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Math Help - a volume integration question

  1. #1
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    a volume integration question

    The area bounded by the parabola y=2x-x^2, the y-axis and the line y=1 is rotated about the X-axis. Find the volume generated.

    Find the volume of the solid generated by the rotating the region bounded by the parabola y=1-x^2 and the lines x=1, y=1 about the X-axis.

    For both of these questions I used the volume of revolution formula and integrated each respective equation and calculated it between x=1 and x=0.
    For BOTH questions i got (8pi)/15 and for both questions, the answer says (7pi)/15. Could someone please point out where i have gone wrong? i realise that this area is a little unusual so maybe i am not working in the right angle. Please help me if you can. Thank You =)
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  2. #2
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    It's almost impossible to point out your mistake unless we see your work. If you type out the integral you got for each volume that would help us help you.
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  3. #3
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    okayy sorry ><""

    for the first question, my equation was 4x^2 - 4x^3 + x^4 and i integrated that to become 4x^3 - x^4 + (x^5) /5 and then when i calculated it from between x = 1 and x=0 , i got 8/15

    for the second question, my equation was 1-2x^2+x^4 and i integrated that to become x-(2x^3)/3+(x^5/5) and again after calculating, it became 8/15

    help anyone please?
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  4. #4
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    Quote Originally Posted by iiharthero View Post
    okayy sorry ><""

    for the first question, my equation was 4x^2 - 4x^3 + x^4 and i integrated that to become 4x^3 - x^4 + (x^5) /5 and then when i calculated it from between x = 1 and x=0 , i got 8/15

    for the second question, my equation was 1-2x^2+x^4 and i integrated that to become x-(2x^3)/3+(x^5/5) and again after calculating, it became 8/15

    help anyone please?
    For number 1 it seems that you didn't account for the y=1. If you integrate \pi \int_{0}^{1}\left(2x-x^2 \right)^2dx, you will get the volume of the parabola and the x-axis. You are taking a chunk out of this by adding in the boundary of y=1. So the actual volume would be \pi \int_{0}^{1}\left(2x-x^2 \right)^2-(1)^2dx
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