide.
So we first integrate with respect to z.
So the antiderivative is just going to be z.
And now I need to find big F of b minus big F of a,
which is just going to be xy minus zero.
So the new integrand function is just xy.
And now we integrate with respect to y,
treating x as a constant.
So the antiderivative is going to be
x times y squared divided by two, or just 1/2 xy squared.
And now we need to find big F of b minus big F of a.
So when y is equal to one, we'd have one squared.
And when y is zero, of course we'd have zero squared.
So the new integrand function is just 1/2 x.
And now we integrate with respect to x.
So we'll have 1/2 times x squared divided by two,
or 1/4 x squared.
So we have 1/4 times the quantity
one squared minus zero squared.
So the exact volume is 1/4 cubic units.
Or of course as a decimal,
we can write this as 0.25 cubic units,
which again going back to the graph,
would be the volume of this bounded solid
below the blue surface
over this square region in the xy plane.
I hope you found this helpful.