Let the new radius be $R$. Write an equation saying: (new surface area) = (original surface area) + $36\pi$, using $4\pi R^2$ and $4\pi r^2$.

From your surface area equation, isolate $R^2$ and then $R$ in terms of $r$. Be careful: the change is in surface area, not directly in radius.

Once you have $R$ written in terms of $r$, substitute it into $\frac{4}{3}\pi R^3$ and simplify the exponent on the $(\dots)$ expression.

In Desmos, define the original radius as r and the new radius as R. Enter the relationship from the surface areas: 4piR^2 = 4pir^2 + 36pi. Then solve this equation for R (you can rewrite it manually as R = sqrt(r^2 + 9)).

Type the volume formula using your expression for R: f(r) = (4/3)pi*(sqrt(r^2 + 9))^3. Let Desmos simplify the exponent; it will show the function in a simplified power form.

Compare the simplified function Desmos shows for f(r) with the four answer choices. Look for the one where the inside is r^2 + 9 and the exponent matches what Desmos displays.

For a sphere with radius $x$:

• Surface area is $4\pi x^2$.
• Volume is $\frac{4}{3}\pi x^3$.

Let the new radius be $R$. The surface area increases by $36\pi$, so

Divide both sides by $4\pi$ to get

Thus $R = \sqrt{r^2 + 9}$ (take the positive root since a radius is positive).

Use the volume formula with the new radius:

Therefore, the function is