Volume of a sphere (solid of revolution)

In summary, the formula for calculating the volume of a sphere is V = (4/3)πr^3, where r is the radius and π is the mathematical constant pi (approximately equal to 3.14). The volume of a sphere is directly proportional to the cube of its radius, meaning that if the radius is doubled, the volume will increase by a factor of 8. The unit of measurement for the volume of a sphere is cubic units, such as cubic meters (m^3) or cubic centimeters (cm^3). The volume of a sphere cannot be negative, as it is a physical quantity and can only have positive values. The volume of a sphere and its surface area are related through the radius,
  • #1
Galadirith
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EDIT: [so sorry guys i thought I had clicked h/w c/w calculus section, if any admin could move this thread there that would be really appreciated, sorry for posting in the wrong sub-forum]

Hi guys, I am having an issue with the volume of a sphere and calculating it. I am using the method where the solid, sphere in this case, is composed of many cylinders, in my case along the x-axis where the center of the sphere is at the origin.

So initially I attempted this by considering each cylinder was of volume:[itex]\pi y^2 \delta x [/itex]
then the formula of the circle in the xy plane of the sphere is [itex]x^2 + y^2 = r^2[/itex] which yields [itex]y^2 = {r^2 - x^2} [/itex] which means the volume of each cylinder becomes [itex]\pi (r^2 - x^2) \delta x[/itex]

I will limit my range to [0,r] as it is symmetrical about the yz plane. So integrating accros the interval [0,r] we get :

[tex]V = \pi \int_{0}^{r} (r^2 - x^2) \delta x [/tex]
[tex] = \pi ({r^2}x - \frac{x^3}{3})\bigg{|}_0^r [/tex]
[tex] = \pi [({r^3} - \frac{r^3}{3})-({r^3}(0) - \frac{0^3}{3})] [/tex]
[tex] = \pi \frac{2r^3}{3}[/tex]

then x2 to get the volume of a sphere to be [itex]{4\pi r^3}{/}{3}[/itex]

now that's all well and good, but I also wanted to see if another method would work, instead of using x as our independent use theta. Now I learned that if you wanted to calculate the surface are of a sphere you can use theta as it makes calculations easier. So using the idea that we can calculate the surface are of a sphere modeling it as composed of many hollow and cylinders with no base or head, each has height [itex]r\delta \theta[/itex]. So each cylindrical surface becomes :

[tex]i. \ A = 2\pi y r \delta \theta[/tex]
[tex]ii. \ y=rsin\theta[/tex]
[tex]iii. \ A = 2r\pi rsin\theta \delta \theta[/tex]

So that's fine, that idea can be taken as can successfully yield the surface area of a sphere successfully. Now I figured if that method can be employed for a surface, it must be adaptable to find the volume of a sphere. so this time I decided to define each solid cylinder as:

[tex]iv. \ V = \pi y^2 r \delta \theta[/tex]
[tex]v. \ V = r\pi (r^2sin^2\theta) \delta \theta [/tex]

so again I approached this as before, this time integrating across the range [0, [itex]\pi / 2[/itex]] where I again will only find the hemisphere for +ve x and theta is the angle from the x axis. So my integration went like this :

[tex]V = \pi r^3\int_{0}^{\frac{\pi}{2}} (sin^2\theta) \delta \theta [/tex]
[tex] = \pi r^3\int_{0}^{\frac{\pi}{2}} \frac{1}{2}(1 - cos2\theta) \delta \theta [/tex]
[tex] = \frac{\pi r^3}{2}(\theta - \frac{1}{2}sin2\theta)\bigg{|}_{0}^{\frac{\pi}{2}}[/tex]
[tex] = \frac{\pi r^3}{2}(\frac{\pi}{2}) = \frac{\pi^2 r^3}{4} [/tex]

then oviously x2, but as is evident it is not the right answer, I really can't see what I have done wrong, I pretty sure none of my actually mathematics is wrong, which means the only part that could be wrong is the [itex]r\delta \theta[/itex] bit and teacher suggested that for some reason it doesn't apply to a solid but does to a shell, but couldn't give a valid reason as to why. I am inclined to agree with him, but again I can't see why. In various texts I have looked though it says things along the lines of for the shell it is tempting to use [itex]\delta x[/itex] but that [itex]r\delta \theta[/itex] is a "better approximation", therefore surely it to is a better approximation for a solid of revolution to.

Can anyone explain why this is, what is the reason the integration can't be done this way, or am have I simply done my maths wrong, thanks guys :D
 
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  • #2


Thank you for sharing your method for calculating the volume of a sphere. Your approach using cylinders is a valid method and it is clear that you have a good understanding of the underlying mathematics. However, I believe the issue lies in your second method using theta as the independent variable.

While it is true that using theta can make calculations easier for finding the surface area of a sphere, it is not applicable for finding the volume. This is because the volume of a sphere is a three-dimensional measurement, while theta only represents an angle in a two-dimensional plane. In other words, theta does not account for the depth or height of the cylinder, which is crucial in finding the volume.

To illustrate this, let's consider a simpler example. Say we have a cylinder with a radius of 2 and a height of 4. Using your second method, we would calculate the volume as follows:

V = 2\pi (2sin\theta)(4\delta \theta) = 8\pi \delta \theta

However, if we use the traditional method of using x as the independent variable, we would calculate the volume as:

V = \pi (2^2)(4\delta x) = 16\pi \delta x

Notice how the results are different, even though both methods use the same formula for the surface area of a cylinder. This is because theta does not account for the depth of the cylinder.

In conclusion, while your approach using theta may work for finding the surface area of a sphere, it is not applicable for finding the volume. To find the volume, we must use a three-dimensional measurement, such as x or y, as the independent variable. I hope this helps clarify the issue for you. Keep up the good work in your mathematical explorations!
 

Related to Volume of a sphere (solid of revolution)

What is the formula for calculating the volume of a sphere?

The formula for calculating the volume of a sphere is V = (4/3)πr^3, where r is the radius of the sphere and π is the mathematical constant pi (approximately equal to 3.14).

How is the volume of a sphere related to its radius?

The volume of a sphere is directly proportional to the cube of its radius. This means that if you double the radius, the volume will increase by a factor of 8.

What is the unit of measurement for the volume of a sphere?

The unit of measurement for the volume of a sphere is cubic units, such as cubic meters (m^3) or cubic centimeters (cm^3).

Can the volume of a sphere be negative?

No, the volume of a sphere cannot be negative. It is a physical quantity and therefore can only have positive values.

How is the volume of a sphere related to the surface area?

The volume of a sphere and its surface area are related through the radius. As the radius increases, the volume increases at a faster rate than the surface area.

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