Question about volume of a sphere

[Chenkel]

Hello everyone,

I was trying to find volume of a sphere by doing some calculus, the area of a circle is ${\pi}r^2$

So I thought I would calculate the volume of one hemisphere and then multiply by two, but I got a different result than the standard formula, the standard formula is $\frac 4 3 {\pi}r^3$

To calculate the volume of a hemisphere I calculate the following integral ${\int_0^r {\pi}r^2 dr}= \frac {{\pi}r^3} {3}$

Then I multiply that by 2 and I get $\frac {2}{3} {{\pi}r^3}$ which I imagine should be the volume of the sphere but it differs from the standard formula.

Where did I go wrong?

Any advice is appreciated!

Thank you in advance!

 

[BvU]

I calculate the following integral ${\int_0^r {\pi}r^2 dr}= \frac {{\pi}r^3} {3}$

How is this related to the volume ?

$\$

 

[Chenkel]

How is this related to the volume ?

$\$

It should be the volume of one of the hemispheres, I create a bunch of circular slices and multiply the area or each slice by $dr$ to get the volume of an infinitesimal slice.

 

[BvU]

and what is the relationship between $h$ and $r$ ?

 

[Hill]

You have calculated the volume of a cone with the radius changing linearly from $r$ to $0$.

 

[Chenkel]

You have calculated the volume of a cone with the radius changing linearly from $r$ to $0$.

That makes sense, thank you!

 

[Chenkel]

View attachment 341008
and what is the relationship between $h$ and $r$ ?

maybe $h^2 + r^2 = R^2$ where $R$ is a constant specifying the unchanging radius of the circle?

$2h{\frac {dh}{dr}} + 2r = 0$

$h{\frac {dh}{dr}} + r = 0$

${\frac {dh}{dr}}= -\frac r h$

$h {dh} = -r {dr}$

$\frac {h^2} 2 = -\frac {r^2} 2 + C$

When h equals 0 at the same time r=R so

$0 = -\frac {R^2} 2 + C$

$C = \frac {R^2} 2$

And that means the solution is

$\frac {h^2} 2 = -\frac {r^2} 2 + \frac {R^2} 2$

 

[dextercioby]

Which is back where you started from.

 

[Chenkel]

Which is back where you started from.

I see what you are saying I think, the equation at the bottom is the same as the one at the top.

 

[mathwonk]

it is not obvious, but if you study the figure in post #4, you may obtain archimedes' result, that the volume of the hemisphere, equals the difference between the volume of the circumscribing cylinder and the volume of the cone. (hint: it uses Pythagoras to prove the analogous fact about the slice areas.)

 

[PeroK]

To calculate the volume of a hemisphere I calculate the following integral ${\int_0^r {\pi}r^2 dr}= \frac {{\pi}r^3} {3}$

Another problem is using $r$ for the variable radius of the circles and for the fixed radius of the sphere. It really should be $\int_0^R \dots$.

 

[PeroK]

You have calculated the volume of a cone with the radius changing linearly from $r$ to $0$.

The volume of a cone is $\frac{\pi r^2 h}3$. That's the volume of a very specific cone whose height equals the radius of the base.

 

[SmartyPants]

You might find it easier use vertical slices (instead of horizontal slices) and integrate along the x-axis. You're already using the "solid of revolution formula $\int\limits_{\scriptsize a}^{\scriptsize b}{\pi\,(f\,(x))^{2}}{\;\mathrm{d}x}$...granted you initially posted it in the form $\int\limits_{\scriptsize 0}^{\scriptsize r}{\pi\,r^{2}}{\;\mathrm{d}r}$.

Of course you first have to solve the implicit equation of a circle for y in terms of x:

$x^{2}+y^{2}=r^{2}\rightarrow y=\pm \sqrt{r^{2}-x^{2}}$

Now all you have to do is revolve either the top half of the circle $y=\sqrt{r^{2}-x^{2}}$ or the bottom half $y=- \sqrt{r^{2}-x^{2}}$ about the x-axis in order to generate a sphere. You can see that the radii of the slices varies as you work your way from -r to r. That changing radius of the slices IS the function $y=\sqrt{r^{2}-x^{2}}$ (or $y=-\sqrt{r^{2}-x^{2}}$). Replace f(x) with either $\sqrt{r^{2}-x^{2}}$ or $-\sqrt{r^{2}-x^{2}}$ in the "solid of revolution" formula:

$\int\limits_{\scriptsize -r}^{\scriptsize r}{\pi\,(\sqrt{r^{2}-x^{2}})^{2}}{\;\mathrm{d}x}$ or $\int\limits_{\scriptsize -r}^{\scriptsize r}{\pi\,(-\sqrt{r^{2}-x^{2}})^{2}}{\;\mathrm{d}x}$.

You can also further simplify the integral by instead integrating from 0 to r and multiplying by 2:

$2\int\limits_{\scriptsize 0}^{\scriptsize r}{\pi\,(\sqrt{r^{2}-x^{2}})^{2}}{\;\mathrm{d}x}$ or $2\int\limits_{\scriptsize 0}^{\scriptsize r}{\pi\,(-\sqrt{r^{2}-x^{2}})^{2}}{\;\mathrm{d}x}$

Evaluate either integral and let us know what you come up with...

 

[Chenkel]

You might find it easier use vertical slices (instead of horizontal slices) and integrate along the x-axis. You're already using the "solid of revolution formula $\int\limits_{\scriptsize a}^{\scriptsize b}{\pi\,(f\,(x))^{2}}{\;\mathrm{d}x}$...granted you initially posted it in the form $\int\limits_{\scriptsize 0}^{\scriptsize r}{\pi\,r^{2}}{\;\mathrm{d}r}$.

Of course you first have to solve the implicit equation of a circle for y in terms of x:

$x^{2}+y^{2}=r^{2}\rightarrow y=\pm \sqrt{r^{2}-x^{2}}$

Now all you have to do is revolve either the top half of the circle $y=\sqrt{r^{2}-x^{2}}$ or the bottom half $y=- \sqrt{r^{2}-x^{2}}$ about the x-axis in order to generate a sphere. You can see that the radii of the slices varies as you work your way from -r to r. That changing radius of the slices IS the function $y=\sqrt{r^{2}-x^{2}}$ (or $y=-\sqrt{r^{2}-x^{2}}$). Replace f(x) with either $\sqrt{r^{2}-x^{2}}$ or $-\sqrt{r^{2}-x^{2}}$ in the "solid of revolution" formula:

$\int\limits_{\scriptsize -r}^{\scriptsize r}{\pi\,(\sqrt{r^{2}-x^{2}})^{2}}{\;\mathrm{d}x}$ or $\int\limits_{\scriptsize -r}^{\scriptsize r}{\pi\,(-\sqrt{r^{2}-x^{2}})^{2}}{\;\mathrm{d}x}$.

You can also further simplify the integral by instead integrating from 0 to r and multiplying by 2:

$2\int\limits_{\scriptsize 0}^{\scriptsize r}{\pi\,(\sqrt{r^{2}-x^{2}})^{2}}{\;\mathrm{d}x}$ or $2\int\limits_{\scriptsize 0}^{\scriptsize r}{\pi\,(-\sqrt{r^{2}-x^{2}})^{2}}{\;\mathrm{d}x}$

Evaluate either integral and let us know what you come up with...

$2{\int_0^r}{{\pi}({r^2 - x^2})dx}=2{\pi}(r^3 - \frac {r^3}{3})=2{\pi}{r^3}(\frac 3 3 - \frac 1 3)=\frac 4 3 \pi r^3$

That gets me the right answer, thank you for the help!

 

[PeroK]

$2{\int_0^r}{{\pi}({r^2 - x^2})dx}=2{\pi}(r^3 - \frac {r^3}{3})=2{\pi}{r^3}(\frac 3 3 - \frac 1 3)=\frac 4 3 \pi r^3$

That gets me the right answer, thank you for the help!

You could also learn about integration in spherical coordinates. The volume element is:
$$dV = r^2\sin \theta d\phi d\theta dr$$ The volume of a sphere can be calculated by integrating this over the sphere:
$$V = \int_0^R \int_0^{\pi}\int_0^{2\pi}r^2\sin \theta \ \ d\phi d\theta dr$$$$= 4\pi\int_0^R r^2 dr = 4\pi \frac {R^3}3$$There's more about spherical and cylindrical coordinates here:

https://web.ma.utexas.edu/users/m40...d our volume element is,makes with the z-axis.

 

[SmartyPants]

It truly is fascinating that there are so many different ways one can go about calculating the volume of a sphere. I can think of 3 methods right off the top of my head that use calculus: solid of revolution, integration over spherical coordinates (as PeroK demonstrated above), and multi-variable calculus via the use of a double or triple integral, depending on where you want to start.

 

[mathwonk]

note that h is also the radius of the circle cut by a horizontal plane from the inverted cone of height and radius R. Hence the formula h^2 + r^2 = R^2, shows the horizontal slice areas of the cone and hemisphere add up to that of the cylinder of height and radius R. Hence the same holds for their volumes. I.e. volume of hemisphere = πR^3 - (1/3)πR^3. This is archimedes' solution, without integrating.

Since a 4-ball is obtained by revolving the (solid) hemisphere, in 4 space, about its base plane, its volume is also the difference between the solids obtained by revolving the cylinder and cone, which can be calculated from knowing their centers of gravity, as archimedes did know.

Specifically, since the center of a cylinder is half way up, and that of an (inverted) cone is 3/4 way up, we get (2π)(R/2)(πR^3) - (2π)(3R/4)(πR^3/3) = π^2R^4 - (1/2)π^2R^4 = (1/2)π^2R^4, for volume of a 4-ball.