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 February 15th, 2012, 06:02 PM #1 Joined: Feb 2012 Posts: 28 Thanks: 0 Work to pump water out of a tank Hi everyone, I looked around on the website and saw some other posts for how to calculate the amount of work it takes to pump water out of a tank but I have a slight variation on these problems that I can't seem to figure out: The ends of a "parabolic" water tank are the shape of the region inside the graph of y = x^2 for 0 ? y ? 9; the cross sections parallel to the top of the tank (and the ground) are rectangles. At its center the tank is 9 feet deep and 6 feet across. The tank is 10 feet long. Rain has filled the tank and water is removed by pumping it up to a spout that is 2 feet above the top of the tank. Set up a definite integral to find the work W that is done to lower the water to a depth of 4 feet and then find the work. [Hint: You will need to integrate with respect to y.] I can't figure out how the "rectangular cross sections" can be calculated. This is where I am so far: W = 62.5pi * integral from 4 to 9 [(9-y^2)(11-y)]dy I know this is wrong (except for the boundaries). Here is where I am headed with the problem: I found the area for a partition of the circle at each y layer but I suppose I need to do it with squares. Would it look something like this: [(9-y^2)]^2? I don't want to submit this yet because I only have one try left so what do you all think? Thanks for your help!
 February 15th, 2012, 06:32 PM #2 Global Moderator     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 428 Math Focus: Calculus/ODEs Re: Work to pump water out of a tank The length of the rectangle is the length of the tank, which is 10 ft. The width of the rectangle is the width of the parabola. Let y be the depth of the water, then the width of the rectangle is $w=2x=2\sqrt{y}$. The rectangular cross section must be pumped up (11 - y) ft. So, we have: $dW=$$62.5\text{ \frac{lb}{ft^3}}$$$$10\text{ ft}$$$$2\sqrt{y}\text{ ft}$$$$(11-y)\text{ ft}$$\,$$dy\text{ ft}$$$ $W=1250\int_4\,^9 11y^{\frac{1}{2}}-y^{\frac{3}{2}}\,dy\text{ lb\cdot ft}$ You should get: $W=\frac{206000}{3}\:\text{lb\cdot ft}$
 February 15th, 2012, 06:47 PM #3 Joined: Feb 2012 Posts: 28 Thanks: 0 Re: Work to pump water out of a tank Ahh, that makes sense. So because the length of the tank is 10 feet you multiply that by the width of the parabola to get the area. Thanks for your help!
 February 15th, 2012, 06:50 PM #4 Global Moderator     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 428 Math Focus: Calculus/ODEs Re: Work to pump water out of a tank Yes, you got it! Glad to help and welcome to the forum!

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# pumping water out of tank cubic problem

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