Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 12.2
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Exercise 142 Page 698

Volume of a cylinder with height h and radius r is 13π r^2 h.

About 164.37 cubic inches.

Practice makes perfect

We know that Martha's original cone had a diameter of 12 inches and a height of 5 inches. Then Matt snatched the cone and cut 2 inches from the top. Then he placed the removed top in such a way shown on the following diagram.

We want to find the amount of water that can fit in such a redesigned cone. The top cone was inverted and placed so that if a water was poured inside, the top cone would obstruct the water. The amount of water that can fit inside is the volume of the truncated cone V_(tr) minus the volume of the top cone V_t. Amount of water = V_(tr) - V_t The volume of the truncated cone V_(tr) is the volume of the original cone V_\text{o} minus the volume of the top cone.

\begin{gathered} {\color{#FF0000}{V_{\text{tr}}}} = {\color{#FF0000}{V_\text{o} - V_{\text{t}}}} \end{gathered} Therefore, we have the following formula states the amount of water that can be poured in. \begin{gathered} \text{Amount of water} = {\color{#FF0000}{(V_\text{o} - V_{\text{t}})}} - V_{\text{t}} \\ \Downarrow \\ \text{Amount of water} = {\color{#A800DD}{V_{\text{o}}}} - 2\textcolor{teal}{V_{\text{t}}} \end{gathered}

Original Cone

The original cone had a diameter of 12 inches, so its radius is 12÷2 = 6 inches. The cone's height was 5 inches. Now, let's recall the formula for the volume of a cone. V = 1/3 π r^2 h Here, r is the cone's radius and h is its height. In our case, r = 6 inches and h = 5 inches. Let's substitute these values into our formula and simplify the result.
V = 1/3 π r^2 h
{\color{#A800DD}{V_\text{o}}} = \dfrac{1}{3} \pi ({\color{#0000FF}{6}})^2({\color{#009600}{5}})
Simplify right-hand side
V_\text{o} = \dfrac{1}{3} \pi (36)(5)
V_\text{o} = \dfrac{1}{3} (180) \pi
V_\text{o} = \dfrac{180}{3} \pi
V_\text{o} = 60\pi
V_\text{o} = 188.495559\ldots
V_\text{o} \approx {\color{#A800DD}{188.496}}
The original cone's volume is about 188.496 cubic inches.

Top Cone

To find the volume of the top cone, let's note that is is similar to the original one. Since Matt cut 2 inches from the top, the top cone's height is 2 inches. The original cone's height was 5 inches, so the linear scale factor between the two is 25. Linear scale factor = 2/5 The ratio of the top cone's volume and the original cone's volume is the volume scale factor and it is a cube of the linear scale factor. Volume scale factor = ( 2/5)^3 = 8/125 Now, let's multiply the original cone's volume by the volume scale factor to get the top cone's volume.
V_t = Volume scale factor * V_()
V_t ≈ 8/125 * 188.496
V_t ≈ 1507.968/125
V_t ≈ 12.063744...
V_t ≈ 12.064
The top cone's volume is about 12.064 cubic inches.

Redesigned Cone

Now, let's substitute the found volume of the original cone and the volume of the top cone into our formula for the amount of water that the redesigned cone can store.
\text{Amount of water} = V_\text{o} - 2V_{\text{t}}

V_\text{o} \approx {\color{#0000FF}{{\color{#A800DD}{188.496}}}}, V_\text{t} \approx {\color{#009600}{\textcolor{teal}{12.064}}}

Amount of water ≈ 188.496 - 2(12.064)
Amount of water ≈ 188.496- 24.128
Amount of water ≈ 164.368
Amount of water ≈ 164.37
About 164.37 cubic inches of water can fit within the redesigned cone.