Geometry (B): READ: Volume of Cylinders Reading

READ: Volume of Cylinders Reading

Volume of a Right Cylinder

You have seen how to find the volume of any right prism.

$V = Bh$

where $B$ is the area of the prism’s base and $h$ is the height of the prism.

As you might guess, right prisms and right cylinders are very similar with respect to volume. In a sense, a cylinder is just a “prism with round bases.” One way to develop a formula for the volume of a cylinder is to compare it to a prism. Suppose you divided the prism above into slices that were $1\;\mathrm{unit}$ thick.

The volume of each individual slice would be given by the product of the area of the base and the height. Since the height for each slice is $1$ , the volume of a single slice would be:

$V \ \text{(single slice)} &= \text{area of base}\cdot \text{height}\\ &= B \cdot 1\\ &= B$

Now it follows that the volume of the entire prism is equal to the area of the base multiplied by the number of slices. If there are $h$ slices, then:

$V \ \text{(whole prism)} &= B \cdot \text{number of slices}\\ &= B h$

Of course, you already know this formula from prisms. But now you can use the same idea to obtain a formula for the volume of a cylinder.

Since the height of each unit slice of the cylinder is $1$ , each slice has a volume of $B \cdot (1)$ , or $B$ . Since the base has an area of $\pi r^2$ , each slice has a volume of $\pi r^2$ and:

$V \ \text{(whole cylinder)} &= B\cdot \text{number of slices}\\ &= B h\\ &= \pi r^2 h$

This leads to a postulate for the volume of any right cylinder.

Volume of a Right Cylinder

The volume of a right cylinder with radius $r$ and height $h$ can be expressed as:

$\text{Volume}=\pi r^2 h$

Example 5

Use the postulate to find the volume of the cylinder.

Write the formula from the postulate. Then substitute in the values and solve.

$V &= \pi r^2 h\\ &= (3.14)(6.5)(6.5) (14)\\ &= 1857.31 \ \text{cubic inches}$

Example 6

What is the radius of a cylinder with height $10 \;\mathrm{cm}$ and a volume of $250\pi$ ?

Write the formula. Solve for $r$ .

$V &= \pi r^2 h\\ 250\pi &= \pi r^2 (10)\\ 250\pi / 10\pi &= r^2\\ 25 &= r^2\\ 5 &= r$

Composite Solids

Suppose this pipe is made of metal. How can you find the volume of metal that the pipe is made of?

The basic process takes three steps.

Step 1: Find the volume of the entire cylinder as if it had no hole.

Step 2: Find the volume of the hole.

Step 3: Subtract the volume of the hole from the volume of the entire cylinder.

Here are the steps carried out. First, use the formula to find the volume of the entire cylinder. Note that since $d$ , the diameter of the pipe, is $6 \;\mathrm{cm}$ , the radius is half of the diameter, or $3 \;\mathrm{cm}$ .

$V &= \pi r^2 h\\ &= (3.14)(3)(3) (5)\\ &= 141.3 \ \text{cubic inches}$

Now find the volume of the inner empty “hole” in the pipe. Since the pipe is $1\;\mathrm{inch}$ thick, the diameter of the hole is $2\;\mathrm{inches}$ less than the diameter of the outer part of the pipe.

$d \ \text{(inner pipe)} &= d \ \text{(outer pipe)} - 2\\ &= 6 - 2\\ &= 4$

The radius of the hole is half of $4$ or $2$ .

$V &= \pi r^2 h\\ &= (3.14)(2)(2) (5)\\ &= 62.8 \ \text{cubic inches}$

Now subtract the hole from the entire cylinder.

$V \text{(pipe)} &= V \ \text{(cylinder)} - V \ \text{(hole)}\\ &= 141.3 - 62.8\\ &= 78.5 \ \text{cubic inches}$

Example 7

Find the solid volume of this cinder block. Its edges are $3 \;\mathrm{cm}$ thick all around. The two square holes are identical in size.

Find the volume of the entire solid block figure. Subtract the volume of the two holes.

To find the volume of the three-dimensional figure:

$V &= l \cdot w \cdot h\\ &= 21 \cdot 21 \cdot 26\\ &= 11,466 \ \text{cubic cm}$

Now find the length of the sides of the two holes. The width of the entire block is $21 \;\mathrm{cm}$ . This is equal to:

$\text{width of block} &= \text{3 edges} + \text{2 holes}\\ 21 &= 3 (3 \ \text{cm}) + 2n\\ 21 &= 9 + 2n\\ 12 &= 2n\\ 6 &= n$

So the sides of the square holes are $6 \;\mathrm{cm}$ by $6 \;\mathrm{cm}$ .

Now the volume of each square hole is:

$V &= l \cdot w \cdot h\\ &= 6 \cdot 6 \cdot 26\\ &= 936 \ \text{cubic cm}$

Finally, subtract the volume of the two holes from the volume of the entire brick.

$V \text{(block)} &= V \text{(solid)} - V \text{(holes)}\\ &= 11,466 - 2(936)\\ &= 9,594 \ \text{cubic cm}$

Last modified: Friday, 2 July 2010, 10:01 AM

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