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.

 

[maxter]

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?

imagine expanding the radius by an infinitesimal dr, then the amount of volume you've added dV is (up to first order) Surface Area * dr, therefor dV/dr = Surface Area

 

[Agent Smith]

If you flip your statement around and try to understand integrals instead of derivatives, you'll figure it out. Hint: infinitesimals, stacking them for a sphere you get what looks like an onion (layers).

Knowing only derivatives and little integration. the only intuition-friendly scenario is the slope of a curve, which doesn't quite map onto areas and volumes. One thing that might help is the rule for differentiation (taking derivatives) is you lower the exponent on the variable by 1 and so cubes (volume) become squares (area) becomes a variable with exponent 1 (a line).