- Topics At a Glance
- The Third Dimension
- Drawings
- Nets
- Surface Area
- Lateral Area of Prisms and Cylinders
- Surface Area of Prisms and Cylinders
- Surface Area of Pyramids
- Surface Area of Cones
- Surface Area of Spheres
**Volume**- Volume of Prisms
- Volume of Cylinders
- Volume of Pyramids
- Volume of Cones
- Volume of Spheres
**Volume and Density**- Congruent and Similar Solids
- The 3D Coordinate System

Volume is a gateway to a lot of other concepts like **density**, which is a fancy word for the ratio between mass and volume. If something has a high density, then it weighs a lot even if the volume is really small.

A tub of water and a tub of chocolate take up the same amount of space (volume), but the chocolate would weigh more than the water (mass) because of density. It would also be a lot tastier than water, but that has nothing to do with density.

We have a rectangular carton filled with strawberry Jell-O (after all, strawberry is the best flavor). If strawberry Jell-O weighs 5 pounds per cubic foot, how much does all the Jell-O weigh?

Here's the game plan: We'll find the volume first, multiply by the density, and find the mass that way. Ready? Break!

*V* = *Bh* = *l* × *w* × *h**V* = (0.7 ft)(0.8 ft)(1 ft)*V* = 0.56 ft^{3}

We have 0.56 ft^{3} of strawberry Jell-O in the carton. To find the mass, we need to multiply the volume by the density. If we do that, we can see that the cubic feet will cancel each other out and we'll be left with pounds.

Mass = *V* × *d*

Mass = (0.56 ft^{3})(5 lbs/ft^{3})

Mass = 2.8 lbs

Our carton of strawberry Jell-O weighs 2.8 lbs. Bon appétit.