Painting a Point

Accumulation 7: An application (of paint)

Suppose you started with a point, the origin to be specific, and painted it. You put on layers and layers of paint until your point grows to a sphere with radius r. Let’s stop and admire your work part way through the job; at this point the radius is ${{x}_{i}}$ and $0\le {{x}_{i}}\le r$ .

How much paint will you need for the next layer?

Easy: you need an amount equal to the surface area of the sphere, $4\pi {{x}_{i}}^{2}$ , times the thickness of the paint. As everyone knows paint is thin, specifically $\Delta x$ thin. So we add an amount of paint to the sphere equal to $4\pi {{x}^{2}}\left( \Delta x \right)$ .

The volume of the final sphere must be the same as the total amount of paint. The total amount of paint must be the (Riemann) sum of all the layers or $\displaystyle \sum\limits_{i=1}^{n}{4\pi {{x}_{i}}^{2}\left( \Delta x \right)}$ . As usual $\Delta x$ is very thin, tending to zero as a matter of fact, so the amount of paint must be

$\displaystyle \underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{4\pi {{x}_{i}}^{2}\left( \Delta x \right)=\int_{0}^{r}{4\pi {{x}^{2}}}dx=\left. \tfrac{4}{3}\pi {{x}^{3}} \right|_{0}^{r}=\tfrac{4}{3}\pi {{r}^{3}}}$ .

A standard related rate problem is to show that the rate of change of the volume of a sphere is proportional to its surface area – the constant of proportionality is dr/dt. So it should not be a surprise that the integral of this rate of change of the surface area is the volume. The integral of a rate of change is the amount of change.

Interestingly this approach works other places as long as you properly define “radius:”

A circle centered at the origin with radius x and perimeter of $2\pi x$ , gains area at a rate equal to its perimeter times the “thickness of the edge” $\Delta x$ : $\displaystyle A=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{2\pi x\Delta x}=\int_{0}^{r}{2\pi x\,dx=\pi {{r}^{2}}}$

A square centered at the origin with “radius” x with sides whose length are s =2 x, gains area at a rate equal to its perimeter (8x) times the “thickness of the edge” $\Delta x$ : $\displaystyle A=\underset{n\to \infty }{\mathop{\text{lim}}}\,\sum\limits_{k=1}^{n}{8x\Delta x=\int_{0}^{x}{8x\,dx}}=4{{x}^{2}}={{(2x)}^{2}}={{s}^{2}}$

A cube centered at the origin with “radius” x and edges of length 2x, gains volume at a rate equal to its surface area, $6(4{{x}^{2}})$ , times the “thickness of the face” $\Delta x$ – think paint again: $V=\underset{n\to \infty }{\mathop{\text{lim}}}\,\sum\limits_{k=1}^{n}{\text{6(4}{{\text{x}}^{2}})\Delta x}=\int_{0}^{x}{24{{x}^{2}}dx}=8{{x}^{3}}={{(2x)}^{3}}={{s}^{3}}$

Teaching Calculus

The pleasure lies not in discovering the truth, but in searching for it.

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