To approximate the volume of the shell using differentials, we start by considering the formula for the volume of a sphere:

$V = \frac{4}{3}\pi r^3$

where $r$ is the radius of the sphere.

Given:

• The outer radius of the sphere $r = 100$ cm.
• The thickness of the shell $a = 0.5$ cm.

To approximate the change in volume ($dV$) due to the thickness of the shell, we use the differential formula:

$dV = \frac{dV}{dr} \cdot dr$

Calculating the derivative of the volume with respect to $r$:

$\frac{dV}{dr} = 4\pi r^2$

Now, we can approximate the change in volume ($dV$) as:

$dV \approx 4\pi r^2 \cdot dr$

Here, $dr = a = 0.5$ cm and $r = 100$ cm. Plugging in these values:

$dV \approx 4\pi (100)^2 \cdot 0.5$

$dV \approx 4\pi \cdot 10000 \cdot 0.5$

$dV \approx 20000\pi$

Approximating $\pi \approx 3.1416$, we get:

$dV \approx 20000 \cdot 3.1416 \approx 62832 \text{ cm}^3$

Thus, the approximate volume of the shell is $62832 \text{ cm}^3$.

Would you like further details or have any questions?

Related Questions:

1. How does using differentials simplify the approximation of volume changes?
2. What is the exact volume of the shell if we use the full volume formula instead of differentials?
3. How would the approximation change if the thickness $a$ were increased to 1 cm?
4. Can differentials be used to approximate the surface area of the sphere's shell? How?
5. What other real-life applications can utilize differentials for approximation?

Tip:

Using differentials is especially helpful for approximations when changes are small, as the error remains minimal.