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Math Help - Volume of solid/washer

  1. #1
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    Volume of solid/washer

    y=2/x
    y=0
     x=1
    x=3
    Rotate about y = -1

    Outside radius = (-1-2/x)^2

    Inside radius = 1

    \int_1^3 ((-1-2/x)^2-1)dx = 22.1831495912

    Is this correct?

    Thanks in advance
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  2. #2
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    Quote Originally Posted by silencecloak View Post
    y=2/x
    y=0
     x=1
    x=3
    Rotate about y = -1

    Outside radius = (-1-2/x)^2

    Inside radius = 1

    \int_1^3 ((-1-2/x)^2-1)dx = 22.1831495912

    Is this correct?

    Thanks in advance
    Outside radius = (2/x + 1) although the square will fix that. Just need a pie

    \pi \int_1^3 ((2/x+1)^2-1)dx = 22.1831495912

    although judging from the answer you put it in.
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  3. #3
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    So I did this problem correct? Or that number is just the correct number for the integral i set up?
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  4. #4
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    Quote Originally Posted by silencecloak View Post
    So I did this problem correct? Or that number is just the correct number for the integral i set up?
    Yes, just make sure you have the outer radius

    <br />
r_o = f(x) - L (for f(x)>L) where  y = f(x) the curve you're rotating and y = L the line you're rotating about. Switch f and L if f is under y= L
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