Why is the surface area of a sphere the derivative of it's volume?

[jddoxtator]

Learning calculus I came across an interesting observation.
The formula for the surface area of a sphere is the derivative of the formula for it's volume.
Logically, this makes sense as the surface of a sphere is by definition tangential to the sphere at all points.
However, as far as I know this far into calculus, the derivative is the instantaneous slope of a curve in a two dimensional sense.
The sphere is a three dimensional object, so we also have the the z axis.
Does that mean the surface area of a sphere is a function of a partial derivative?
Is this something I will understand in multi-variable calculus?

 

[logistic_guy]

Learning calculus I came across an interesting observation.
The formula for the surface area of a sphere is the derivative of the formula for it's volume.
Logically, this makes sense as the surface of a sphere is by definition tangential to the sphere at all points.
However, as far as I know this far into calculus, the derivative is the instantaneous slope of a curve in a two dimensional sense.
The sphere is a three dimensional object, so we also have the the z axis.
Does that mean the surface area of a sphere is a function of a partial derivative?
Is this something I will understand in multi-variable calculus?

Great question. There are many different answers to your questions.

Mine is:

Think of the sphere like if it is a cube or a cylinder, or any basic shape where the volume of that shape is known to be the area of its base \(\displaystyle \times\) its height.

\(\displaystyle \text{Volume of a Sphere} = \text{area} \times \text{height}\)

Now let us work with symbols instead of names.

\(\displaystyle V = A \times h\)

Now let us take a very tiny volume of that sphere.

\(\displaystyle dV = A \times dh\)

We know that \(\displaystyle A\) is the surface area of the sphere, then \(\displaystyle dh\) must be a very tiny height. In other words, a very tiny height of a sphere is just a very tiny radius, so \(\displaystyle dh \rightarrow dr\).

Then,

\(\displaystyle dV = A \times dh = 4\pi r^2 \ dr\)

Take the integral of both sides.

\(\displaystyle \int dV = \int 4\pi r^2 \ dr\)

\(\displaystyle V = \frac{4}{3}\pi r^3 \ \ \ \ \ \) (We didn't write the constant of integration here because the volume and the radius will start from zero. In other words, \(\displaystyle c = 0\).)

The First Fundamental Theorem of Calculus essentially says that:

If the integral of \(\displaystyle f\) gives you \(\displaystyle F\), then the derivative of \(\displaystyle F\) gives you back \(\displaystyle f\).

Or

If the integral of \(\displaystyle A\) gives you \(\displaystyle V\), then the derivative of \(\displaystyle V\) gives you back \(\displaystyle A\).

 

[blamocur]

Sphere's volume depends on a single variable, i.e. radius, i.e., no need for partial derivatives.
A hand-waving explanation: look at the change from R to R + dR. It adds a thin layer to the surface of the sphere, so the volume of this thin layer is S x dR.

 

[logistic_guy]

The sphere is a three dimensional object, so we also have the the z axis.
Does that mean the surface area of a sphere is a function of a partial derivative?

Yes, in a way. If you want to go deeper with the surface area of a sphere taking into account the \(\displaystyle z\)-axis, write down the formula of a sphere in three dimensions:

\(\displaystyle x^2 + y^2 + z^2 = r^2\)

Parametrize it in spherical coordinate as:

\(\displaystyle x = r\sin\phi\cos\theta\)
\(\displaystyle y = r\sin\phi\sin\theta\)
\(\displaystyle z = r\cos\phi\)

\(\displaystyle \bold{r}(\theta,\phi) = (x(\theta,\phi), y(\theta,\phi), z(\theta,\phi)) = (r\sin\phi\cos\theta,r\sin\phi\sin\theta,r\cos\phi)\)

The surface area \(\displaystyle S\) of a sphere is:

\(\displaystyle S = \int dS = \int\int |\bold{n}| \ d\theta \ d\phi\)

where \(\displaystyle \bold{n} = \bold{r}_\theta \times \bold{r}_\phi \ \ \ \) (You will take the partial derivatives and the cross product.)

You will get:

\(\displaystyle \int\int |\bold{n}| \ d\theta \ d\phi = \int\int r^2\sin\phi \ d\theta \ d\phi = \int_{0}^{2\pi}\int_{0}^{\pi} r^2\sin\phi \ d\phi \ d\theta = 4\pi r^2\)

Or you can work with this second approach:

\(\displaystyle x^2 + y^2 + z^2 = r^2\)

\(\displaystyle z = \pm \sqrt{r^2 - x^2 - y^2} \)

If we take the top part of the sphere then we have this formula \(\displaystyle z = \sqrt{r^2 - x^2 - y^2} \).

Let \(\displaystyle f(x,y) = z = \sqrt{r^2 - x^2 - y^2}\).

\(\displaystyle S = \int dS = \int\int |\bold{n}| \ dx \ dy = \int\int \sqrt{(f_x)^2 + (f_y)^2 + 1} \ dx \ dy = \int_{-r}^{r}\int_{-\sqrt{r^2 - y^2}}^{\sqrt{r^2 - y^2}}\frac{r}{\sqrt{r^2 - x^2 - y^2}} \ dx \ dy\)

Remember this is only the half surface area of the sphere, so the complete surface area \(\displaystyle S_c\) is:

\(\displaystyle S_c = 2S = 2\int_{-r}^{r}\int_{-\sqrt{r^2 - y^2}}^{\sqrt{r^2 - y^2}}\frac{r}{\sqrt{r^2 - x^2 - y^2}} \ dx \ dy = 4\pi r^2\)

While this integral cannot be solved with normal techniques, the method used here is very useful to solve many surface areas.

Does that mean the surface area of a sphere is a function of a partial derivative?

The main point is yes, partial derivatives are involved to calculate the surface area of a sphere. (It is not really a sphere is a function of a partial derivative, but it acts like that in some sense.)

 

[jddoxtator]

Interesting, I have not learned integrals yet, so I don't quite fully understand that part, but the derivatives formulas are familiar.
I get that all the dimensions of the sphere are expressible in the single variable of radius, so in that way it is a special case which can be found with simpler techniques.
In mathematics so far, I have only ever seen the surface area of a sphere expressed as a curved 2D plane. Is it ever considered with depth for the purposes of mapping? I assume it must be. Say the elevation map of a dome for instance.