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Math Help - area of a circle

  1. #1
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    Unhappy area of a circle

    I am having a hard time with this one.
    A horizontal cylinder has D = 2m and L = 4m. The drum is slowly filled with benzene (density =.879 g/cm^3). Derive a formula for the weight of the benzene as a function of the depth (h) of the liquid in centimeter. So basically it wants the wight of the benzine as a function of the height of the liquid as it rises in the drum. I though about using the circle formula in cartesian coordinates and the integrating but I am not sure what the limits would be.

    I have (x-1)^1 + y^2 = r^2. This is like the drum is on its side and it is being filled from left to right if you can image it on an xy diagram. The point of filling would be at the origin, cut the circle in half and only take the left hand side of the circle, so you have x= -sqrt(r^2-y^2)-1. Then just divide by 2 to just worry about just the top portion. So you basically have a curve increasing in the x and y direction in the first quadrant. Then multiply by two at the end to get the whole thing. I know that sound very confusing but if someone could at least give me an idea, that would be great.

    Thanks
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  2. #2
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    Here is a picture of what I mean. Hope this helps.
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  3. #3
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    any chance you could post the problem out of the book? im afraid your explanation is a little tricky for me
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  4. #4
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    This is exactly what the book says:
    A horizontal cylindrical drum is 2 m in diameter and 4 m long. the drum is slowly filled with benzene (density = .879 g/mol). Derive a formula for W, the weight in newtons of the benzene in the tank, as a function of h, the depth of the liquid in centimeters. They have this diagram shown:


    Don't worry about the density and the actual formula, I just need to know how to integrate to get the area of the section liquid as seen from the diagram as a function of h by integrating.
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  5. #5
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    ill be honest its been a long time since ive done a problem like this... although i remember the setup being similar to this....hope it helps

    use trig
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  6. #6
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    Is there any way to do this using calc?
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  7. #7
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    you need to use integration. the line makes the base of the two triangles represents dh
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  8. #8
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    I tried doing it this way. I used the area of a sector formula,
    A = theta*pi*r^2/360. To use this I cut the moon shaped area in half, so I have 90 degress for theta, and for example I used r = 8 in (.667 ft). I got .698 ft^2. Would this be okay?
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  9. #9
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    call the longest side of each triangle x. (the side that is not r and not 1-h)
    to find the area at each moment you would use 2x * dh. you use the sum of these smaller areas from 0 to your upper limit (2r if you want to fill the whole circle), aka the integral.

    at the very bottom, the angle of theta is 0, and it increases as you increase height. use the cosine to find the length of x, put this all in terms of h and solve.
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  10. #10
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    Can you please set up the integral for me? I am totally lost right now.
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  11. #11
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    Quote Originally Posted by nertil1 View Post
    I am having a hard time with this one.
    A horizontal cylinder has D = 2m and L = 4m. The drum is slowly filled with benzene (density =.879 g/cm^3). Derive a formula for the weight of the benzene as a function of the depth (h) of the liquid in centimeter. So basically it wants the wight of the benzine as a function of the height of the liquid as it rises in the drum. I though about using the circle formula in cartesian coordinates and the integrating but I am not sure what the limits would be.

    I have (x-1)^1 + y^2 = r^2. This is like the drum is on its side and it is being filled from left to right if you can image it on an xy diagram. The point of filling would be at the origin, cut the circle in half and only take the left hand side of the circle, so you have x= -sqrt(r^2-y^2)-1. Then just divide by 2 to just worry about just the top portion. So you basically have a curve increasing in the x and y direction in the first quadrant. Then multiply by two at the end to get the whole thing. I know that sound very confusing but if someone could at least give me an idea, that would be great.

    Thanks
    I don't have the time to give details now but this thread: http://www.mathhelpforum.com/math-he...cylinders.html

    contains the ideas you need to get a formula.
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  12. #12
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    OKay, I using h = 8 inches, r = 15 in, I got Area = 22.578 and the volume (using 60 inches for length) I got 18.81 ft^3 for the liquid. Can someone please check that.

    Thanks
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