# Volume of a sphere (solid of revolution)

In summary, the conversation is about finding the volume of a sphere using different methods. The first method involves dividing the sphere into cylinders and integrating along the x-axis. The second method uses theta as the independent variable and involves calculating the surface area of the sphere. However, when trying to use this method to find the volume, there is an error and it is not the correct answer. The conversation concludes with a suggestion to include the Jacobian from the coordinate transformation into polar coordinates to get a more accurate result.
Hi guys, I did post this over on the h/w c/w forum last week, but I think actually it is out of the scope of that, plus its not actually h/w c/w, its my general wondering :D

So 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:$\pi y^2 \delta x$
then the formula of the circle in the xy plane of the sphere is $x^2 + y^2 = r^2$ which yields $y^2 = {r^2 - x^2}$ which means the volume of each cylinder becomes $\pi (r^2 - x^2) \delta x$

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 :

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

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

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 $r\delta \theta$. So each cylindrical surface becomes :

$$i. \ A = 2\pi y r \delta \theta$$
$$ii. \ y=rsin\theta$$
$$iii. \ A = 2r\pi rsin\theta \delta \theta$$

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:

$$iv. \ V = \pi y^2 r \delta \theta$$
$$v. \ V = r\pi (r^2sin^2\theta) \delta \theta$$

so again I approached this as before, this time integrating across the range [0, $\pi / 2$] 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 :

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

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 $r\delta \theta$ 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 $\delta x$ but that $r\delta \theta$ 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

If I let $\theta$ be the azimuthal component then I get for the height of the cylinder $r \cos\theta$ with $\theta$ running from 0 to pi.
Also you need to include the Jacobian from the coordinate transformation into polar coordinates, which I think (since r is fixed) is just sin(theta) - you should check this.
Then I find
$$\pi r^3 \int_{0}^{\pi} \cos^2\theta (\sin\theta \delta \theta) = \frac{2 \pi r^3}{3}$$
for half the volume.

I hope that gets you started.

## 1. What is the formula for finding the volume of a sphere?

The formula for finding the volume of a sphere is V = (4/3)πr3, where r is the radius of the sphere.

## 2. Can the volume of a sphere be calculated using calculus?

Yes, the volume of a sphere can be calculated using calculus by using the concept of a solid of revolution. This involves rotating a 2-dimensional shape around an axis to create a 3-dimensional object, in this case, a sphere.

## 3. What is the relationship between the volume of a sphere and its radius?

The volume of a sphere is directly proportional to the cube of its radius. This means that if the radius is doubled, the volume will increase by a factor of 8. Similarly, if the radius is halved, the volume will decrease by a factor of 1/8.

## 4. Can the volume of a sphere be negative?

No, the volume of a sphere cannot be negative. Since volume is a measure of space, it can only have positive values. In the formula V = (4/3)πr3, even if the radius is negative, the cube of the radius will result in a positive value, making the volume positive as well.

## 5. How is the volume of a hollow sphere calculated?

The volume of a hollow sphere can be calculated by subtracting the volume of the inner sphere from the volume of the outer sphere. The formula for the volume of a hollow sphere is V = (4/3)π(R3 - r3), where R is the outer radius and r is the inner radius.

• Calculus
Replies
4
Views
979
• Calculus
Replies
16
Views
1K
• Calculus
Replies
29
Views
1K
• Calculus
Replies
2
Views
786
• Calculus
Replies
8
Views
694
• Calculus
Replies
3
Views
1K
• Calculus
Replies
4
Views
2K
• Calculus
Replies
3
Views
2K
• Calculus
Replies
5
Views
523
• Calculus
Replies
5
Views
3K