McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
1. Expressions and Formulas
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Exercise 42 Page 9

Practice makes perfect
a

Let's draw two cylinders where neither the radius nor the height is the same. Note, to illustrate the cylinder as a 3D solid, you need to draw an oval instead of a circle for the top and bottom.

b

Even though the drawing is only a 2D representation of a 3D solid, we can measure the height along a vertical edge. We can also measure the radius from the midpoint to the end of the longest diameter of the oval.

The table below summarizes the measurements.

Radius Height
1in. 3in.
1.5in. 2in.

Let's use the formula to find the volume of the cylinder on the left.

V=Ï€ r^2h
V=Ï€* 1^2( 3)
V=Ï€* 3
V=3Ï€

We can find the volume of the other cylinder similarly. Let's include a column in the table for the volumes.

Radius Height Volume
1in. 3in. π* 1^2*3 = 3πin.^3
1.5in. 2in. π* 1.5^2*2 = 4.5πin.^3

c

The difference can be found by subtracting the volume of the smaller cylinder from the volume of the larger cylinder.

d

If we call the volume of the smaller cylinder V_S and the volume of the larger cylinder V_L, then we can use subtraction to express the difference.

V_L-V_S For the cylinders in Part A, the smaller volume is V_S=3Ï€, and the larger volume is V_L=4.5Ï€. Let's do the subtraction to find the difference.

V_L-V_S
4.5Ï€- 3Ï€
(4.5-3)Ï€
1.5Ï€

The difference of the volumes of the cylinders sketched in Part A is 1.5Ï€ cubic inches.