## blackstreet23 one year ago 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.

1. blackstreet23
2. blackstreet23

I found a picture of it in that website

3. phi

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4. 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 $1000 \int_{-2}^2 \sqrt{4-y^2}\ (3-y) \ dy$

5. phi

to integrate that, write it as two separate integrals the first one requires some kind of trig substitution the second is more straightforward.

6. blackstreet23

I do not understand how you found the width :s

7. 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?

8. 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"

9. blackstreet23

Thanks a lot !! Sorry I didn't answer before I have been busy.