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Homework Help: Solid of revolution

  1. Aug 26, 2005 #1
    Could someone please explain how to do this question.

    Find the volume of the solid formed when the area between y=√x and y=x^4 is rotated about the line x=2.

    I know how to do this when it's rotated about the x and y axes, but I'm not sure how to do it with a different line.

    Thank you.
     
  2. jcsd
  3. Aug 26, 2005 #2

    AKG

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    Option 1: compute the area bounded by those curves, then find the x co-ordinate of its centroid and find the circumference of the circle whose radius is the distance from this centroid to the line x=2. Multiply the area by the circumference to get your volume (Pappus's Second Theorem).

    Option 2: Notice why this problem is the same as the problem of finding the volume of the solid formed when the area between y = √(x + 2) and y = (x + 2)4 are rotated about the x-axis, and use this to solve the problem as you already know how to.
     
  4. Aug 26, 2005 #3
    Do you mean about the y-axis? If the original problem says to rotate the solid about the line x=2.
     
  5. Aug 26, 2005 #4

    AKG

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    Yes, you're right, the y-axis (the line x = 0).
     
  6. Aug 28, 2005 #5
    Thanks for your help :smile:
     
  7. Aug 29, 2005 #6
    I can't seem to get the answer. I'm using the 2nd method. I'm meant to be rotating about the y-axis, aren't I?

    In that case,

    x^2=(y^2-2)^2------------- x^2=((y^1/4)-2)^2
    =y^4-4y^2+4 --------------= y-(4y1/4)+4

    Then you integrate these? The bounds are y=0, y=1. But I end up with completely the wrong answer.
     
    Last edited: Aug 29, 2005
  8. Aug 29, 2005 #7

    TD

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    By using [itex]y = \sqrt {x+2}[/itex] and [itex]y = (x+2)^4[/itex], you 'moved' the functions two places to the left. Your bounds have to move then as well, so you integrate from y = -2 to y = -1.

    The volume of a solid of revolution, about the y-axis, is the given by:

    [tex]\pi \int\limits_a^b {f\left( y \right)^2 dy} [/tex]
     
  9. Aug 29, 2005 #8
    Okay, the bounds move, but that results in an undefined answer, if my expansion above is correct.

    Shouldn't it be y= 2,1. You can't get a negative answer from those equations.
     
  10. Aug 29, 2005 #9

    TD

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    Your expansion isn't really clear to me...
     
  11. Aug 29, 2005 #10
    I'm not sure how I can make it clearer, but when you rearrange [itex]y = \sqrt {x+2}[/itex], you get [itex]x^2 = y^4-4y^2+4[/itex].

    When you rearrange [itex]y = (x+2)^4[/itex], you get [itex]x^2 = y-4y^(1/4)+4[/itex]

    For the latter, that is the 4th root for the 2nd term.
     
  12. Aug 29, 2005 #11

    TD

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    If you have 2 functions g(x) and h(x) and you want to find the solid of revolution of their difference (f(x) = h(x) - g(x)), the function f(x)² in the previous mentioned formula isn't h(x)²-g(x)² but (h(x) - g(x))² which isn't the same.
     
  13. Aug 29, 2005 #12
    okay, [itex]y = \sqrt {x+2}[/itex], is rearranged to [itex]x = y^2-2[/itex].

    [itex]y = (x+2)^4[/itex] is rearranged to [itex]x = y^(1/4)-2[/itex]

    Then, [itex][(y^2-2)-(y^(1/4)-2)]^2[/itex]

    Then you get [itex]y^4-y^2+y^(1/16)[/itex]

    But shouldn't you just integrate from the same bounds because your only changing the values of the x-axis.
     
  14. Aug 29, 2005 #13

    TD

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    I'm sorry, I think I confused you and my last answer wasn't entirely correct.

    Using f(x) = h(x) - g(x) doesn't change the area, but it does move it so the volume of the solid of revolution doesn't stay the same. Therefore, it would be indeed better to calculate the entire V of the solid of revolution of h(x) and then substract the one of g(x).
     
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