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Math Help - region between the curve

  1. #1
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    Post region between the curve

    The region between the curve (y=1/x 2) and the x-axis from x=1/2 to x=2 is revolved about the y-axis to generate a solid. Find the volume of the solid.

    I know the answer is pi ln16

    i tried drawing it out but i am confused witht the "revolved about the Y-AXIS" i don't see it. i see x-axis though...

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  2. #2
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    so you draw out the region, shade it, and then imagine it as being pasted onto a glass door, where the hinges of the door is the Y axis and spin the door.

    The shape ends up looking a little like the red piece on the right side.
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  3. #3
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    okay so i go to v= pi ∫ 1/x^4 dx

    so i need to do du/u... and if so du= 3x^3 so do i add a negative on the outside?
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  4. #4
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    yes because of power rule.

    eg.

    \frac{d}{dx}(\frac{1}{x^2})

    is:

    -2*\frac{1}{x^3}
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  5. #5
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    so i got v= pi [-3x^3 (lnx^4)] from .5 to 2 (right?)
    then do i include the .5 to 2 with the /3x^3?
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  6. #6
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    Yes, when you do your substitution you substitute it for all the x's in your equation thing.

    just to be clear your starting function is:

    y=\frac{1}{x^2} right?
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  7. #7
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    Quote Originally Posted by genlovesmusic09 View Post
    okay so i go to v= pi ∫ 1/x^4 dx

    so i need to do du/u... and if so du= 3x^3 so do i add a negative on the outside?
    No, that is the wrong integral. That would be the volume if the region were rotated around the x axis. Since you are rotating around the y-axis, a small segment above the x-axis, with height y= 1/x^2 and thickness dx, rotated around the y-axis, will form a ring with 2\pi x, the circumference of the circle, height 1/x^2 and thickness dx and so volume (2\pi x)(1/x^2)(dx). The entire volume is the "sum" of those which becomes the integral
    2\pi\int_{x=1}^2 \frac{dx}{x}.
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