Transcript 12.8 Notes

Multiple Integrals
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12.8
Triple Integrals in Cylindrical and Spherical Coordinates
Cylindrical Coordinates
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Cylindrical Coordinates
Recall that the cylindrical coordinates of a point P are
(r, , z), where r, , and z are shown in Figure 1.
Figure 1
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Cylindrical Coordinates
Suppose that E is a type 1 region whose projection D onto
the xy-plane is conveniently described in polar coordinates
(see Figure 2).
Figure 2
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Cylindrical Coordinates
In particular, suppose that f is continuous and
E = {(x, y, z) | (x, y)  D, u1(x, y)  z  u2(x, y)}
where D is given in polar coordinates by
D = {(r,  ) |     , h1( )  r  h2( )}
We know that
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Cylindrical Coordinates
But we also know how to evaluate double integrals in polar
coordinates. In fact, combining Equation 1 with the equation
below,
we obtain
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Cylindrical Coordinates
Formula 2 is the formula for triple integration in
cylindrical coordinates.
It says that we convert a triple
integral from rectangular to
cylindrical coordinates by writing
x = r cos , y = r sin , leaving z
as it is, using the appropriate limits
of integration for z, r, and ,
and replacing dV by r dz dr d.
Figure 3
Volume element in cylindrical
coordinates: dV = r dz dr d
(Figure 3 shows how to remember this.)
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Cylindrical Coordinates
It is worthwhile to use this formula when E is a solid region
easily described in cylindrical coordinates, and especially
when the function f(x, y, z) involves the expression x2 + y2.
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Example 1 – Finding Mass with Cylindrical Coordinates
A solid E lies within the cylinder x2 + y2 = 1, below the plane
z = 4, and above the paraboloid z = 1 – x2 – y2. (See
Figure 4.) The density at any point is proportional to its
distance from the axis of the cylinder. Find the mass of E.
Figure 4
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Example 1 – Solution
In cylindrical coordinates the cylinder is r = 1 and the
paraboloid is z = 1 – r2, so we can write
E = {(r, , z) | 0    2, 0  r  1, 1 – r2  z  4}
Since the density at (x, y, z) is proportional to the distance
from the z-axis, the density function is
f(x, y, z) = K
= Kr
where K is the proportionality constant.
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Example 1 – Solution
Therefore, from Formula
cont’d
, the mass of E is
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Example 1 – Solution
cont’d
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Spherical Coordinates
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Spherical Coordinates
We have defined the spherical coordinates (, , ) of a
point (see Figure 6) and we demonstrated the following
relationships between rectangular coordinates and spherical
coordinates:
x =  sin  cos 
y =  sin  sin 
z =  cos 
Figure 6
Spherical coordinates of P
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Spherical Coordinates
In this coordinate system the counterpart of a rectangular
box is a spherical wedge
E = {(, , ) | a    b,     , c    d }
where a  0 and  –   2. Although we defined triple
integrals by dividing solids into small boxes, it can be shown
that dividing a solid into small spherical wedges always
gives the same result.
So we divide E into smaller spherical wedges Eijk by means
of equally spaced spheres  = i, half-planes  = j, and
half-cones  = k.
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Spherical Coordinates
Figure 7 shows that Eijk is approximately a rectangular box
with dimensions , i  (arc of a circle with radius i,
angle ), and i sin k  (arc of a circle with radius
i sin k, angle  ).
Figure 7
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Spherical Coordinates
So an approximation to the volume of Eijk is given by
()  (i )  (i sin k ) = i2 sin k   
Thus an approximation to a typical triple Riemann sum is
But this sum is a Riemann sum for the function
F(, , ) = f( sin  cos ,  sin  sin ,  cos )  2 sin 
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Spherical Coordinates
Consequently, the following formula for triple integration
in spherical coordinates is plausible.
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Spherical Coordinates
Formula 4 says that we convert a triple integral from
rectangular coordinates to spherical coordinates by writing
x =  sin  cos 
y =  sin  sin 
z =  cos 
using the appropriate limits of
integration, and replacing dV
by  2 sin  d d d.
This is illustrated in Figure 8.
Figure 8
Volume element in spherical
coordinates: dV =  2 sin  d d d
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Spherical Coordinates
This formula can be extended to include more general
spherical regions such as
E = {(, , ) |     , c    d, g1(, )    g2(, )}
In this case the formula is the same as in (4) except that the
limits of integration for  are g1(, ) and g2(, ).
Usually, spherical coordinates are used in triple integrals
when surfaces such as cones and spheres form the
boundary of the region of integration.
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Example 3
Evaluate
where B is the unit ball:
B = {(x, y, z) | x2 + y2 + z2  1}
Solution:
Since the boundary of B is a sphere, we use spherical
coordinates:
B = {(, , ) | 0    1, 0    2, 0    }
In addition, spherical coordinates are appropriate because
x2 + y2 + z2 =  2
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Example 3 – Solution
cont’d
Thus (4) gives
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