# Work Problem

1. Feb 2, 2004

### eddo

I apologize that I don't know how to make the math equations.

Alright it's going to be kind of complicated trying to describe this in words, but I'll do my best. There is a tank shaped like a right cylinder on it's side. The length of the tank (or height of the cylinder) is 6m, and the radius is 1.5m. There is a pipe sticking 1m out of the top of the tank. How much work is required to empty the tank, remembering that the density of water is 1000kg/m^3 and g=9.81m/s^2.

I know the answer, because it is the same as the work done moving all the water from the center point, or lifting it all 2.5m. But we are required to solve it using integration, by taking the work to move a thing horizontal slice, and summing up infinitely many infinitely small slices.

So here's what I've done so far. set x axis vertically, origin at the bottom of the tank. A horizontal slice will create a rectangle with area 6 x w, w being the width of the slice. To find w in terms of x, i made a triangle with the two radii and w, and the distance from the centre of the circle to the centre of w being r-x. This give w to be 2SQRT(3x-x^2), and so the area is 12SQRT(3x-x^2). So now the volume is the area times delta x. If p is density, the force is than:
F=mg=Vpg=pg12SQRT(3x-x^2)
this must be lifted a distance of 4-x. so the work is:
W=pgSQRT(3x-x^2)(4-x)
Now this just has to be integrated between x=0 and x=3, but I don't know how to do this. Is there an easier way? Have I made a mistake along the way? Or how do I integrate this? BTW we haven't done integration by parts yet, so there should be a way to do this without using that, we've only learned the substitution method.
Thank you for any and all help.

2. Feb 2, 2004

### HallsofIvy

Staff Emeritus
Actually, you've done very well. I would have been inclined to put x=0 at the center of the circle rather than at the bottom. Then w= &radi;(1.52-x2) rather than &radic;(3x-x2) and the integration is a little easier. Have you had "trigonometric substitution" yet? Since cos2(&theta;)= 1- sin2, the substitution x= 1.5 sin(&theta;) makes &radi;(1.52-x2) into 1.5&radic;(1- sin2(&theta;))= 1.5cos(&theta;). Also dx= 1.5 cos(&theta;).

With your pg&radic;(3x-x^2)(4-x) you might want to complete the square: 3x- x^2= (9/4)-(9/4)+ 3x- x^2= (9/4)- (x- 3/2)^2. Let u= x- 3/2, x= u+ 3/2 so 3-x= 4- u- 3/2= 5/2- u and the integrand is &radic;(9/4- u^2)(5/2- u)du. Now break it into two integrals: (5/2)&radic(9/4- u^2)du can be done by u= 3/2sin(&theta;) and -(9/4-u^2)udu can be done by v= 9/4- u^2 (so dv= -2u du or udu= -1/2 dv).

Last edited: Feb 3, 2004