# work problem - tank of water

• Jul 8th 2013, 06:52 PM
baldysm
work problem - tank of water
I am trying to understand work and water in a tank being pumped out.

I have a problem with the frustrum of a cone. It is a 6' radius on top, 3' radius on the bottom, and it's 8' tall.

In calculating the volume of the tank, they used the equation of the line of the side of the tank for the radius, which is $\displaystyle r=-\frac{3}{8}y+6$

If you think of the tank as laying on it's side, with the origin at the center of the 6' radius, I understand how that equation is gotten, no problem, and it gives a correct answer. If I know that's the equation of the line, I'm good with the whole problem.

My question is - how do I know to orient the tank that way to get the equation of the line?

If I don't reorient the tank, and set y0 at the center of the top of the tank, the 2 points used to get the equation of the line are (6,0) and (3,-8). The slope is $\displaystyle \frac{8}{3}$ and a very different equation for the radius, which gives an incorrect answer.

How do I make sure I don't use the wrong equation?

Thanks!
• Jul 8th 2013, 08:08 PM
ibdutt
Re: work problem - tank of water
in fact it is just immaterial as to how one orients the object, in this case the frustum. it is always sensible to keep the axis of frustum as one of the coordinate axis and the origin on the top / bottom face of the frustum for easy calculation and understanding. ideally speaking if i was to do this question i would keep the axis of the frustum as x axis and the origin at the center of the bottom face.