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16 MULTIPLE INTEGRALS

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SECTION <strong>16</strong>.7<br />

TRIPLE <strong>INTEGRALS</strong> IN CYLINDRICAL COORDINATES<br />

Move the bottom surface of S up to<br />

z = g (r, θ), as pictured at right.<br />

The volume now becomes<br />

∫ β<br />

∫ h2 (θ) ∫ f (r,θ)<br />

α h 1 (θ) g(r,θ)<br />

rdzdrdθ. The basic<br />

volume element is given in Figure 3 of<br />

the text. Conclude with the situation<br />

where we have h (r, θ, z) defined on S.<br />

Then the triple integral of h on S is<br />

∫ β ∫ h2 (θ) ∫ f (r,θ)<br />

α h 1 (θ) g(r,θ)<br />

h (r, θ, z) rdzdrdθ.<br />

WORKSHOP/DISCUSSION<br />

• Describe in terms of cylindrical coordinates the surface of rotation formed by rotating z = 1/x about the<br />

z-axis, noting that there is an axis of symmetry. Point out that the equations come from simply replacing<br />

x by r.<br />

• Develop a straightforward example such as the region depicted below:<br />

The Capped Cone<br />

Set this volume up as a triple integral in cylindrical coordinates, and then find the volume. (The<br />

computation of this volume integral is not that hard, and can be assigned to the students.) Conclude<br />

by setting up the volume integral of h (r, θ, z) = rz over this region.<br />

GROUP WORK: A Partially Eaten Sphere<br />

Notice that you are removing “ice cream cones” both above and below the xy-plane.<br />

Answers:<br />

1. 36π − 2 ∫ 2π<br />

0<br />

2. 36π − 2 ∫ 2π<br />

0<br />

∫ √ 3 ∫ √ (<br />

6<br />

0 −r √ 2 rdzdrdθ = π 36 − 10 √ )<br />

6<br />

∫ √ 3 ∫ √ 6<br />

0 −r √ 2 zr2 sin θ dzdr dθ = 36π<br />

913

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