We want to evaluate the triple integral
over the region e,
where the region e is a solid region
enclosed by the paraboloid z equals three, plus x squared,
plus y squared, the cylinder x squared
plus y squared equals two, and the xy plane.
Let's begin by looking at the region e in space.
So the region e is a solid region here,
bounded by the paraboloid graphed here in blue,
the cylinder graphed here in purple,
and the xy plane graphed here in yellow.
Because this region is bounded by the cylinder,
it's going to be easier to use
cylindrical coordinates,
rather than rectangular coordinates
to evaluate the triple integral.
If we look down on the xy plane,
we can see the xy trace.
Notice how the xy trace is this circle,
x squared, plus y squared, equals two.
So going back to our work,
try the triple integral given in rectangular coordinates.
In cylindrical coordinates,
we have the write the integrating function
f[x,y,z] as a function of r theta and z,
and differential V is equal to r,
dZ, dr, d theta.
So using cylindrical coordinates
just like polar coordinates,
we're going to have an extra factor of r
in the integrand function.
So starting with the given triple integral,
to write this in cylindrical coordinates,
e to the z is already a function of z,
so there's no conversion to perform with this function,
but we are going to have an extra fact of r
in the integrand function,
so the integrand function is going to be
r times e raised to the power of z,
and then we'll use the order of integration
given by dz dr d theta.
And now we need to find the limits of integration
for z, r, and theta.
Well the solid region of integration
is bounded below by the xy plane
which has an equation z equals zero,
and above by the paraboloid given by
z equals three plus x squared plus y squared.
But we need to express z in terms of r and theta,
so we'll perform a substitution
for x squared plus y squared.
X squared plus y squared is equal to r squared,
so the region of integration is bounded above
by z equals three plus r squared.
Which means the limits of integration for z
are going to be from zero to three plus r squared.
And now to help us find the
limits of integration for r and theta,
we'll take a look at the xy trace.
The xy trace is going to be the circle here
given by the equation x squared plus y squared equals two,
which in polar or cylindrical form,
would be the equation r squared equals two,
or r equals plus or minus the square root of two.
So the xy trace is the circle here.
So using the circle, the limits of integration for r,
are going to be from r equals zero
to r equals the square root of two.
The limits of integration for theta to trace out the circle
are going to be from theta equals zero
all the way around the circle,
to theta equals two pi radians.
So we have from zero to two pi radians.
And now we first integrate with respect to z,
treating r as a constant,
and therefore the antiderivative
is just going to be r times e to the z.
And now I need to find big F of b minus big F of a
by performing substitution for z.
So we're going to have r times the quantity e
raised to the power of three plus r squared
minus e to the zero.
Well e to the zero is equal to one,
so distributing the new integrand function
is going to be r times e
raised to the power of three plus r squared
and then minus r times one, so minus r.
And now we integrate with respect to r.
Notice how to integrate this first term with respect to r,
we do have to perform u substitution,
where u is equal to the exponent of three plus r squared,
and therefore differential u is equal to
two r differential r,
so if we divide both sides by two,
notice how 1/2 differential u was equal to rdr.
Which means we can think of r times
e raised to the power of three plus r squared differential r
as 1/2 e to the u differential u.
So when we find the antiderivative of this first term,
we're just going to have 1/2 times e
raised to the power of three plus r squared,
and then we have minus the antiderivative r
with respect to r is going to be r squared divided by two
or 1/2 r squared.
And now we need to find big F of b minus big F of a
by performing substitution for r.
So when r is equal to the square root of two,
we're going to have 1/2 times e
raised to power of three plus,
the square root of two is just two
minus 1/2 times the square root of two squared is just two,
minus when r is zero we're going to have
1/2 times e raised to power of three,
and then just minus zero.
And now simplifying.
Let's write this as 1/2 e to the fifth.
We have a minus one here,
but we also have minus 1/2 e to the third.
Let's write minus 1/2 e to the third first
and then minus one.
Notice how each of these terms are just constants.
So now when we integrate with respect to theta,
we're just going to have this constant
1/2 e to the fifth minus 1/2 e to the third
minus one times the antiderivative of theta.
And finally performing substitution here.
We just have the constant 1/2 e to the fifth
minus 1/2 e to the third, minus one times,
we just have two pi minus zero.
So the exact value of the triple integral
is going to be 1/2 e to the fifth
minus 1/2 e to the third, minus one times two pi,
or as a decimal approximation,
this comes out to approximately 396.8699.
I hope you found this helpful.