Find the mass of the solid bounded below
by the circular paraboloid z equals x squared
plus y squared and above by the circular paraboloid
z equals four minus x squared minus y squared
if the density is given by the density function
rho of x comma y comma z.
We can determine the mass of a solid
in a three-dimensional region r
by evaluating this triple integral.
Before we do this, let's first look at this
in three dimensions.
Our goal is to find the mass of the solid
bounded between these two paraboloids,
which is this solid here.
If you look down on the xy-plane,
we can see the intersection of these two paraboloids
is a circle, and because of this,
we will use cylindrical coordinates for the triple integral
to determine the mass.
Remember, when using cylindrical coordinates,
dV is equal to r dz dr d theta,
and therefore this is the triple integral
that we need to set up in order to determine the mass.
To help us determine the limits of integration, though,
let's graph the intersection of these two paraboloids
on the xy-plane, which is already shown here,
but let's determine the equation of the circle.
To do this, we solve the two equations of the paraboloids
as a system of equations for x and y.
So let's substitute x squared plus y squared for z
in this equation here,
which gives us x squared plus y squared
equals four minus x squared minus y squared.
From here, let's add x squared and add y squared
to both sides of the equation, which gives us two x squared
plus two y squared equals four,
and now we'll divide both sides of the equation
by two, which gives us x squared plus y squared equals two.
We should recognize this equation as a circle
centered at the origin where r squared is equal to two,
and therefore r, the radius,
is equal to the square root of two.
'Kay, and so record this on the circle.
Because we are using cylindrical coordinates,
we need to write the density function
as a function of z r theta.
Rho of x comma y comma z is equal to the square root
of the quantity x squared plus y squared,
and remember, x squared plus y squared is equal
to r squared, which is equal to the square root
of r squared, which is equal to r,
so the density function here is equal to r.
We have another factor of r here, from differential V,
and therefore the integrand function
is going to be r squared.
Now let's work on determining the limits of integration.
We first need to determine the limits of integration for z,
and they must be functions of r and or theta.
Looking at the three-dimensional graph,
we can see the solid is bounded below by the paraboloid
that opens up, and above by the paraboloid
that opens down.
The paraboloid that opens up
is z equals x squared plus y squared,
which must be expressed using cylindrical coordinates,
and we know x squared plus y squared is equal to r squared,
so z equals r squared is the lower limit
of integration for z.
The paraboloid that opens down
is z equals four minus x squared minus y squared.
Let's write that as minus the quantity x squared
plus y squared, which is equal to four minus r squared,
which gives us the upper limit of integration.
And now we'll use the intersection of these two paraboloids
projected onto the xy-plane graphed here
to determine the limits of integration for r and theta.
Because the radius of the circle is square root two,
the limits of integration for r are from
zero to square root two, and we need to go all the way
around the circle, and therefore the limits of integration
for theta are from zero to two pi.
Let's evaluate this on the next slide.
We first integrate with respect to z,
treating r squared as a constant.
The antiderivative is r squared z.
And now we find big F of b minus big F of a,
and we substitute the limits of integration for z, not r.
This gives us r squared times the quantity four
minus r squared minus r squared.
Simplifying here, we have four minus r squared
minus r squared, which is four minus two r squared,
and then we distribute r squared,
which gives us four r squared minus two r to the fourth.
And now we integrate with respect to r,
which would be four times r to the third
divided by three, or four thirds r cubed
minus two times r to the fifth divided by five,
or minus two fifths r to the fifth.
And now we find big F of b minus big F of a.
The cube of square root two is going to be
two square root two, so here we have four thirds
times two square root two minus two fifths
times the square root of two raised to the fifth power
is four square root two, then we have minus,
when r is zero, both terms are zero.
Let's continue on the next slide.
Simplifying here, here we have eight thirds square root two
minus eight fifths square root two.
Common denominator is 15.
Looking at the numerators, 40 minus 24 is equal to 16.
The integrand function simplifies
to 16 fifteenths square root two.
And now we integrate with respect to theta,
which gives us 16 fifteenths square root two theta,
which gives us 16 fifteenths square root two
times two pi minus zero.
Nothing simplifies.
We have 32 fifteenths square root two pi
as the mass of the solid,
so going back to our graph, we have now found the mass
of the solid in this bounded region here.
I hope you found this helpful.