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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/12-y. This is incorrect, because you get a y value, and a constant. The exercise uses just plain old point-slope, 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 12-y. 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: 3,002 Thanks: 1588 work =$\displaystyle \int$ WALT W = weight density A = cross-sectional 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 y-axis. $\displaystyle r = x = \frac{y}{8}$ $\displaystyle A = \pi x^2 = \pi \left(\frac{y}{8}\right)^2$ $\displaystyle L = 12-y$ work = $\displaystyle W\pi \int_0^{12} \left(\frac{y}{8}\right)^2(12-y) \, dy$ Last edited by skeeter; October 22nd, 2014 at 02:15 PM. Tags cone, confusion, inverted, points, pulling, water Search tags for this page

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