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  1. Subscriber AThousandYoung
    Do ya think?
    18 Dec '08 19:28 / 3 edits
    I'm working on deriving the volume of a sphere by taking circular slices of it with volume V = A*dx. However I'm having trouble, because volumes have three dimensions, and the example methods I've seen do not use three variables:

    http://www.math.hmc.edu/calculus/tutorials/volume/

    It seems to me that A = pi*r^2, and r^2 = y^2 + z^2. So, I should be integrating from -R to R

    pi*(y^2 + z^2) dx

    Where R is the radius of the sphere and r is the radius of the circular slice.

    I'm getting stuck here. Help!

    EDIT - The next step seems to be to state that r = y, which is true where z = 0, but this bothers me...why can this step be assumed?
  2. Standard member PBE6
    Bananarama
    18 Dec '08 20:26
    Originally posted by AThousandYoung
    I'm working on deriving the volume of a sphere by taking circular slices of it with volume V = A*dx. However I'm having trouble, because volumes have three dimensions, and the example methods I've seen do not use three variables:

    http://www.math.hmc.edu/calculus/tutorials/volume/

    It seems to me that A = pi*r^2, and r^2 = y^2 + z^2. So, I shoul ...[text shortened]... e that r = y, which is true where z = 0, but this bothers me...why can this step be assumed?
    Well, if you're adding up slices of a sphere in the x-direction you don't need to worry about any differences in the y- and z-directions, because it's symmetrical. You can simplify things by just looking at a circle on the x-y plane.

    The radius of each slice is simply the distance from the x-axis (y=0) to the y-coordinate of the surface of the sphere (y=SQRT(R^2 - x^2)). Therefore the volume of each slice is:

    dV = pi*r^2 dx = pi*(R^2 - x^2) dx

    If you integrate this expression from -R to R, you should end up with the formula for a sphere (4/3)pi*R^3.
  3. Standard member Agerg
    The 'edit'or
    18 Dec '08 20:30 / 2 edits
    Originally posted by AThousandYoung
    I'm working on deriving the volume of a sphere by taking circular slices of it with volume V = A*dx. However I'm having trouble, because volumes have three dimensions, and the example methods I've seen do not use three variables:

    http://www.math.hmc.edu/calculus/tutorials/volume/

    It seems to me that A = pi*r^2, and r^2 = y^2 + z^2. So, I shoul e that r = y, which is true where z = 0, but this bothers me...why can this step be assumed?
    See...you're summing discs of infinetisemal width dx, area pi*y^2...now the value of y is dependent upon x, (the value of z is irrelevent btw since we are dealing wth circles )

    So can you find the relationship between y and x and plug this in for y^2?

    Finally your limits are fine


    *edit PBE6 got here whilst I typed this
  4. Subscriber AThousandYoung
    Do ya think?
    18 Dec '08 20:45 / 1 edit
    Originally posted by PBE6
    Well, if you're adding up slices of a sphere in the x-direction you don't need to worry about any differences in the y- and z-directions, because it's symmetrical. You can simplify things by just looking at a circle on the x-y plane.

    The radius of each slice is simply the distance from the x-axis (y=0) to the y-coordinate of the surface of the sphere (y=SQR ...[text shortened]... te this expression from -R to R, you should end up with the formula for a sphere (4/3)pi*R^3.
    If you look at a circle in the x-y plane then you should integrate with respect to z! The volume of the slices will be pi*r^2, with r a function of x and y, and your infinitisimal value will be dz. Right?
  5. Standard member Agerg
    The 'edit'or
    18 Dec '08 20:53 / 2 edits
    Originally posted by AThousandYoung
    If you look at a circle in the x-y plane then you should integrate with respect to z! The volume of the slices will be pi*r^2, with r a function of x and y, and your infinitisimal value will be dz. Right?
    nope..it will be dx, forget about z...it's value takes whatever value y takes since we are dealing with [u]circles[/u]. you wish to evaluate the integral from -R to R of pi*y^2. (or if you are really really bothered about what happened to that pesky z, your area is pi*y*z where z = y)

    It would be handy if you could express y in terms of x, again...what is this relationship?
  6. Subscriber AThousandYoung
    Do ya think?
    18 Dec '08 21:04 / 1 edit
    Originally posted by Agerg
    nope..it will be dx, forget about z...it's value takes whatever value y takes since we are dealing with circles. you wish to evaluate the integral from -R to R of pi*y^2. (or if you are really really bothered about that pesky z, your area is pi*y*z where z = y)

    It would be handy if you could express y in terms of x, again...what is this relationship?
    I think I got it. The x^2 + y^2 is actually not from the equation of the circle slice!

    You make a triangle with the radius of the sphere R, making sure to keep the radius you're going to use in the xy plane. This gives you a triangle with R being the hypotenuse and two legs - x and y. You use the Pythagorean Theorem to get the R^2 = x^2 + y^2, instead of the equation of the circle slice, which is r^2 = y^2 + z^2! r, the radius of the circle slice, is the same length as this y, as can be seen if r is oriented in the y direction.

    Rearranged, we have

    y = (R^2 - x^2)^(1/2)

    Which is simply a function of x, and then we integrate from -R to R

    pi*r^2 dx
    pi*y^2 dx
    pi*(R^2 - x^2) dx
  7. Standard member Agerg
    The 'edit'or
    18 Dec '08 21:08
    Originally posted by AThousandYoung
    I think I got it. The x^2 + y^2 is actually not from the equation of the circle slice!

    You make a triangle with the radius of the sphere R, making sure to keep the radius you're going to use in the xy plane. This gives you a triangle with R being the hypotenuse and two legs - x and y. You use the Pythagorean Theorem to get the R^2 = x^2 + y^2, i ...[text shortened]... circle slice, is the same length as this y, as can be seen if r is oriented in the y direction.
    rock on!
  8. Standard member sasquatch672
    Don't Like It Leave
    03 Jan '09 23:50
    Originally posted by AThousandYoung
    If you look at a circle in the x-y plane then you should integrate with respect to z! The volume of the slices will be pi*r^2, with r a function of x and y, and your infinitisimal value will be dz. Right?
    If you do that you get a cylinder, not a sphere. Ooh, the math of circles on a Saturday night...
  9. 04 Jan '09 19:29
    Originally posted by sasquatch672
    If you do that you get a cylinder, not a sphere. Ooh, the math of circles on a Saturday night...
    No, the radius of the circles varies with respect to z. You will get a circle if you write down the correct relation between z and the radius of the circle in the xy plane.
  10. 06 Jan '09 11:13 / 1 edit
    Originally posted by KazetNagorra
    No, the radius of the circles varies with respect to z. You will get a circle if you write down the correct relation between z and the radius of the circle in the xy plane.
    Oops, I meant sphere, not circle.