Translating word problems to 3d geometry formulae

A 3D word problem is really asking you to pick the right formula.
First identify the shape: sphere, cube, cylinder, or cone.
Then decide whether the question is about volume or surface area.

Surface area is the size of the outside skin.
Choose it whenever something covers the outside — painting a ball, wrapping a box, or coating a surface.

Volume is the amount of space inside.
Choose it whenever something fills the object — water in a tank, drink in a cup, or the capacity of a container.

Then use the matching formula.

Sphere: $V = \frac{4}{3}\pi r^3$ and $SA = 4\pi r^2$.
Cylinder: $V = \pi r^2 h$.
Cone: $V = \frac{1}{3}\pi r^2 h$.
Cube: $V = s^3$ and $SA = 6s^2$.

For example, to find how much paint covers a spherical ball, you need the outside, so use the surface area of a sphere, $4\pi r^2$.
To find how much drink fills a cylindrical tank, you need the inside, so use the volume, $\pi r^2 h$.

A spherical balloon has radius $5$, and its entire outside will be coated. How much surface must be coated?
Coating the outside means surface area, so use $SA = 4\pi r^2$.
With $r = 5$: $4\pi (5)^2 = 100\pi$.

A cylindrical tank has radius $3$ and height $10$. How much water does it hold when full?
Holding water means volume, so use $V = \pi r^2 h$.
With $r = 3$ and $h = 10$: $\pi (3)^2 (10) = 90\pi$.

A cone-shaped cup has radius $6$ and height $15$. What is its capacity?
Capacity means volume, so use the cone volume $V = \frac{1}{3}\pi r^2 h$.
With $r = 6$ and $h = 15$: $\frac{1}{3}\pi (36)(15) = 180\pi$.