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A tank, shaped like a cone has a height of $15$ meters and a base radius of $6$ meters. The cone is standing up, with the circular part as its base. It is full of water, and we want to pump all of it out through a pipe that is at the tip of the cone. Assume that the density of water is $\rho = 1000~\text{kg}/\text{m}^3$ and the acceleration due to gravity is $g = 9.8~\text{m}/\text{s}^2$ .

Set up the integral of the work done to pump all the water out of the tank. DO NOT EVALUATE THE INTEGRAL. [Note: You can use ρ and g in your answer; you do not have to use 1000 and 9.8.]

My attempt

$$\int _{ 0 }^{ 15 }{ (1000)(9.8)\left(\frac { 4\pi y^ 2 }{ 25 }\right)(15-y)dy }$$

but the answer was $$\int _{ 0 }^{ 15 }{ (1000)(9.8)\left[\frac { 2(15-y) }{ 5 }\right]^ 2(15-y)dy }$$

Can anyone please explain this?

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  • $\begingroup$ Can you show what you have tried to get your answer? We can help you find which part you made a mistake. $\endgroup$ – user061703 May 7 '18 at 16:33
  • $\begingroup$ It looks like there is a factor of $\pi$ missing from the answer. $\endgroup$ – N. F. Taussig May 7 '18 at 16:47
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It looks like the $A(y)$ seems to be the problem

The tank is a cone, so the radius of the cross section increases linearly with $h$

At $y=0$, $r=6$ and at $y=15$, $r=0$

So

$$r(y)=\frac{6}{15}(15-y)$$

$$A(y)=\pi \left[\frac{6(15-y)}{15}\right]^2=\pi \left[\frac{2(15-y)}{5}\right]^2$$

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  • $\begingroup$ Can you please explain me briefly $\endgroup$ – tien lee May 7 '18 at 17:19
  • $\begingroup$ Sure, the $6$ in the fraction $\frac{6}{15}$ comes from what the radius is at maximum (at the base when $y = 0$), and at $y=0$, in order to return $r$ to 6, we have to divide (modulate) by 15. $r(y)$ is now linearized in $y$ and satisfies what we want it to do, namely $r(0) = 6$ and $r(15)=0$ $\endgroup$ – AEngineer May 7 '18 at 17:39

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