Deriving formulae for spheres

[Amaton]

One can derive the formula for the area enclosed by a circle using the elementary method with integration.

 

I thought about applying the same idea to spheres, except this time I assumed the final formulae to be true, and then try to get the differentials separated to see how it works out conceptually. Now that I've gotten the differentials, I'm wondering if there's any intuitive visualization behind these identities.

 

First, surface area.

 

[math]\mbox{A}_{\mbox{s}} = 4 \pi r^2[/math]

 

[math]\mbox{A}_{\mbox{s}} = 8 \pi\,\frac{1}{2}r^2[/math]

 

[math]\dfrac{\mbox{dA}_{\mbox{s}}}{\mbox{dr}}= 8 \pi r[/math]

 

[math]\mbox{dA}_{\mbox{s}} = 8 \pi\,r \, \mbox{dr}[/math]

 

Now, it doesn't make much sense for the radius to be multiplied by its change length. One could instead consider [math]C=2\pi r[/math] in regards to the great circle of the sphere. Where [math]\mbox{C}_{\mbox{p}}[/math] denotes the circumference of the great circle, this gives:

 

[math]\mbox{dA}_{\mbox{s}} = 4\mbox{C}_{\mbox{p}}\,\mbox{dr}[/math]

 

Or in other words, a change in surface area is equal to 4 times the circumference of the great circle multiplied by the change in radius.

 

 

As seen here, this is the same as 4 times the area between the circumferences of the two spheres. I'm still wondering if there's any intuitive sense behind this. (likewise with volume, but that's later) Thoughts?

[mathematic]

For surface area I believe you are on the wrong track. The differential you need is r²dΩ, where Ω is solid angle.

[Amaton]

For surface area I believe you are on the wrong track. The differential you need is r²dΩ, where Ω is solid angle.

 

There is indeed that, and it's likely more insightful, but I'm just looking at this particular method.

 

I derived [math]\mbox{dA}=4\mbox{C}_{\mbox{p}}\mbox{dr}[/math], which (hopefully) is true. The question now is how one can make intuitive sense of it, preferably by some visualization.

Edited June 15, 2013 by Amaton

[daniton]

Integrate the circumference with boundary -R to R then you can find the surface area just by adding the circumference of all the circles that made the sphere.This is what I see behind your work.

Actually you trying to see sth just by visualizing the relationship between the area and the circumference by inserting one equation on the other. I don't think that is going anywhere.

For the volume just integrate the area of circle with the same boundary.

remark R isn't the radius of sphere but the radius of the circles.

Edited June 16, 2013 by daniton

[Daedalus]

Now, it doesn't make much sense for the radius to be multiplied by its change length.

 

That's not necessarily true as it depends on the coordinate system you are working with. For instance, when dealing with double integrals, we normally work with rectangular coordinates where [math]dA = dx\,dy[/math]. However, when dealing with polar coordinates we find the transformation from rectangular to polar involves [math]dA = r\,dr\,d\theta[/math]. This gives us the following relationship:

 

Double Integrals over Polar Rectangular Regions

 

Let [math]f[/math] be continuous in the [math]xy[/math]-plane [math]R=\{\,(r,\theta): 0\le a\le r\le b,\,\,\alpha\le\theta\le\beta\}[/math], where [math]\beta-\alpha\le2\pi[/math]. Then

 

[math]\iint\limits_R f(r,\theta)\,dA=\int_{\alpha}^{\beta}\int_{a}^{b} f(r,\theta)\,r\,dr\,d\theta[/math]

 

Of course, we can integrate a function in any coordinate system by using a change of variables.

 

The Jacobian determinant is used when making a change of variables when evaluating a multiple integral of a function over a region within its domain.

 

This gives us a generalized method, which can be easily extended to triple integrals, that makes it easier to work in different coordinate systems such as spherical coordinates.

 

Change of Variables for Double Integrals

 

Let [math]T:\,x=g(u,v),\,y=h(u,v)[/math] be a transformation that maps a closed bounded region [math]S[/math] in the [math]uv[/math]-plane onto a region [math]R[/math] in the [math]xy[/math]-plane. Assume that [math]T[/math] is one-to-one on the interior of [math]S[/math] and that [math]g[/math] and [math]h[/math] have continuous first partial derivatives there. If [math]f[/math] is continuous on [math]R[/math], then

 

[math]\iint\limits_R f(x,y)\,dA=\iint\limits_S f(g(u,v),\,h(u,v))\,\left|J(u,v)\right|\,dA[/math]

 

Jacobian Determinant of a Transformation of Two Variables

 

[math]J(u,v)=\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{vmatrix}=\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}[/math]

 

If we use the equation for a circle, we can confirm that the Jacobian of the polar to rectangular transformation yields [math]\left|J(u,v)\right|dA=r\,dr\,d\theta[/math]:

 

[math]T:\,\,\,\,x=g(r,\theta)=r\,\text{cos}\,\theta,\,\,\,y=h(r,\theta)=r\,\text{sin}\,\theta[/math]

 

[math]J(u,v)=\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix}\text{cos}\,\theta&-r\,\text{sin}\,\theta\\\text{sin}\,\theta&r\,\text{cos}\,\theta\end{vmatrix}=r(\text{cos}^2\,\theta+\text{sin}^2\,\theta)=r[/math]

 

Thus giving us [math]dA=r\,dr\,d\theta[/math]. Now if we let [math]f(x,y)=1[/math], then we can derive the area of a circle using our new found knowledge by setting [math]\alpha=0,\,\beta=2\pi,\,a=0,\,\text{and}\,b=r[/math]:

 

[math]\int_{0}^{2\pi}\int_{0}^{r} r\,dr\,d\theta=\int_{0}^{2\pi}\frac{r^2}{2}\,\,d\theta=\pi\,r^2[/math]

 

The above example translates directly into finding the area of polar regions

 

Area of Polar Regions

 

The area of the region [math]R=\{(r,\theta):0\le g(\theta)\le r\le h(\theta),\,\,\alpha\le \theta\le \beta\}[/math], where [math]\beta-\alpha \le 2\pi[/math], is

 

[math]A=\iint\limits_R dA=\int_{\alpha}^{\beta}\int_{g(\theta)}^{h(\theta)}\,r\,dr\,d\theta[/math]

 

The reason why I introduced this into the discussion is because, when dealing with spheres, it is much easier to work in spherical coordinates where

 

[math]x=\rho\,\text{sin}\,\varphi\,\text{cos}\,\theta[/math]

[math]y=\rho\,\text{sin}\,\varphi\,\text{sin}\,\theta[/math]

[math]z=\rho\,\text{cos}\,\varphi[/math]

 

[math]\rho[/math] is the radius where [math]0\le\rho<\infty[/math]

[math]\varphi[/math] is the colatitude where [math]0\le\varphi\le\pi[/math]

[math]\theta[/math] is the polar angle where [math]0\le\theta\le 2\pi[/math]

 

Note: Physicist may reverse the role of [math]\varphi[/math] and [math]\theta[/math] - "Calculus Early Transcendentals - Briggs & Cochran" (pg 924)

 

Also because this line of thinking might give you some new insights on deriving formulas for spheres. Now let's compute the Jacobian for the equation of our sphere so that we can derive the equation for it's volume.

 

[math]J(u,v,w)=\frac{\partial(x,y,z)}{\partial(u,v,w)}=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}&\frac{\partial x}{\partial w}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}&\frac{\partial y}{\partial w}\\\frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}&\frac{\partial z}{\partial w}\end{vmatrix}=\begin{vmatrix}\text{sin}\,\varphi\,\text{cos}\,\theta&\rho\,\text{cos}\,\varphi\,\text{cos}\,\theta&-\rho\,\text{sin}\,\varphi\,\text{sin}\,\theta\\\text{sin}\,\varphi\,\text{sin}\,\theta&\rho\,\text{cos}\,\varphi\,\text{sin}\,\theta&\rho\,\text{sin}\,\varphi\,\text{cos}\,\theta\\\text{cos}\,\varphi&-\rho\,\text{sin}\,\varphi&0\end{vmatrix}=\rho^2\,\text{sin}\,\varphi[/math]

 

This give us the relationship that [math]dV[/math], which in rectangular coordinates is equal to [math]dx\,dy\,dz[/math], is equal to [math]dV=\rho^2\,\text{sin}\,\varphi\,d\rho\,d\varphi\,d\theta[/math] in spherical coordinates. This allows us to find volumes in spherical coordinates by using triple integrals.

 

Triple Integrals in Spherical Coordinates

 

Let [math]f[/math] be continuous over the region

 

[math]D=\{(\rho,\,\varphi,\,\theta):\,g(\varphi,\,\theta)\le\rho\le h(\varphi,\,\theta),\,\,a\le\varphi\le b,\,\,\alpha\le\theta\le\beta\}.[/math]

 

Then [math]f[/math] is integrable over [math]D[/math] and the triple integral of [math]f[/math] over [math]D[/math] in spherical coordinates is

 

[math]\iiint\limits_D f(\rho,\varphi,\theta)\,dV=\int_{\alpha}^{\beta}\int_{a}^{b}\int_{g(\varphi,\theta)}^{h(\varphi,\theta)}\,f(\rho,\varphi,\theta)\,\rho^2\,\text{sin}\,\varphi\,d\rho\,d\varphi\,d\theta[/math]

 

If we let [math]f(\rho,\varphi,\theta)=1[/math], then we can derive the formula for the volume of a sphere:

 

[math]\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{r}\rho^2\,\text{sin}\,\varphi\,d\rho\,d\varphi\,d\theta=\int_{0}^{2\pi}\int_{0}^{\pi}\frac{1}{3}r^3\,\text{sin}\,\varphi\,d\varphi\,d\theta=\int_{0}^{2\pi}\frac{2}{3}\,r^3\,d\theta=\frac{4}{3}\pi\,r^3[/math]

 

Furthermore, we can derive the surface area of the sphere by modifying the above integral such that we no longer integrate along the radius. We'll also use [math]r[/math] instead of [math]\rho[/math]:

 

[math]\int_{0}^{2\pi}\int_{0}^{\pi}r^2\,\text{sin}\,\varphi\,d\varphi\,d\theta=\int_{0}^{2\pi}2\,r^2\,d\theta=4\,\pi\,r^2[/math]

 

Although this post might not be along the same line as what you intended, I do hope that this information proves helpful in deriving formulas for spheres

Edited June 17, 2013 by Daedalus

[Amaton]

 

That's not necessarily true as it depends on the coordinate system you are working with. For instance, when dealing with double integrals, we normally work with rectangular coordinates where [math]dA = dx\,dy[/math]. However, when dealing with polar coordinates we find the transformation from rectangular to polar involves [math]dA = r\,dr\,d\theta[/math].

 

You bring up an interesting point. I hadn't considered non-Cartisian coordinates (though I've done some basic stuff with polar equations on conic sections).

 

Although this post might not be along the same line as what you intended, I do hope that this information proves helpful in deriving formulas for spheres

 

Thanks for that. I'm admittedly not very far into college level mathematics, as I only know the basics of multivariable calculus and DE's. I sort of glazed over your quote sources, but I was able to follow your final integrals and the derivation of the formulae from them.

 

I'm still toying with the idea of making a construction in rectangular coordinates so that one can relate the differentials in just the right way, like the simple area of a circle method. Nonetheless, this does seem more sensible using spherical coordinates.

[Daedalus]

You bring up an interesting point. I hadn't considered non-Cartisian coordinates (though I've done some basic stuff with polar equations on conic sections).

...

Nonetheless, this does seem more sensible using spherical coordinates.

 

Another thing to consider is the relationship between a sphere's surface area and volume.

 

This allows us to find volumes in spherical coordinates by using triple integrals.If we let [math]f(\rho,\varphi,\theta)=1[/math], then we can derive the formula for the volume of a sphere:

 

[math]\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{r}\rho^2\,\text{sin}\,\varphi\,d\rho\,d\varphi\,d\theta=\int_{0}^{2\pi}\int_{0}^{\pi}\frac{1}{3}r^3\,\text{sin}\,\varphi\,d\varphi\,d\theta=\int_{0}^{2\pi}\frac{2}{3}\,r^3\,d\theta=\frac{4}{3}\pi\,r^3[/math]

 

Furthermore, we can derive the surface area of the sphere by modifying the above integral such that we no longer integrate along the radius. We'll also use [math]r[/math] instead of [math]\rho[/math]:

 

[math]\int_{0}^{2\pi}\int_{0}^{\pi}r^2\,\text{sin}\,\varphi\,d\varphi\,d\theta=\int_{0}^{2\pi}2\,r^2\,d\theta=4\,\pi\,r^2[/math]

 

Since we can arrange the order of integration to however we see fit, we can choose to integrate along the radius last, which means that we are using surface area to determine the volume of a sphere.

 

Deriving the Surface Area of a Sphere

 

This integral takes care of all the spherical coordinate transformations

 

[math]A_s=\int_{0}^{2\pi}\int_{0}^{\pi}r^2\,\text{sin}\,\varphi\,d\varphi\,d\theta=\int_{0}^{2\pi}2\,r^2\,d\theta=4\,\pi\,r^2[/math]

 

Using the Surface Area to Derive the Volume

 

We are back to using rectangular coordinates to integrate along the radius of the sphere

 

[math]V_s=\int_{0}^{r}A_s\,dr= \int_{0}^{r}4\,\pi\,r^2\,dr=\frac{4}{3}\pi\,r^3[/math]

 

If we take the derivative of both sides we arrive at the following relationship:

 

[math]dV_s=A_s\, dr[/math]

Edited June 18, 2013 by Daedalus

[ecstaticdancer]

Amaton

 

You might find this useful

 

 

needs a little tightening up but has the intuitive visualization you are looking for...

Edited January 3, 2015 by ecstaticdancer