Calculating Volume of a Hollow Sphere w/ Differentials

In summary, the first method gives a slightly lower result than the second because it makes the assumption that the inner surface area is the same as the outer surface area.
  • #1
Dethrone
717
0
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?
 
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  • #2
Rido12 said:
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 $\frac 4 3 \pi (r+\Delta r)^3$.
 
  • #3
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
 

FAQ: Calculating Volume of a Hollow Sphere w/ Differentials

1. How do you calculate the volume of a hollow sphere?

The volume of a hollow sphere can be calculated by subtracting the volume of the inner sphere from the volume of the outer sphere. The formula for the volume of a sphere is 4/3 * pi * r^3, so for a hollow sphere, the formula would be (4/3 * pi * R^3) - (4/3 * pi * r^3), where R is the radius of the outer sphere and r is the radius of the inner sphere.

2. What is the purpose of using differentials in calculating the volume of a hollow sphere?

Differentials are used to approximate the volume of a hollow sphere when the radius of the sphere is very small. By using differentials, we can break the sphere into smaller, more manageable parts and calculate the volume of each part separately.

3. Is it necessary to use calculus to calculate the volume of a hollow sphere with differentials?

Yes, calculus is necessary to calculate the volume of a hollow sphere with differentials because it involves finding the limit of the sum of the volumes of the smaller parts as the number of parts approaches infinity.

4. Can the volume of a hollow sphere be calculated without knowing the radius of the inner sphere?

Yes, the volume of a hollow sphere can still be calculated without knowing the radius of the inner sphere. In this case, the formula would be (4/3 * pi * R^3) - (4/3 * pi * (R-d)^3), where R is the radius of the outer sphere and d is the thickness of the hollow part.

5. How can the volume of a hollow sphere with differentials be used in real life?

The volume of a hollow sphere with differentials has many practical applications, such as calculating the volume of hollow objects in engineering and manufacturing, estimating the amount of liquid or gas that can be stored in a spherical container, and determining the size and shape of biological cells and particles.

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