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Math Help - Solid of revolution

  1. #1
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    Solid of revolution

    We have a raindrop that is obtained by revolving the profile curve:

    f(x)=sqrt(x)*(x-C)^2 about the x-axis for 0<=x<=C, with C a positive constant.

    a) Sketch the profile curve and solid of revolution.
    b) For what value of C will the volume be 1? What are the dimensions, length and diameter, of such a raindrop?

    How can I sketch the curve when the C varies, this is a little confusing to me. I don't know how to produce this sketch.

    I think to do the volume I just need to set integral of the function pi* (f(x))^2) dx from 0 to C equal to 1 and solve for C. I get (30/pi)^1/6 doing this. I am not sure if this is correct though, and don't know how to find the dimensions. Thanks.
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  2. #2
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    For part (a), just try using a couple values for C and you should be able to identify the general behavior of the graph. What happens when x=C? (Remember that we only care about the part of the graph from 0 to C, so ignore the rest of the graph.)

    Here are a couple example graphs:

    f(x)=sqrt(x)*(x-10)^2 - Wolfram|Alpha

    f(x)=sqrt(x)*(x-5)^2 - Wolfram|Alpha

    For part (b), your answer is correct.
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  3. #3
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    Ok, gotcha, thanks! So I would just draw the graph kind of like in the sketches, but just replace the number with a C on the x-axis since it pretty much seems to look the same behavior-wise.

    Also, I have the volume, but how do I get the dimensions (length and diameter)?
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  4. #4
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    Well, the length you can infer from part (a). Notice how the raindrop shape always extends from x=0 to x=C. It should be easy to see that the length of the raindrop is equal to C, right?

    As for the diameter, I'm assuming they mean at the widest point of the raindrop. You need to use the derivative of f(x) to find the point where f(x) reaches its maximum value (between 0 and C). Once you know the x-value, you can just calculate f(x) to get the radius.
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  5. #5
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    Hello twittytwitter
    Quote Originally Posted by twittytwitter View Post
    We have a raindrop that is obtained by revolving the profile curve:

    f(x)=sqrt(x)*(x-C)^2 about the x-axis for 0<=x<=C, with C a positive constant.

    a) Sketch the profile curve and solid of revolution.
    b) For what value of C will the volume be 1? What are the dimensions, length and diameter, of such a raindrop?

    How can I sketch the curve when the C varies, this is a little confusing to me. I don't know how to produce this sketch.

    I think to do the volume I just need to set integral of the function pi* (f(x))^2) dx from 0 to C equal to 1 and solve for C. I get (30/pi)^1/6 doing this. I am not sure if this is correct though, and don't know how to find the dimensions. Thanks.
    I agree with your answer C = \left(\frac{30}{\pi}\right)^{\frac16} to give a volume of 1.

    So that's the length of the raindrop.

    Then differentiate to get the values of x for which f'(x) = 0. This comes out as x = \frac15C and x = C.

    The diameter is then twice the value of f(x) when x = \tfrac15C, which is \frac{32\sqrt5c^{\frac52}}{125}, if my arithmetic is correct.

    Grandad
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