Welcome to the first video
on Triple Integrals and Volume.
We can determine the volume of a solid region V
by using a triple integral over the region V
integrated with respects to V.
Which means if we let the integrand equal
the most basic function f[x,y,z] equals one
and rewrite differential V as dxdydz
or some other order of these three differentials
and then let the limits of integration represent the solid V
this will give us the volume of the solid.
And this is very similar to how we use a double integral
to determine area.
We can now use a triple integral to determine volume.
And then remember because there are three differentials,
there are six possible orders of integration.
In the intro video for triple integrals
we looked at this definition to evaluate triple integrals.
We're going to use this same principle to determine volume
but f[x,y,z] will equal one.
Let's take a look at an example.
Here we're going to determine the volume of the solid
bounded by y equals -2x plus four, z equals three
with x, y, and z greater than or equal to zero.
So the most challenging part
about determining this triple integral
is going to be determining the limits of integration
with respect to x, y, and z.
And to do that it's often helpful to first
take a look at the xy trace, xz trace, and yz trace.
And this example is pretty straightforward.
The xy trace would be when z is equal to zero
so we'd have the equation y equals -2x plus four
so we'd have a y-intercept of four and a slope of -2
so we go down two, right one.
And we also know that x is greater than or equal to zero
and y is also greater than or equal to zero.
So the xy trace would be this two-dimensional region here.
Now let's take a look at the xz trace.
If we let this equal the z axis and equal the x axis
we'll go through each equation and let y equal zero.
This first equation here for y is equal to zero
we can solve this for x, x would be equal to two,
that would be this vertical line here,
z equals three would be here,
and again because y and z are positive
this would be the region in the xz plane.
And then for the yz trace
if we let this equal the z axis and this equal the y axis
if x is zero we would have y equals four,
that would be this line here.
Again this is the y axis, z is equal to three,
and again both y and z are positive,
this would be the area of the yz trace.
And these will help us determine the limits of integration.
But next we need to decide
in which order we want to integrate.
Remember if we integrate first with the respect to y,
the limits of integration must be in terms of x and z
and since this equation here is already solved for y,
we'll start with differential y
and the order of the remaining two don't really matter,
let's go ahead and write this as dxdz.
So the interval for y will be from zero
all the way to this line, which is -2x plus four.
And then for x we'll go from zero all the way to two.
And then lastly for z
we can look at either of these two traces,
z will be from zero all the way to three.
These intervals for y, x, and z form the region V
and therefore will give us the volume of the solid.
If we take a look at V in three dimensions
the volume that we're determining
will be the volume of this triangular prism here.
So let's go ahead and determine this volume.
We'll first integrate with respects to y.
Now we'll evaluate this at the limits of integration.
So we'll have -2x plus four minus zero
and now we'll go ahead and integrate with respects to x,
we'll go ahead and do that on the next slide.
So we'll have -2 times x squared over two,
that'll be -x squared plus 4x
and then when x is two we're going to have -4 plus 8,
that'll be four and then when x is zero
they'll both be zero.
And now we'll integrate with respects to z,
we'll have 4z, so we'll have four times three, zero,
that'll give us 12 cubic units.
Let's go back and take a look at that region in 3D.
This distance here was two, this distance here was four
and the height was three.
So if we had a rectangular prism
it would have length times width times height
or two times four times three, that would give us 24,
and we can see geometrically that the volume that we wanted
was half of that, 24 divided by two would be 12 cubic units.
And most of the problems we look at,
we won't be able to use one of our basic geometric formulas
but we could have on this one
and this does verify your answer.
Let's go ahead and take a look at another one
that's a little bit more involved.
We've already determined the volume
bounded by these three equations.
Let's go ahead and take a look at our traces.
Let's start with the xy trace, so we'll set z equal to zero.
Notice if z is equal to zero here we'd have y equals zero,
that would be this line here.
Here we have the graph of y equals one minus x squared,
that would be a parabola that opens down shifted up one unit
so it looks something like this.
So the bounded region in the xy plane is this region here.
So this was y equals zero
and this was y equals 1 minus x squared.
Let's look at the xz trace now.
So we'll call this z and we'll call this x.
So y would be equal to zero.
If y is zero we have z equals zero, this line here.
If y is zero here we'd have,
if y was equal to zero here we could add x squared
to both sides, we'd have x squared equals one
so x is equal to plus or minus one.
So we'd have this vertical line here
and this vertical line here.
And then lastly for the yz trace
we set x equal to zero, so we have the line z equals y,
that would be this line here, slope of one.
If x equals zero here we'd have y equals one,
and this is the y axis here so it'd look like this
and again z is equal to zero so we'd have this line here.
So we have this triangular region in the yz plane.
So our volume would be equal to the triple integral
over the region v with respect to v.
So now we need to determine the order of integration
as well as the limits of integration.
This time I think we'll start with integration
with respects to v, y, and then x.
So if the limits of integration with effect to z
it must be expressed in terms of y or x.
So let's look at the yz trace.
Z is bounded by zero in this line z equals y.
So it will be from zero to y
and then for the limits of integration with respects to y
let's go ahead and take a look at this first trace.
Y is bounded by zero and this curve
which is one minus x squared.
So it will be from zero to one minus x squared
and then lastly for x, x will be from -1 to +1
as we see here as well as what we see here.
And this triple integral will give us the volume
of the solid.
Let's go ahead and take a look at that solid.
So the volume that we're determining is this region in here
and as we angle this you can see each of the traces.
Let's go ahead and find that volume.
Let's go ahead and start this on the next slide.
So we're going to go with respects to z first.
If we place z with y, that'll give us y and then with zero
so we'll just have y
and now we'll integrate with respects to y,
so we'll have y squared over two or 1/2y squared
and replace y with 1 minus x squared
we have to square this or foil it.
So we're going to have 1/2 and then we'll have times one
minus two x squared plus x to the fourth.
And now we'll integrate with respects to x
we'll have x minus two times x to the third over three
plus x to the fifth over five.
So we'll have 1/2 times x minus 2/3x cubed
plus 1/5x to the fifth.
Let's go ahead and finish this on the next slide.
X is equal to one, we'll have one minus 2/3,
that's 1/3 plus 1/5.
1/3 plus 1/5 will give us 18/15
and then when x is -1 we're going to have -1 plus 2/3,
it'll be -1/3 plus a -1/5, that'll be -8/15.
So we have 1/2 times this will be 16/15
so we'll have 16/30 which simplifies to 8/15.
And again this gives us the volume of region V.
So 8/15 is the volume of this solid region here
bounded by this green plane and this blue plane
and this gray surface.
We'll take a look at some additional examples in part two.