McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
6. Surface Areas and Volumes of Spheres
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Exercise 39 Page 854

Practice makes perfect
a

Let's analyze the given diagram.

Here, AC=x and AB=r. We are asked to find the radius BC of circle ⊙ C. Let's use the Pythagorean Theorem for the right triangle, △ ABC.

AC^2+BC^2=AB^2
x^2+BC^2= r^2
â–¼
Solve for BC
BC^2=r^2-x^2
BC=sqrt(r^2-x^2)

Therefore, the radius of the disc from the sphere in terms of its distance x above the sphere's center is sqrt(r^2-x^2).

b

From Part A we know that the radius of the disc is R=sqrt(r^2-x^2). Since the thickness of the disc is y, its height is h=y, too. Now, let's use the formula for the volume of a cylinder.

V=Ï€ R^2h
V=Ï€( sqrt(r^2-x^2))^2 y
V=Ï€(r^2-x^2)y

This tells us that the volume of the disc is π (r^2-x^2)y.

c

Let's analyze the cylinder and then later analyze the cones to find the radii of the discs at level x.

The disc from the cylinder at any level has radius r.

Since â–³ ABC is isosceles, the radius of the disk at level x is also R=x, too. Now, let's find the volumes of the discs from the cylinder and from the cones with the thickness of y.

Disc from Cylinder Disc from Cones
Radius r x
Height y y
Volume V=Ï€* Radius^2* Height
V_1=Ï€ r^2 y V_2=Ï€ x^2 y

Now we will analyze the hollowed-out cylinder.

The volume of hollowed-out disc with thickness of y units, V_\text{hol.}, is equal to the difference of the volume of the discs from the cylinder and from the cones. \begin{gathered} V_\text{hol.}=V_1-V_2 \\ \Downarrow \\ V_\text{hol.}=\pi r^2y-\pi x^2y=\pi(r^2-x^2)y \end{gathered} This is the same as the volume from Part B.

d

Let's start by recalling what Cavalieri's Principle states.

Cavalieri's Principle

If two solids have the same height and the same cross-sectional area at every level, then they have the same volume.

From Part B and Part C, we know that the expressions for the volume of the discs at the same height are the same. Since we used the same thickness of y when finding their volumes, the discs also have the same area.

Disc From Sphere Hollowed-Out Disc
Volume π(r^2-x^2)y
Area π(r^2-x^2)

Notice also that the hollowed-out cylinder and sphere have the same height, 2r. Therefore, by Cavalieri's Principle it is guaranteed that the volume of the sphere is the same as the volume of the hollowed-out cylinder. \begin{aligned} V_{\text{sphere}} = V_\text{hol.} \end{aligned}

e

To find the volume of the sphere, we will subtract the two volumes of the cone from the volume of the cylinder. We can use the formula for the volume of a cylinder and for the volume of a cone.

Cylinder Cone
Radius r r
Height 2r r
Volume V_\text{cylinder}=\pi\cdot{\color{#0000FF}{\text{Radius}}}^2\cdot{\color{#009600}{\text{Height}}} V_\text{cone}=\dfrac{1}{3}\pi\cdot{\color{#0000FF}{\text{Radius}}}^2\cdot{\color{#009600}{\text{Height}}}
\textcolor{darkorange}{V_\text{cylinder}}=\pi {\color{#0000FF}{r}}^2({\color{#009600}{2r}})=\textcolor{darkorange}{2\pi r^3} \textcolor{darkviolet}{V_\text{cone}}=\dfrac{1}{3}\pi {\color{#0000FF}{r}}^2({\color{#009600}{r}})=\textcolor{darkviolet}{\dfrac{1}{3}\pi r^3}

Finally, let's find the volume of the sphere!

V_\text{sphere}=\textcolor{darkorange}{V_\text{cylinder}}-2\cdot \textcolor{darkviolet}{V_\text{cone}}
â–¼
Substitute values and evaluate
V_\text{sphere}=\textcolor{darkorange}{2\pi r^3}-2\cdot \textcolor{darkviolet}{\dfrac{1}{3}\pi r^3}
V_\text{sphere}=2\pi r^3-2\cdot\dfrac{\pi r^3}{3}
V_\text{sphere}=2\pi r^3-\dfrac{2\pi r^3}{3}
V_\text{sphere}=\dfrac{3\cdot 2\pi r^3}{3}-\dfrac{2\pi r^3}{3}
V_\text{sphere}=\dfrac{6\pi r^3}{3}-\dfrac{2\pi r^3}{3}
V_\text{sphere}=\dfrac{4\pi r^3}{3}

We proved that the volume of the sphere with the radius of r is V_\text{sphere}=\frac{4\pi r^3}{3}.