Calculating Volume of a Hollow Sphere w/ Differentials

[Dethrone]

This is probably an elementary question, but I stumbled upon it while thinking about total differentials. One of their many applications is calculating the error in a volume, for example, given uncertainties in its dimensions.
I'm not in the mood to tackle a 3D problem, so let's revert to a 2D problem :D

Suppose you have a (hollow) sphere, made of some metal with radius 21 cm. The metal is 0.05 cm thick. Calculus the amount of metal.

Using differentials, we have $dV=4 \pi r^2 \,dr=4\pi (21)^2(0.05)$

I was thinking...couldn't this problem be also solved this way?
$$\int_r^{r+\Delta r}4\pi r^2 \,dr=\int_{21}^{21.05}4 \pi r^2 \,dr$$
The first method gives me $277.08 \text{ cm}^3$ where as the second gives me $277.75\text{ cm}^3$. Why are they different?

 

[I like Serena]

This is probably an elementary question, but I stumbled upon it while thinking about total differentials. One of their many applications is calculating the error in a volume, for example, given uncertainties in its dimensions.
I'm not in the mood to tackle a 3D problem, so let's revert to a 2D problem :D

Suppose you have a (hollow) sphere, made of some metal with radius 21 cm. The metal is 0.05 cm thick. Calculus the amount of metal.

Using differentials, we have $dV=4 \pi r^2 \,dr=4\pi (21)^2(0.05)$

I was thinking...couldn't this problem be also solved this way?
$$\int_r^{r+\Delta r}4\pi r^2 \,dr=\int_{21}^{21.05}4 \pi r^2 \,dr$$
The first method gives me $277.08 \text{ cm}^3$ where as the second gives me $277.75\text{ cm}^3$. Why are they different?

Hey Rido! (Wave)

In your first method, you have the surface of the inner sphere times its thickness $\Delta r$.
It makes the assumption that the inner surface area is the same as the outer surface area.
Since the outer surface area is actually a little bit bigger, the result is slightly too low.
It is only an approximation.

In your second method, you do effectively the same thing, but with an infinitesimal thickness $dr$ instead of the actual thickness $\Delta r$. Then this infinitesimal volume gets summed over the ever larger sphere areas.
This way, the volume is correct.

The alternative interpretation of the second method, is that you take the total volume of the outer sphere $\frac 43 \pi (r+\Delta r)^3$ and subtract the total volume of the inner sphere $\frac 43 \pi r^3$. With this method there is no approximation error.

Btw, in this case the approximation error of your first method is:
$$4\pi r (\Delta r)^2 + \frac 4 3 \pi (\Delta r)^3$$
This can be found by expanding $\frac 4 3 \pi (r+\Delta r)^3$.

 

[Dethrone]

Hey ILS! (Wave)

Thanks for the awesome response! I guess I was distracted by the notation rather than thinking about it geometrically. I also don't recall the book mentioning that those application questions involving differentials were approximations, so I was thinking they were exact results. Now that I think about it, my prof. last semester did talk very briefly about the error associated with it, but he had already lost most of the class by then. :p