692
Chapter 6 | Applications of Integration
W = ∑ i =1 n
W i ≈ ∑ i =1 n
156,800 πx i * Δ x .
This is a Riemann sum, so taking the limit as n →∞, we get
→∞ ∑ i =1 n
W = lim n
156,800 πx i * Δ x
10
=156,800 π ∫ 2
xdx
⎤ ⎦ | 2
⎡ ⎣ x 2 2
10 = 7,526,400 π ≈ 23,644,883.
=156,800 π
The work required to empty the tank is approximately 23,650,000 J. For pumping problems, the calculations vary depending on the shape of the tank or container. The following problem- solving strategy lays out a step-by-step process for solving pumping problems. Problem-Solving Strategy: Solving Pumping Problems 1. Sketch a picture of the tank and select an appropriate frame of reference. 2. Calculate the volume of a representative layer of water. 3. Multiply the volume by the weight-density of water to get the force. 4. Calculate the distance the layer of water must be lifted. 5. Multiply the force and distance to get an estimate of the work needed to lift the layer of water. 6. Sum the work required to lift all the layers. This expression is an estimate of the work required to pump out the desired amount of water, and it is in the form of a Riemann sum. 7. Take the limit as n →∞ and evaluate the resulting integral to get the exact work required to pump out the desired amount of water.
We now apply this problem-solving strategy in an example with a noncylindrical tank.
Example 6.26 A Pumping Problem with a Noncylindrical Tank
Assume a tank in the shape of an inverted cone, with height 12 ft and base radius 4 ft. The tank is full to start with, and water is pumped over the upper edge of the tank until the height of the water remaining in the tank is 4 ft. How much work is required to pump out that amount of water? Solution The tank is depicted in Figure 6.54 . As we did in the example with the cylindrical tank, we orient the x -axis vertically, with the origin at the top of the tank and the downward direction being positive (step 1).
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