Volume of a Sphere - Calculus

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?

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?

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?

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.

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?

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?

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.

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?

Originally posted by sasquatch672
If you do that you get a cylinder, not a sphere. Ooh, the math of circles on a Saturday night...

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.