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We have a cylindrical tank with radius 2m and length 10m filled with water. How much work does it tank to pump the water out of the tank from the top?

My attempt at the problem goes as follow.

$g$ is gravity $9.8\, m/s^2$, density of water is $1000\, kg/m^3$.

Let $x$ be the distance from the top of the tank. The width of a general cross section of the tank is $w = 2\sqrt{4 - (2 - x)^2} = 2\sqrt{4x - x^2}$. Thus the volume of the cross section is $V^* = 20\sqrt{4x - x^2}\Delta x$.

The work required to pump this section is $W^* = (20,000g)x\sqrt{4x - x^2}\Delta x$.

Thus the total work to pump the tank should be

$$W = 20,000g \int_0^4 x\sqrt{4x - x^2}dx.$$

Wolfram alpha tells me the integral here is $4\pi$, which seems like a nice enough answer. However, my calculus is very rusty. A simple u-sub does not seem to work very well here. Is there another technique of integration I should be using? Thanks.

Edit: (Thanks to André Nicolas) Treating everything as the center of mass is much easier, but for completeness I went ahead and solved the integral. Thanks for the help.

$$W = \int_0^4 x\sqrt{4 - (2 - x)^2} dx$$ Let $u = 2 - x$, so $du = -dx$. $$W = \int_{-2}^2 (2 - u)\sqrt{4 - u^2} du = 2\int_{-2}^2 \sqrt{4 - u^2}du + \int_{-2}^2 u\sqrt{4 - u^2}du$$

For the first integral we let $u = 2\sin v$, so $du = 2\cos vdv$. $$2\int_{-2}^2\sqrt{4 - u^2} = 8 \int_{-\pi/2}^{\pi/2} \sqrt{1 - \sin^2 v}\cos v dv = 8 \int_{-\pi/2}^{\pi/2} \cos^2 v dv = 8\int_{-\pi/2}^{\pi/2} \frac{1 + \cos(2v)}{2}dv$$ $$ = 8(\frac{v}{2} + \frac{\sin(2v)}{4} |_{-\pi/2}^{\pi/2}) = 4\pi.$$

For the second integral, notice it is an odd function. Since we are integrating from $-2$ to 2, the intervals $(-2, 0)$ and $(0, 2)$ cancel eachother out, so we get $$\int_{-2}^2 u\sqrt{4 - u^2}du = 0.$$

Thus $W = 20,000g (4\pi)\,J$, or $80,000\pi g\, J$.

Edit 2:

Alternatively for the first integral, we can avoid the trig sub by noticing $\int_{-2}^2 \sqrt{4 - x^2} dx$ is equal to half of the area under a circle of radius 2, thus $2\int_{-2}^2 \sqrt{4 - x^2} dx = 2 \cdot (1/2)\cdot (\pi 2^2) = 4\pi$.

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Completing the square, we get $4x-x^2=4-(x-2)^2$. It is now natural to let $x-2=2u$. Then things become familiar.

One can make things simpler to begin with by choosing the origin at the centre of the circle of cross-section.

Remark: In general, for a work problem of this type, we can imagine the whole mass as concentrated at the centre of mass. That point of view yields a one-line solution.

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  • $\begingroup$ If I treat it as one mass, I get $V = 40\pi$, so the mass is $M = 40,000 \pi$ (density of water). Thus the force is $F = 40,000 \pi g$. We are displacing the mass by $2m$, so its $W = 80,000\pi g$ where $g$ is gravity. That sounds good thanks. $\endgroup$
    – zrbecker
    Commented Aug 8, 2013 at 7:10
  • $\begingroup$ If this is a calculus exercise, you are probably supposed to do it the integration way. $\endgroup$ Commented Aug 8, 2013 at 7:16
  • $\begingroup$ I do not believe they are at the point of the book where they are doing trig subs. But I will present it both ways for completeness. Thanks. $\endgroup$
    – zrbecker
    Commented Aug 8, 2013 at 7:35
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    $\begingroup$ You don't need trig sub. The integral breaks up into two parts, one of them a sub $t=4-u^2$, while the other can be viewed as the area of a circle. $\endgroup$ Commented Aug 8, 2013 at 7:43
  • $\begingroup$ Haha, wow. I can't believe I didn't see that. Thanks. $\endgroup$
    – zrbecker
    Commented Aug 8, 2013 at 7:44

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