Calculus Volume 1

694

Chapter 6 | Applications of Integration

From properties of similar triangles, we have

r i 12− x i * =

4 12 = 1 3 3 r i = 12− x i * r i = 12− x i * 3 = 4− x i * 3 .

Then the volume of the disk is

⎛ ⎝ 4−

⎞ ⎠

2

x i * 3

V i = π Δ x (step 2). The weight-density of water is 62.4 lb/ft 3 , so the force needed to lift each layer is approximately

⎛ ⎝ 4−

⎞ ⎠

2

x i * 3

F i ≈62.4 π

Δ x (step 3).

Based on the diagram, the distance the water must be lifted is approximately x i * feet (step 4), so the approximate work needed to lift the layer is

⎛ ⎝ 4−

⎞ ⎠

2

x i * 3

W i ≈62.4 πx i *

Δ x (step 5).

Summing the work required to lift all the layers, we get an approximate value of the total work: W = ∑ i =1 n W i ≈ ∑ i =1 n 62.4 πx i * ⎛ ⎝ 4− x i * 3 ⎞ ⎠ 2 Δ x (step 6).

Taking the limit as n →∞, we obtain

⎛ ⎝ 4−

⎞ ⎠

2

→∞ ∑ i =1 n

x i * 3

W = lim n

62.4 πx i *

Δ x

8

⎛ ⎝ 4− x 3

⎞ ⎠

2

= ∫

62.4 πx

dx

0

8 ⎛ ⎝ 16 x − 8 x

⎞ ⎠ dx

x ⎛

⎞ ⎠ dx =62.4 π ∫ 0

8

x 3 9

2 3 +

x 2 9

=62.4 π ∫

⎝ 16− 8 x

3 +

0

⎤ ⎦ | 0 8

⎡ ⎣ 8 x 2 − 8 x

3 9 +

x 4 36

=62.4 π =10,649.6 π ≈ 33,456.7. It takes approximately 33,450 ft-lb of work to empty the tank to the desired level.

6.26 A tank is in the shape of an inverted cone, with height 10 ft and base radius 6 ft. The tank is filled to a depth of 8 ft to start with, and water is pumped over the upper edge of the tank until 3 ft of water remain in the tank. How much work is required to pump out that amount of water?

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