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Unformatted text preview: l coordinates: z = ρ cos φ and r = ρ sin φ. This allows us to make
the connection to Cartesian coordinates: x = r cos θ = ρ sin φ cos θ
(4) y = r sin θ = ρ sin φ cos θ
z= z = ρ cos φ Of course, we’ll need the Jacobian of this transformation:
xρ xθ xφ
∂ (x, y, z )
y y y
= ρ
θ
θ
∂ (ρ, θ, φ)
zρ zθ z φ
sin φ cos θ ρ cos φ cos θ −ρ sin φ sin θ
= sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ
cos φ
−ρ sin φ
0
6
= ... = ρ2 sin φ.
We’ll let you ﬁll in the missing lines of the calculation on your own. We can use the result
as a formula; whenever we express a triple integral in spherical coordinates we may write
dV = ρ2 sin φ dρ dφ dθ.
Example 4: Recalling the discussion of the previous section, we should be able to calculate
ˆ
the volume of a sphere of radius a as
dV . Now, if D is a sphere, then spherical coordinates
D are the most sensible tool. To describe the entire sphere, we need the full range for both
angles. The volume is V= ˚ dV = Dxyz = ρ2 sin φ dρ dθ dφ Dρθφ ˆ π ˆ π 0 = ˚ ˆ 2π 0 ˆ a ρ2 sin φ dρ dθ dφ 0 sin φ dφ 0 ˆ 2π dθ 0 = (2)(2π ) ˆ a ρ2 dρ 0 a3
3 4
= π a3 ,
3
which is the result we should have been expecting.
x2 + y 2 + z 2 over the region lying
above the xy plane, inside the sphere x2 + y 2 + z 2 = 4, and below the cone z = x2 + y 2 .
Example 5: Find the average value of f (x, y, z ) = Solution: The domain is illustrated (from two diﬀerent perspectives) in Figure 8; we’ll
call it E . We are looking for favg ˝ ˝
f dV
E f dV
=
=˝
.
Volume(E )
E dV
E 7 Figure 8:
In spherical coordinates, the integrand
x2 + y 2 + z 2 is simply ρ, and dV = ρ2 sin φ dρ dφ dθ. We also need to describe the boundaries of the domain in this system: the sphere has the simple
equation ρ = 2, but what about the cone? Well,
z = x2 + y 2
ρ cos φ = ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ means = ρ sin φ,
so tan φ = 1, meaning that φ = π /4. Upon reﬂection this should have been obvious; any
cone will have equation φ = κ, for some constant κ, and our cone opens up at an angle of
45◦ from the z axis. If you think about that for a moment, it should also become clear that
the spherical equation of our third boundary, the xy plane, is φ = π /2. Putting all of this
information to use, then, we have
˚ E ˆ
x2 + y 2 + z 2 dV = 2π 0 = ˆ ˆ 2π dθ 0 π
2
π
4 ˆ ˆ 0 π
2
π
4 2 ρ · ρ2 sin φ dρ dφ dθ sin φ dφ ˆ 2 ρ3 dρ 0 √
√
2
= (2π )
(4) = 4 2π .
2
Meanwhile,
˚ dV = E ˆ 2π ˆ = π
2 2π 0 0 ˆ π
4 dθ ˆ 2 ρ2 sin φ dρ dφ dθ 0 ˆ π
2
π
4 sin φ dφ ˆ 2 ρ2 d ρ 0 √
√
2
8
8 2π
= (2π )
=
.
2
3
3 8 √
4 2π
3
2 + y 2 + z 2 within E is √
Therefore the mean value of x
=.
2
8 2π /3
Suggestion: think about what the stated problem was here, and convince yourself that our
result is realistic. Then go back and do the same for Example 2. 9...
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This document was uploaded on 03/30/2014.
 Winter '13
 Calculus, Integrals

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