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Homework Help: Finding volumes by rotating around an axis of revolution

  1. Aug 25, 2010 #1
    1. The problem statement, all variables and given/known data

    k so here is the equation i need help with that will find me the volume of a sphere
    2*pi*y*sqrt(25-(y-1)^2) dy - 5*pi from 0 to 6

    the 5 pi is the volume of a cylinder


    2. Steps
    so first i subbed u=y-1
    took the 2 pi out of the integral
    that got me 2 integrals u*sqrt(25-(u)^2) du + sqrt(25-(u)^2) du
    the first integral = (2*pi*24^(3/2))/3

    the second i used a trig sub and my final answer is
    (25pi^2)/2 +
    -25*asin(-.2)+sqrt(24)



    3. Answer

    the answer is suppose to be 25pi^2 + 500pi/3

    i got 25pi^2 + (-75*asin(-.2)+3*sqrt(24)+2*(24)^(3/2) -15)/3
    which is approx
    25pi^2 + 499.89pi/3

    how do i get the exact answer
     
  2. jcsd
  3. Aug 25, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    You are trying to find the volume of a sphere of radius 5 with center at (0, 1, 0)? Projected onto a x, y plane, gives the circle [itex]x^2+ (y- 1)^2= 25[/itex] and taking only the right half-plane, [itex]x= \sqrt{25- (y-1)^2}[/itex]. That is the figure being rotated around the line y= 1.

    And, from the way you are doing this, I take it you are using "cylinders" (I would have been inclined to use disks). Each "cylinder" of radius y has length
    [tex]\sqrt{25- (y- 1)^2}- (-\sqrt{25- (y-1)^2})= 2\sqrt{25- (y- 1)^2}[/tex]

    and so surface area [itex]\pi r^2h= \pi (y- 1)^2(2\sqrt{25- (y-1)^2})[/itex]. With "thickness" dy, each cylinder will have volume [itex]2\pi (y- 1)^2\sqrt{25- (y-1)^2} dy[/itex] so the volume of the entire sphere will be the integral, as y goes from 1 to 1+ 5= 6,
    [tex]2\pi \int_1^6 (y- 1)^2\sqrt{25- (y- 1)^2}dy[/tex]

    A good first step, just as you say, would to be let u= y- 1 so that du= dy, when y= 1, u= 0, when y= 6, u= 5 and the integral becomes
    [tex]2\pi\int_0^5 u^2\sqrt{25- u^2}du[/tex]

    Yes, you can let [itex]u= 5sin(\theta)[/itex] so that [itex]du= 5 cos(\theta)d\theta[/itex] and [itex]\sqrt{25- u^2}= \sqrt{25- 25sin^2(\theta)}= 5\sqrt{cos^2(\theta)}= 5 cos(\theta)[/itex]. Further, when u= 0, [itex]5sin(\theta)= 0[/itex] so [itex]\theta= 0[/itex] and when [itex]u= 5[/itex], [itex]5 sin(\theta)= 5[/itex] so [itex]\theta= \pi/2[/itex].

    The integral becomes
    [tex]2\pi\int_0^{\pi/2}(25 sin^2(\theta))(5 cos(\theta))(5 cos(\theta))d\theta[/itex]

    I do not know why you are using "y" outside the square root rather than "y- 1" nor do I see why it is not squared. I had to guess at what you really intended since you did not say what the radius or center of the sphere were. I also do not know why you subtracted [itex]5\pi[/itex], the "volume of a cylinder" when you said you were finding the volume of a sphere.
     
  4. Aug 25, 2010 #3
    no no... sorry the circle is being revolved around the x-axis and y2= 1 which is why so the 5pi exists because i am subtracting the volume from y2
     
  5. Aug 25, 2010 #4
    oh and yes amd the question asked me to revolve the circle around the x-axis in terms of y, which is longer by the way.
     
  6. Aug 25, 2010 #5
    [tex]
    2\pi \int_0^6 y\sqrt{25- (y- 1)^2}dy + 5\pi
    [/tex]

    that's the integral I want solved and the answer i got is (25pi^2)/2 + (500*pi)/3 intstead of 25pi^2 + 499.89pi/3
     
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