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Read on to understand my question at the end.

Problem 1: How much work to lift cable?

A machine lifts a 180ft steel cable weighing 4.5lb/ft. Assume when the cable is at maximum length, y = 0. Work done by the machine?

W = $\int_{0}^{180} 4.5(180 - y) \hspace{.3cm}dy$

Problem 2: How much work to drain tank by pumping water to ground level?

A rectangular tank, with it's top at ground level, and volume of 2400$ft^3$, is filled to the brim with water weighing 62.4lb/$ft^3$. The height of the tank is 20ft. Work to drain tank by pumping water to ground level?

At first, I thought this was the answer, based on problems similar to the first one:

W = $\int_{0}^{20} 62.4(2400 - y) \hspace{.3cm}dy$

However, the solution is:

W = $\int_{0}^{20} (62.4*2400)y \hspace{.3cm}dy$

Can someone explain the difference in logic between the two problems?

1 Answers 1

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There is no real definition in logic. Pay attention to what you're doing. You have the integrand is $(2400-y).$ $2400$ is the volume of the tank, not its height. The height is $20.$ Thus you should have an integral of $\int_0^{20}(20-y)dy$ which is equal to $\int_0^{20}ydy$ by symmetry.

The factor of $62.4*2400$ out front is a conversion factor of weight per length that is analogous to the $4.5$ out front of the first problem. The answer to the first problem could be written $\int_0^{180}4.5ydy$

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    So I could have written the integrand as: 62.4*2400(20-y)? – 2017-02-14
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    @darylnak Yes, that's right. – 2017-02-14