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Calculus- Early Transcendentals, 2021a

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504 Multiple Integration<br />

Example 14.10: Volume of a Box<br />

Compute the volume of the box with opposite corners at (0,0,0) and (1,2,3).<br />

Solution. We use an integral to compute the volume of the box:<br />

∫ 1 ∫ 2 ∫ 3<br />

0 0 0<br />

dzdydx =<br />

∫ 1 ∫ 2<br />

0 0<br />

z| 3 0 dydx = ∫ 1<br />

∫ 2<br />

0 0<br />

3dydx =<br />

∫ 1<br />

0<br />

∫ 1<br />

3y| 2 0 dx = 6dx = 6.<br />

0<br />

♣<br />

Of course, this is more interesting and useful when the limits are not constant.<br />

Example 14.11: Volume of a Tetrahedron<br />

Find the volume of the tetrahedron with corners at (0,0,0), (0,3,0), (2,3,0),and(2,3,5).<br />

Solution. The whole problem comes down to correctly describing the region by inequalities: 0 ≤ x ≤ 2,<br />

3x/2 ≤ y ≤ 3, 0 ≤ z ≤ 5x/2. The lower y limit comes from the equation of the line y = 3x/2 that forms one<br />

edge of the tetrahedron in the xy-plane; the upper z limit comes from the equation of the plane z = 5x/2<br />

that forms the “upper” side of the tetrahedron; see Figure 14.12. Now the volume is<br />

∫ 2 ∫ 3<br />

0<br />

3x/2<br />

∫ 5x/2<br />

0<br />

dzdydx =<br />

=<br />

=<br />

∫ 2 ∫ 3<br />

0 3x/2<br />

∫ 2 ∫ 3<br />

0<br />

∫ 2<br />

0<br />

∫ 2<br />

3x/2<br />

z| 5x/2<br />

0<br />

dydx<br />

5x<br />

2 dydx<br />

∣<br />

5x ∣∣∣<br />

3<br />

2 y dx<br />

3x/2<br />

15x<br />

2 − 15x2<br />

=<br />

0 4 dx<br />

= 15x2<br />

4 − 15x3 2<br />

12 ∣<br />

0<br />

= 15 − 10 = 5.<br />

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