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Sphere:
A sphere is
3-dimensional (solid) figure
in which all of it's points are the same distance from one
fixed point in space,
which is called it's center.
Don't get confused! A sphere
is not a circle. A circle is 2 -
dimensional, or flat. A circle is also known as a
plane figure.
A line segment which joins any point on the surface of the
sphere with it's center is called it's
radius.
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The Volume formula:
Unlike many other formulas for finding the volume of a
3-dimensional, or solid figure, we need only one dimension to
calculate the volume of a sphere....it's radius!
The formula for the volume of a sphere in which it's radius is
"r" is:
V = 4/3 Π
r3 |
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For all of the model
problems below let Π = 3.14
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Model
Problems |
Let's start with
an easy example.
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An official handball
has a diameter of 4.8 centimeters. Find it's volume.
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First
you must recognize that the dimension you have been given is the
diameter of the sphere.
However, as was discussed earlier in this lesson, the dimension
needed is the radius.
That's easy to fix!
Simply cut the diameter in half (or divide by 2).
If d = 4.8, then
r = (4.8)/2 = 2.4!
Then, plug the value of "r" into the formula:
V = 4/3 Π r3
V = 4/3 (3.14)(2.4)3
V = 4/3(3.14)(13.824)
V = 4/3(43.40736)
The easiest way to multiply a number like
this
by a fraction is simply multiply the number (43.40736)
by 4 (the numerator), and then divide
the result by 3 (the denominator).
First multiply by 4
4(43.40736)
173.62944
Then divide by 3
(173.62944)/3
57.87648
V = 57.87648 cubic centimeters!
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Got the idea?
Let's try another..
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A red rubber ball
needs air before the class plays kickball. After it
has been properly inflated for the game, it's radius has
increased from 8 inches to 11 inches in length.
What was the change in the ball's volume, to the the nearest
cubic inch?
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To answer this problem we
will need to find the volume twice. Once for the ball in it's
deflated form (radius = 8"),
then a second time after it has been filled
with air (radius = 11 inches).
Then to find the change we will subtract the 2 volumes.
Deflated ball:
V = 4/3Π r3
V = 4/3 (3.14)(8)3
V = 4/3(3.14)512)
V = 4/3(1607.68)
V = 2143.573333
V = 2144 cubic inches.
Inflated ball:
V = 4/3(3.14)(11)3
V = 4/3(3.14)(1331)
V = 4/3(4179.34)
V = 5572.453333
V = 5572 cubic inches
The change in volume:
5572 - 2144
3428 cubic inches!
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Now let's look
at one more..
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A ball having a
diameter of 10 cm is being shipped in a box shaped like a
cube whose height, width, and length are also exactly 10
cm.. How many cubic centimeters of packing material
are needed to hold the ball firmly in place inside the box?
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Can you
picture the problem? Can you see a ball inside a cube, so
that when the top is closed the ball fits snugly in the box, but
there are these areas of open space where the ball and box don't
touch. That's the volume we will be solving for.
First let's find the volume of the cube.
A cube is a rectangular prism in which all of the dimensions are
the same. In this case l = 10, w = 10 and h = 10.
V = (l)(w)(h)
V = (10)(10)(10)
V = 1000 cubic cm.
Now for the volume of the
ball:
d = 10, so r = 5
V = 4/3 Π r3
V = 4/3(3.14)(5)3
V = 4/3(3.14)(125)
V = 4/3(392.5)
V = 523.333... cubic cm
The
amount of "empty space" is the difference
between these 2 volumes.
Empty
space = 1000 - 523 (to nearest cubic cm)
Empty space = 477 cubic cm!
Did
that answer surprise you??
If you think about it, the ball only fills up a little bit more
than half of the volume of the cube!
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