14.7 Triple Integrals in Cylindrical and Spherical Coordinates
Triple Integrals in Cylindrical Coordinates
Many simple solid regions yield difficult triple integrals. Historically,
the other coordinates systems were developed just to overcome difficulties
in integration. Recall first that the conversion formulae are 
for
rectangular into cylindrical coordinates. The easy way to remember this is
just to think of the xy plane is given in polar coordinates and the z axis
is unchanged. If we can imagine the region Q defined as follows:
Q = {(x,y,z): (x,y) is in R and 
}
and
R = {(r,q):
}
If f is continuous over the solid region Q, then the triple integral of f over Q can be written as
Since the double integral over the region R is to be evaluated in polar coordinates, we can write
To visualize what the volume element dV is, look at the small polar rectangle in the sketch here:
This is contrasted with the volume element in rectangular coordinates in the sketch here:
Ex 1 Find the volume of the solid region Q cut from
the sphere 
by
the cylinder 
Note we can write the equation of the sphere as
which
allows us to write the z limits of integration as 
.
The r and q
limits are found by examining the region given here of the xy plane in polar
coordinates for the function.
If u = 4 - r2 and du = -2rdr
Ex 2 Find the center of mass of the solid bounded
by the paraboloid 
and
the plane 
if
the density at each point is proportional to the distance between the point
and the z axis.
The x and y coordinates of the center of mass are both 0 from the symmetry of the region. We need the mass and the moment about the xy plane to determine the z coordinate of the center of mass.
Switching to cylindrical coordinates, note 
.
Then the integral for the mass becomes
The center of mass is 
Triple Integrals in Spherical Coordinates
The volume element for spherical coordinate triple integrals is shaped like a square of the skin of a basketball as the drawing shows:
The dV in a spherical coordinate integral is r2sinf dr df dq. In the drawing above, r = rsinf since r = rcos(p/2-f) is the right triangle relationship between r and r. Hence the triple integral can be solved in spherical coordinates using the following:
Keep in mind some regions are very simply defined in spherical coordinates, even though the integration formula may look a bit intimidating.
Ex 3 Find the volume of the solid region bounded
below by the upper half of the cone 
and
above by the sphere 
.
The sphere has equation r
= 3. Substituting into the cone equation
the formulas for conversion to spherical coordinates 
yields
the equation 
. Simplifying
this leaves us with 
so
. The integral giving
the volume we seek is:
Ex 4 Find the centroid of the region in the above example.
We can assume the density functions are constant
and will cancel out of the fraction representing 
.
Also the x and y coordinates are obviously both 0, since the figure is centered
on the z axis and symmetrical. The integral we now must solve to get the moment
about the xy plane is this one below. Note the substitution rcos(f) for
z:
Assignment Pg 1040# 1,3,9,11,13,19,33,35