We want to use a triple integral
to determine the volume of the solid
bounded by z equals zero, z equals x,
and x equals four minus y squared.
Let's first look at this graphically.
So the graph of z equals zero is the yellow plane,
the graph of z equals x is the blue plane,
and the graph of x equals four minus y squared
is this green cylinder.
Notice how the bounded solid would be this solid here.
Notice how it's bounded below by z equals zero,
and above by the blue plane, z equals x.
We can also determine the xy trace
by looking down on the xy plane.
Notice how the xy trace would be
the area bounded by the parabola here and this line here.
Again, our goal is to find the volume of the solid
of this region here.
So going back to our work, we can determine
the volume of a solid by evaluating this triple integral
where the volume is equal to the triple integral
over the region d, where the region d
is the solid we're trying to find the volume of.
And the differential v is equal to dzdydx
or any of the six possible orders of integration.
So again, we're going to find the volume
by setting up the triple integral, with the region d.
Notice how the integrated function would be one.
Let's include one, and then we have differential v.
So in our case, we're going to have the volume.
And now we need to decide on the order of integration.
Let's first integrate with respect to z.
Then with respect to x, then with respect to y.
And I'll explain in a moment why we're integrating
with respect to x first and then y.
Let's first determine the limited integration for z.
And looking at our equations here,
the solid is bounded below by z equals zero,
and bound above by z equals x.
And again, looking at this graphically,
from this view we can see the solid
is bounded below by the yellow plane
and above by the blue plane.
Where we have z equals zero here, and z equals x here.
So these will be the limited integration for z.
We'll have from zero to x.
And then to help us find the limited integration
for x and y, we do want to determine the xy trace
which is shown here, but let's explain where this came from.
To find the xy trace, we set z equal to zero.
So looking at our equations again,
notice how if z is zero, we'd have the equation
x equals zero here, and here we'd have
the equation x equals four minus y squared.
So here is x equals zero, and here is the graph
of x equals four minus y squared.
So this area here is the xy trace.
And this is the reason why we're integrating
with respect to x first and then y.
If we integrate with respect to x first,
we need to determine how this area
is bounded to the left and to the right,
which would be easier than determining
how this area is bounded below and above.
From left to right, this area is bounded by x equals zero,
and x equals four minus y squared.
Which gives us the limits of integration for x.
Again, we integrate from zero to four minus y squared.
If we did integrate with respect to y first,
we'd have to take this equation here and solve it for y,
which would take more work.
It could still be done, it would just take some extra work.
And now with respect to y,
we integrate from y equals negative two
all the way up to y equals two.
So this triple integral will give us the volume
of the solid bounded by these three equations.
Let's evaluate this on the next slide.
So we first integrate with respect to z,
which will give us a double integral,
the integral derivative is just going to be z,
and the limited integration
with respect to z from zero to x.
So big f of b minus big f of a
is just going to be x minus zero,
where the integrated function is now just x,
and we have dxdy.
So now when you integrate with respect to x,
which will just give us x squared
divided by two or 1/2, x squared.
And now we need to determine big f of b
minus big f of a, so we're going to have
1/2 times big f of b would be
the quantity of four minus y squared, squared,
and then minus big f of a is just going to be zero squared.
So expanding here, we're going to have integral
from negative two to two of 1/2
times, you're going to have 16 minus eight y squared
plus y to the fourth.
Let's go ahead and distribute the 1/2.
So the integrated function is eight minus four y squared
plus 1/2 y to the fourth.
Now we integrate with respect to y, so we'll have eight y,
then we're going to have minus four times y to the third
divided by three, or minus 4/3 y to the third.
And then we have plus 1/2 times y to the fifth
divided by five, which would be 1/10 y to the fifth.
And now we need to evaluate this.
So when y is two, we have eight times two
minus 4/3 times two cubed
plus 1/10 times two to the fifth.
And then when y is negative two,
we have eight times negative two minus 4/3
times negative two to the third plus 1/10
times negative two to the fifth.
Simplifying this comes out to 128/15
minus negative 128/15,
which equals 256/15,
which again is the volume, so now we know
the volume of the solid is equal to
256/15 cubic units,
Which as a decimal would be approximately 17.0667,
again cubic units.
Which again would be the volume
of this bounded solid region here.
I hope you found this helpful.