Question 178·Hard·Advanced Math

Test the allowed radius $r=4$. Type $t=4$ in Desmos, using $t$ for this test radius; Desmos shows $4$. Choosing a nonzero radius matters because some different formulas agree when $r=0$.

Let $R$ be the new radius. A sphere’s surface area is $4\pi$ times its radius squared. The coating adds $36\pi$ to the old area, so:

Add the increase to the area, not to the radius.

Type $4\pi x_1^2\sim4\pi t^2+36\pi\{x_1>0\}$. Here $x_1$ is the new radius, $\sim$ asks Desmos to find its value, and the restriction keeps it positive, as a radius must be. Desmos reports $x_1=5$ under PARAMETERS.

The sphere’s volume is $\frac{4}{3}\pi R^3$. Type $\frac{4}{3}x_1^3$ to find the number multiplying $\pi$. Desmos shows about $166.667$; its fraction button shows $\frac{500}{3}$. So the test volume is $\frac{500\pi}{3}$ cubic centimeters.

Each choice has a factor of $\pi$, so type its number multiplying $\pi$, with $t$ in place of $r$. In order, Desmos shows about $457.333$, $166.667$, $62.496$, and $20{,}833.333$. Only the formula with $(r^2+9)^{3/2}$ matches $\frac{500}{3}$. A rule for every radius must work at $4$, so the other three cannot define the new sphere’s volume. The answer, in cubic centimeters, is $f(r)=\frac{4}{3}\pi(r^2+9)^{3/2}$. Choice B.

Instead of testing one radius, let $R$ be the new radius and keep the given $r$. The surface area of a sphere is $4\pi$ times its radius squared. New area equals old area plus the increase, so:

Divide both sides by $4\pi$ to leave $R^2$ alone:

The increase is $9$ in the squared-radius term because $36\pi\div4\pi=9$.

Take the square root to undo the square:

Use the nonnegative square root because a radius is a length. The $9$ was added to $r^2$, not to $r$.

A sphere’s volume is $\frac{4}{3}\pi$ times its radius cubed. For the new sphere, the radius is $R$, so:

This uses volume, not surface area, because $f$ gives cubic centimeters.

Substitute $R=\sqrt{r^2+9}$, keeping the whole square root inside the cube:

A square root is a power of $\frac12$, so cubing it gives a power of $\frac32$:

That expression gives the new sphere’s volume in cubic centimeters for any radius $r$. Choice B.