We're going to use the triple integral
to determine the volume of the tetrahedron
with the following four vertices.
Let's first take a look at this in space.
If we were to plot these four points it would form this
tetrahedron bonded by the blue plane
and the three yellow planes,
which are the three coordinate planes.
So, our goal here is to find the volume of this tetrahedron.
Now, if we didn't have graphing software
we'd still want to graph this.
So, using the x, y and z axes
we would plot these our points.
So, we have this point here at the origin.
The x-intercept is three, comma, zero, comma, zero here.
The y-intercept is zero, comma, four, comma, zero here,
and the z-intercept is zero, comma, zero, comma, five,
which would be here.
So, this would be the tetrahedron.
And, again, that we're trying to find the volume of.
Notice how it's bounded by the first octant
with the three coordinate planes, and this plane here,
where we know the three intercepts.
The volume of a solid region, D, is given by the
triple integral where we integrate over the region, D,
which, again, is the closed solid region.
Notice how the integrand function is just one
and differential V can be any of these six
possible orders of integration.
So, let's go ahead and set this up as the volume, V,
equals the triple integral.
And let's go ahead and include one as an integrand function.
And let's use the order of integration of integrating
with respect to z first, then y, and then x.
Any order would work, but let's go ahead and use
this order of integration.
So, now we'll determine the limits of integration for z.
We'll notice how the solid is bounded by
the xy plane, which would be z equals zero,
and it's bounded above by this blue plane.
So, now we need to define the equation of this blue plane,
which we can do because we know the intercepts.
But, again, we know the lower limit of integration
with respect to z is going to be zero.
And the equation of the plane with these three intercepts
is given by this equation.
So, because the x-intercept is three we're going to have
x divided by three, plus the y-intercept is four,
so, we have y divided by four, plus the z-intercept is five,
so, we have z divided by five equals one.
And, now if we solve this equation for z
we can determine the upper limit integration for z.
Now, if we integrate with respect to a different
variable first we'd have to solve this equation for
whichever variable we integrate with respect to first.
But we'll solve this for z.
So, we have z divided by five equals one,
minus x divided by three, minus y divided by four.
Now we'll multiply three by five.
So, we have z equals five, minus 5/3 x,
minus 5/4 y.
Again, this is going to be the
upper limit integration with respect to z.
And notice how we have z expressed in terms of x and y.
So, we have five, minus 5/3 x,
minus 5/4 y.
And, now, we need to find the limits of
integration for y and x.
To do this we'll want to have the xy trace.
In general, to determine the xy trace,
we set z equal to zero.
So, notice that if z is equal to zero for these four points
we'd only have three points to consider:
zero, zero,
three, zero,
and zero, four.
And I've already plotted these three points in the xy plane,
so, we can determine the limits of integration for y and x
using this triangular region here.
I also included the yz and xz traces.
We don't need those for this example
because of the order of integration that we're using,
but if we change order of integration,
depending on the order,
we'd have to use one of these traces instead.
We're going to need the equation of the line
that contains this side of the triangle,
but there's two ways of doing this.
We could set z equal to zero and solve this equation for y,
which would take some work.
It's much easier to recognize that the vertical intercept
is four, and the slope would be down four, right three,
so the slope is negative 4/3.
So, the equation of the line containing this side
of the triangular region would be y equals four,
the vertical intercept, minus 4/3 x.
So, with respect to y we integrate from y equals zero
to y equals four, minus 4/3 x.
So, again, we have from zero to four, minus 4/3 x.
Notice how we express y in terms of x.
And, again, these are limits of integration for y
because this area in the xy plane is bounded below by y
equals zero, and bounded above by y equals four minus 4/3 x.
And finding the limits of integration for x would be
from x equals zero to x equals three.
So, this triple integral will give us
the volume of the tetrahedron.
Let's evaluate this on the next slide.
So, we first integrate one with respect to z,
which would just be z.
So, we got the double integral.
And, again, the antiderivative is z.
Notice when we evaluate big f of b, minus big f of a,
we're just going to have this quantity minus a zero.
Which means for the double integral
the integrand function is just going to be
the quantity five, minus 5/3 x, minus 5/4 y.
Now we integrate with respect to y
treating x as a constant.
So, we'd have five y, minus 5/3 xy,
and then we'll have minus 5/4 times y squared,
divided by two, or 5/4 times 1/2 y squared.
That'd be 5/8 y squared.
Now, this starts to get messy.
We need to find big f of b, minus big f of a.
So, we'd have the integral from zero to three of.
We substitute four, minus 4/3 x, for y,
so we'd have five times the quantity four, minus 4/3 x,
minus 5/3 x, times four,
minus 4/3 x, and then minus 5/8,
times the quantity four, minus 4/3 x squared.
There's big f of b, then for big f of a, of course,
each of these has a factor of y, so we'll just get zero.
Now, let's go ahead and simplify all of this.
Let's go ahead and distribute here and here,
and then we'll go ahead and square the quantity
four, minus 4/3 x.
And, now, we'll distribute here and then combine like terms.
Now, let's combine like terms.
So, we have 20 minus 10, which is 10.
And let's get the x terms.
These two x terms are opposites,
so we just have minus 20/3 x.
Looking at the x squared terms we have 20/9 x squared,
minus 10/9 x squared, which would be positive 10/9,
or plus 10/9 x squared.
And then, finally, we integrate with respect to x.
So, we'll have 10 x,
and then minus 20/3 times x squared,
divided by two, that's going to be minus 20/6 x squared.
And then plus 10/9 times x to the third, divided by three,
which is going to be 10/27 x cubed.
So, when x is three, we have 10 times three, minus,
I know 20/6 simplifies to 10/3,
but I'm going to leave it as 20/6, times three squared,
plus 10/27, times three cubed.
So, all of this minus...
When x is zero notice all the terms would be zero
because each has a factor of x.
And this simplifies perfectly to 10.
Which means the volume of the tetrahedron
would be 10 cubic units.
Which, again, would be the volume
of the tetrahedron shown here, bounded by the blue plane,
and the three coordinate planes.
I hope you found this helpful.