The problem involves a spherical cell with:

• Volume: $V = \frac{4}{3} \pi r^3$
• Surface Area: $S = 4 \pi r^2$

You are asked to:

1. Express $V$ as a function of $S$.
2. Determine what happens to $r$ if $S$ is quadrupled.

Step 1: Expressing $V$ as a Function of $S$

From the surface area formula:

$S = 4 \pi r^2$

Solve for $r^2$:

$r^2 = \frac{S}{4 \pi}$

Take the square root to find $r$:

$r = \sqrt{\frac{S}{4 \pi}}$

Substitute this expression for $r$ into the volume formula $V = \frac{4}{3} \pi r^3$:

$V = \frac{4}{3} \pi \left( \sqrt{\frac{S}{4 \pi}} \right)^3$

Simplify:

$V = \frac{4}{3} \pi \cdot \left( \frac{S}{4 \pi} \right)^{3/2}$

Factor out powers of $\pi$ and constants:

$V = \frac{4}{3} \cdot \frac{1}{(4 \pi)^{3/2}} \cdot S^{3/2}$

Simplified further:

$V = \frac{1}{6 \sqrt{\pi}} S^{3/2}$

Step 2: What Happens to $r$ When $S$ is Quadrupled?

If $S$ becomes $4S$, substitute $4S$ into the formula for $r$:

$r = \sqrt{\frac{S}{4 \pi}}$

For $S = 4S$:

$r = \sqrt{\frac{4S}{4 \pi}} = \sqrt{4} \cdot \sqrt{\frac{S}{4 \pi}} = 2 \cdot \sqrt{\frac{S}{4 \pi}}$

So, $r$ doubles when $S$ is quadrupled.

Final Answers:

1. $V = \frac{1}{6 \sqrt{\pi}} S^{3/2}$
2. If $S$ is quadrupled, $r$ doubles.

Let me know if you'd like any clarifications or additional details!

Related Questions:

1. How does the volume $V$ change when $S$ is doubled instead of quadrupled?
2. Derive the relationship between surface area $S$ and radius $r$.
3. Explore the physical meaning of $S^{3/2}$ in the context of spheres.
4. Can you generalize this relationship for non-spherical shapes?
5. What happens to $V$ if the radius $r$ triples?

Tip:

When solving for one variable in terms of another, always simplify expressions step-by-step to avoid mistakes with exponents and roots.