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October 22nd, 2014, 01:06 PM  #1 
Member Joined: Oct 2013 Posts: 38 Thanks: 2  Pulling up water from an Inverted Cone  a few points of confusion
Hi all, Given an inverted cone tank of water (full), the radius is 1.5m at the top, and the length, from top to bottom is 12m  We need to measure the work from pulling up all the water. Density and gravity are just standard here (1000, 9. eight [posted a smiley face when I put "eight"]). I have two points of confusion here: 1) When finding out the radius, I set up similar triangles. 1.5/12 = r/12y. This is incorrect, because you get a y value, and a constant. The exercise uses just plain old pointslope, with the bottom of this tank at the origin. I see what the correct method is, but I don't understand why my use of sim. triangles here is incorrect. *The correct answer for the radius is y/8. 2) Once I have the area for the circle (pi y^2 / 64), my instinct is to sum up the areas of all the circles, when in fact we have to multiply that by 12y. I am confused as to why we can't sum up the circles with pi y^2 / 64, as well as, obviously, the constants. Thanks for looking. Last edited by leo255; October 22nd, 2014 at 01:09 PM. 
October 22nd, 2014, 01:40 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,880 Thanks: 1503 
work =$\displaystyle \int$ WALT W = weight density A = crosssectional area of a horizontal slice of liquid w/r to y L = lift distance of a slice in terms of y T  slice thickness (dy) using a set of coordinate axes, sketch a line from the origin (bottom of cone) to the point (1.5,12) ... equation of this line is $\displaystyle y = 8x \implies x = \frac{y}{8}$ this line forms the cone when rotated about the yaxis. $\displaystyle r = x = \frac{y}{8}$ $\displaystyle A = \pi x^2 = \pi \left(\frac{y}{8}\right)^2$ $\displaystyle L = 12y$ work = $\displaystyle W\pi \int_0^{12} \left(\frac{y}{8}\right)^2(12y) \, dy$ Last edited by skeeter; October 22nd, 2014 at 02:15 PM. 

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