MTH 174
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I've been working on some practice test questions and was hoping you might clarify one or two things if you can. What is a billiard ball model exactly? The question in the box below also has me confused: what does it mean by 'lower side'? And if it is laying on its side, with the altitude being horizontal,how would it be pumped out? There is also a question that says to prove TRAP(n) = LEFT(n) + RIGHT(n) but isn't defined to be TRAP(n)=(LEFT(n)+RIGHT(n))/2?
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Use Riemann Sums to obtain the integral required to solve the following, and evaluate the integral: A tank is in the shape of a right circular cylinder with base diameter 3 meters and altitude 13 meters, and is lying on its side. The tank is originally full of a fluid with weight density 14200 Newtons / m^3. How much work will be required to pump all the fluid to a height of 10 meters above the lower side of the container?
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I know you may not be able to answer all of these questions because it is a test after all, but whatever you can do to help steer me in the right direction would be greatly appreciated.
You haven't seen the billiard ball model (it requires a program that doesn't consistently run in recent versions of Windows), so if it appears on the test the question will be omitted.
You are correct that TRAP = (LEFT + RIGHT) / 2
With respect the work problem:
If you slice the tank with a horizontal plane at height y, you will get a rectangle. Find the dimensions and area of this rectangle.
Now consider a thin horizontal slice of the tank, thickness `dy. Let y_i be the height of any point in this slice. Write down the expression for the area of the rectangular slice at height y_i (just like your expression for the area of the slice at height y).
Note that we have a 'thin slice', which has volume, an a rectangular slice, which has an area but no volume.
The cross-sectional area of the thin slice varies a little as you move up and down, but if `dy is small it won't vary much. So the volume of the thin horizontal slice is very nearl equal to any of its cross-sectional areas, multiplied by `dy. Using the expression for the area of the rectangular slice at y_i, you get the expression for the approximate volume of the thin slice. Multiplying this by the weight density you get an expression for the weight of the thin slice.
Now, how far do you need to raise this thin slice to get it up to that 10-meter height? A simple expression in terms of y_i gives you that distance.
Multiply the weight by the distance it needs to be raise, and you get the work.
You get an expression for the work required to raise the thin slice:
`dW = sqrt( (3 m)^2 - (y_i - 1.5 m)^2 ) * 13 m * 14200 N / m^3 * (10 - y_i) * `dy
Summing all such contributions and taking the limit as `dy -> 0 we get the integral
int( 14 200 N / m^2 * sqrt( 9 m^2 - (y - 1.5m)^2) * 13 m * (10 - y) dy, y from the bottom of the tank to the top).
I'll be glad an answer further questions.