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Math Help - another volume integral

  1. #1
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    another volume integral

    Hi everyone,

    Could someone please help me with this? I'm trying to find the volume under the curve: y=x, sqare root x; about x=2

    The graphs cross at (0,0) and at (1,1)

    I then would take the integral from 0 to 1 pie(sq. x squared)-pie(x squared) dx, which is 1/6pie. Is that correct???????

    Thank you very much
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  2. #2
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    Quote Originally Posted by chocolatelover View Post
    Hi everyone,

    Could someone please help me with this? I'm trying to find the volume under the curve: y=x, sqare root x; about x=2

    The graphs cross at (0,0) and at (1,1)

    I then would take the integral from 0 to 1 pie(sq. x squared)-pie(x squared) dx, which is 1/6pie. Is that correct???????

    Thank you very much
    Please note: \pi is spelled "pi."

    Your question makes no sense. Are you asking for the volume generated by rotating the area between y = x and y = \sqrt{x} rotated about the line x = 2? Please be more specific.

    -Dan
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