We are working on surface area and volume formulae for various solids.  Today we looked at spheres and pyramids.

To calculate the volume of a sphere, we used displacement.  We submerged a tennis ball in a displacement tank.  The water displaced overflowed into a juice concentrate container that has been cut so that the height and diameter are equal.  The juice concentrate container was selected as it has the same diameter as the tennis ball.  We can substitute 2r in the place of h in the formula for the volume of a cylinder.

Since the displaced water (the volume of the sphere) is 2/3 the volume of the cylinder, through a bit of calculating we can derive the formula for the volume of a sphere.

We also looked at the area of a sphere.  We used an orange to help us.

The diameter of the orange is close to 8cm.

We then peeled the orange and put the peels into a rectangle.  We know how to calculate the area of a rectangle.

The next step is to make a relationship between the area calculated, and the diameter squared.  We have to compare square units to square units.

The factor we found was always around 3.  This value should be pi.  Since the sphere is not exact, and neither is the rectangle, we don’t expect our pi approximation to be exact either.

We then used some algebra and exponents to derive the formula for the surface of the sphere.

Our pyramid exploration was open ended.  The restrictions were that they needed a square base, and the height needed to be exactly 8 cm.

In order to make the pyramid actually 8cm tall we need to calculate how tall each triangle face should be.  We use the pythagorean theorem to do this.