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Math Help - Volume between curves

  1. #1
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    Volume between curves

    Hello, here is my problem.

    Find the volume of the solid generated by revolving the region bounded by the curves y = -x^2+6 and y = 3x+2, about the line y=8. Now I'm only suppose to setup but I just want to make sure I'm doing this correctly. I can use any method and in this case I picked the washer method.

    So, here is what I have R(x) = 8 - (3x+2), and r(x) = 8-(-x^2+6) and my integral is

    Pi * integral[(6-3x)^2-(3+x^2)^2]dx from -4, to 1. Is this the correct way to go about doing this?
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  2. #2
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    What do you get for the final result?

    Did you deliberately do anything incorrectly?

    It really is a little silly to write (-x^2+6) when (6-x^2) is so much less likely to be misinterpreted. It's okay on this one, since you are about to subtract it and change both signs.

    Why did you bother to define R(x) and r(x) if you were not planning to use them?

    Do we inherently know what R(x) and r(x) mean? Not every notation is generally accepted. It MAY be used by only one author. In this case, it may be pretty clear to most that you mean an inner radius (r(x)) and an outer radius (R(x)), but you probbly shoudl define that explicitly.

    You r(x) seems to have changed by the time you put it in the integral expression. 8 - 6 = 2

    Why did you pick washers? It is a bit more challenging as shells. Did you know that?
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  3. #3
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    Yes, I apologize, I do have 2 different books and they are using the same notation so I just assumed. R(x) is my outer radius and r(x) is my inner radius, in which case I have

    R(x) = 8 - (3x+2) = 6-3x
    r(x) = 8 - (-x^2+6) = 2+x^2



    Pi * integral[R(x)^2 - r(x)^2]dx from -4 to 1. Which goes to



    Pi*integral[(6-3x)^2 - (2+x^2)^2]dx from -4 to 1.

    I'm told to not evaluate this integral and I'm simply asking if I set it up correctly.
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  4. #4
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    Right. You said that and I missed it. Good work. I like it. Use shells and do it again.
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