Lecture32and33

# We can use the result as a formula whenever we

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 03/30/2014.

Ask a homework question - tutors are online