A cylindrical tank lying on its side is filled with liquid weighing 50lb/ft^3. Find the work required to pump all the liquid to a level 1 ft above the top of the tank.The diameter of the tank is 4 feet.The depth of the tank is 10 feet. NOTE: Please remember the tank is lying on its SIDE, it's NOT upright. A hint my professor gave is that the answer should be a number multiplied by pi. I'm not really sure where my slice goes and how to set up my integral because it's not upright.

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A cylindrical tank lying on its side is filled with liquid weighing 50lb/ft^3. Find the work required to pump all the liquid to a level 1 ft above the top of the tank.The diameter of the tank is 4 feet.The depth of the tank is 10 feet. NOTE: Please remember the tank is lying on its SIDE, it's NOT upright. A hint my professor gave is that the answer should be a number multiplied by pi. I'm not really sure where my slice goes and how to set up my integral because it's not upright.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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http://www.chegg.com/homework-help/questions-and-answers/cylindrical-tank-lying-side-filled-liquid-weighing-50lb-ft-3-find-work-required-pump-liqui-q5973459
I found a picture of it in that website
  • phi
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  • phi
The work to lift x lbs up y ft is x*y ft-lbs Each thin "plate" is lifted from its "y" to y=3 and the distance is 3-y the weight of the place is density * volume or in this case 50 lbs/ft^3 * 2*x*10*dy ft^3 and it is lifted 3-y feet thus \[ \int_{-2}^2 50\cdot 2 \cdot x \cdot 10 \cdot (3-y) \ dy \] because we are integrating over y, replace x with the equivalent \[ x= \sqrt{4-y^2} \] we have \[ 1000 \int_{-2}^2 \sqrt{4-y^2}\ (3-y) \ dy \]
  • phi
to integrate that, write it as two separate integrals the first one requires some kind of trig substitution the second is more straightforward.
I do not understand how you found the width :s
  • phi
a cylinder has a circle as its base and (in this case) a height of 10 tip the cylinder on its side, and look at its base (a circle) I put the circle with radius 2 (i.e. diameter/2) at the origin the equation of a circle at the origin is x^2 + y^2 = r^2 follow?
  • phi
if we solve for x, we get \[ x = \sqrt{4-y^2} \] if we take the positive value for the square root, then for a given y , we have found the point on the circle on the right side of the origin, in other words, the distance from the y-axis to the circle on its right side. double that value to get the entire width of the "plate"
Thanks a lot !! Sorry I didn't answer before I have been busy.

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