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| 14 Theory slides |
| 10 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
VSphere=34πr3
The volume of a sphere with radius r is four-thirds the product of pi and the radius cubed.
Cavalieri's Principle will be used to show that the formula for the volume of a sphere holds. For this purpose, consider a hemisphere and a right cylinder with a cone removed from its interior, each with the same radius and height.
Draw a right triangle with height x, base y, and hypotenuse r. Here, x is the distance between the center of the base of the hemisphere and the center of the cross sectional circle, y is the radius of the cross sectional circle, and r is the radius of the hemisphere.
Substitute values
a=33⋅a
ca⋅b=ca⋅b
Subtract fractions
Last weekend, Tearrik installed a spherical water tank for his house.
If the tank has a radius of 2.15 feet, what is the maximum amount of water it can hold? Round the answer to two decimal places.
Find the volume of the tank.
The volume of the Earth is approximately 259875159532 cubic miles.
Assuming the Earth is spherical, what is the Earth's radius? Round the answer to one decimal place.
Solve the volume formula of the sphere for the radius and substitute the given volume.
LHS⋅3=RHS⋅3
LHS/4π=RHS/4π
Rearrange equation
3LHS=3RHS
3a3=a
V=259875159532
Use a calculator
Round to 1 decimal place(s)
Find the volume of the bucket and the volume of each balloon. Then, divide the volume of the bucket by the volume of a balloon.
To determine the minimum number of balloons needed to fill up the bucket, first solve for the volume of the bucket and the volume of each balloon. To do so, recall the formulas to find the volume of a cylinder and a sphere.
Volume of a Cylinder | Volume of a Sphere |
---|---|
VC=πr2h | VS=34πr3 |
Cross out common factors
Cancel out common factors
b/ca=ba⋅c
Multiply
Calculate quotient
Ramsha bought a standard pencil whose radius is 4 millimeters and the length, not including the eraser, is 180 millimeters. After a good sharpening, the tip turned into a 12 millimeter high cone.
Assuming the eraser is half of a sphere, what is the volume of the pencil? Round the answer to two decimal places.
The pencil is made of a cone, a cylinder, and a hemisphere, all with the same radius. Therefore, the volume of the pencil equals the sum of the volumes of each solid.
Volume of a Cone | Volume of a Cylinder | Volume of a Hemisphere |
---|---|---|
V1=31πr2h | V2=πr2h | V3=32πr3 |
Consider the fact that baseballs commonly have a radius of 1.43 inches. With this information, the volume of a baseball can be calculated, right? What about the surface area? How much leather is needed to make the lining of a baseball?
To answer these question, the following formula can be used.
The surface area of a sphere with radius r is four times pi multiplied by the radius squared.
To find the surface area of a sphere, this informal justification will use the areas and volumes of known figures. It includes approximating the ratio between a sphere's surface area and its volume. In considering a known figure, suppose that a sphere is filled with n congruent pyramids. The area of the base of each pyramid is B.
Substitute expressions
b/ca=ba⋅c
Cross out common factors
Cancel out common factors
Cancel out common factors
Simplify quotient
Substitute values
LHS⋅Vsphere=RHS⋅Vsphere
Vsphere=34πr3
Multiply fractions
ba=b/ra/r
ba=b/3a/3
Find the amount of leather needed to make the lining of a baseball that has a radius of 1.43 inches. Round the answer to one decimal place.
Use the formula to find the surface area of a sphere.
r=1.43
Calculate power
Use a calculator
Round to 1 decimal place(s)
Notice that the surface area of a sphere with radius r is four times the area of a circle with the same radius as the sphere. This relationship can be roughly seen in the covering of a baseball.
In the following applet, calculate either the volume or the surface area of the given sphere and round the answer to two decimal places.
In talking about spheres, there is a fascinating case called the Napkin Ring Problem.
Consider two solids that have spherical shapes, for example, a soccer ball and the planet Earth. Each is represented in the diagram, along with its respective diameter.
The interesting fact about these two napkin rings is that, since they have the same height, they have exactly the same volume.
From the diagram, the radius of the inner circle is AB, which is equal to OD. Since △ODE is a right triangle, by the Pythagorean Theorem, OD can be found.
OD=r2−(2h)2=AB
Also, AC is the radius of the outer circle and applying the Pythagorean Theorem to the right triangle AOC, an expression for it can be deducted.
AC=r2−y2
AC=r2−y2, AB=r2−(2h)2
(a)2=a
Factor out π
Distribute -1
Subtract terms
Commutative Property of Addition
Briefly describe a circle and a sphere. Write the definitions side by side.
Circle | Sphere |
---|---|
A circle is the set of all the points in a plane that are equidistant from a given point. | A sphere is the set of all points in the space that are equidistant from a given point called the center of the sphere. |
As can be seen, the two definitions are identical except for the fact that a circle is a two-dimensional figure while a sphere is three-dimensional. But this is not all. There is one more gnarly relationship. To see it, consider a circle and draw a line containing one diameter.
A spherical cap is a portion of a sphere cut off by a plane.
To calculate the volume of the spherical cap, we need to know the height h and the distance a. V=π h/6(3 a^2+ h^2) We already know the value of a, so we need to determine the height h.
From the exercise, we have only been given the value of a and the radius of the sphere. Let's add this information to the diagram. We will also label the distance between the center of the sphere and the center of the spherical cap as x.
Notice the right triangle with a hypotenuse of 5 and legs of 4 and x. This is a Pythagorean Triple. In this particular triangle, the shorter leg is 3 feet long. Let's add this information to the diagram.
Now we can determine the value of h as the difference between the sphere's radius and 3. h=5-3= 2
Now we have everything we need to calculate the volume of the spherical cap.
The volume of the spherical cap is 52π3 cubic feet.
This time we know the height of the spherical cap. Since we also know the sphere's radius, we can find the distance between the center of the sphere and the center of the spherical cap. 17-9 = 8 ft Let's add this information to the diagram.
Again, we see a right triangle. This time, it has a hypotenuse of 17 and legs of 8 and a. This is also a Pythagorean Triple with legs of 8 and 15 feet. Let's add this information to the diagram.
Now we have everything we need to calculate the volume of the spherical cap.
The volume of the spherical cap is 1134π cubic feet.
The following sphere has a diameter of 4(x+3) centimeters and a surface area of 1024π square centimeters.
From the exercise, we know that the sphere has a surface area of 1024 π square centimeters. Let's recall how we calculate the surface area of a sphere. S=4π r^2 We also know from the exercise that the sphere has a diameter of 4(x+3) centimeters. However, we need the radius which is half the diameter. We can find it by dividing the given diameter by 2. r=4(x+3)/2=2(x+3) To find x, we will substitute the radius and surface area into the formula for the surface area of a sphere and then solve for x.
Since 4(x+3) is the diameter of a sphere, it must be nonnegative. Therefore, the value of x must be 5.
A sphere is inscribed in a cube with a volume of 216 cubic inches.
What is the sphere's volume? Leave your answer in terms of π.To determine the volume of a sphere, we use the following formula. V=4/3π r^3 As we can see, we need to know the sphere's radius. We know that the sphere is inscribed in the cube. This means that the side of the cube equals the diameter of the sphere. If we do a cross-section across the midpoint of opposite sides of one of the cube's surfaces, it would look something like below.
Therefore, to find the radius of the sphere, we must first find the side of the cube.
Remember that we have been given the cube's volume. To calculate the volume of a cube, we cube its side length. V=s^3 By substituting 216 for V and solving for s, we can determine the cube's side.
The side of the cube is 6 inches.
Since we know the side of the cube, we also know the diameter of the sphere. With this information, we can calculate the radius of the sphere. r=6/2=3 in Now we have everything we need to find the volume of the sphere.
The volume is 36π cubic inches.
Find the radius of a sphere with the given properties.
Let r be the radius of a sphere where the surface area of the sphere equals its volume in cubic meters.
Let's substitute the formulas for finding the surface area and volume of a sphere in the equation S=V and solve for r.
The radius of a sphere with equal surface area in square meters and volume in cubic meters is 3 meters.
This time, we want to determine the radius of a sphere where the ratio of the surface area to the volume is 1:6. This means that if we divide the surface area by the volume, we should get a ratio of 16.
As in the previous section, we will substitute the formulas used to find the surface area and the volume into this proportionality and solve for r.
Therefore, the radius of the sphere is 18 meters.