Chapter 6.1, Problem 87E

Conical Cup A conical cup is made from a circular piece of paper with radius 6 cm by cutting out a sector and joining the edges as shown below. Suppose θ = 5π/3.

1. (a) Find the circumference C of the opening of the cup.
2. (b) Find the radius r of the opening of the cup. [Hint: Use C = 2πr.]
3. (c) Find the height h of the cup. [Hint: Use the Pythagorean Theorem.]
4. (d) Find the volume of the cup.

To determine

The circumference $C$ of the opening of the cup.

Answer to Problem 87E

The circumference $C$ of the opening of the cup is $\underset{¯}{31.4\mathrm{cm}}$.

Explanation of Solution

Given:

Radius of the circular piece of paper is $6\mathrm{cm}$.

Angle of the remaining circle is $\frac{5\pi }{3}$.

Formulas used:

Circumference of circle is $2\pi R$ and arc length is $s={\theta }_{1}R$, where $R$ is the radius of the circle and ${\theta }_1$ is the central angle of the sector.

Calculations:

The angle of a circle is $2\pi$. That is, the angle of the sector is ${\theta }_1=2\pi -\theta$.

Substitute $\frac{5\pi }{3}$ for $\theta$ in ${\theta }_1=2\pi -\theta$:

$\begin{array}{c}{\theta }_1=2\pi -\frac{5\pi }{3}\\ =\frac{\pi }{3}\end{array}$

The angle of a sector is $\frac{\pi }{3}\mathrm{rad}$.

Formula for arc length is $s=R{\theta }_1$.

Substitute $\frac{\pi }{3}$ for ${\theta }_1$ and $6\mathrm{cm}$ for R in $s=r{\theta }_1$.

$\begin{array}{c}s=\frac{\pi }{3}\times 6\\ =2\pi \end{array}$

The arc length of the sector that has been cut out of the circle is $2\pi$.

Circumference of the circle is $2\pi R$. That is,

$\begin{array}{c}2\pi R=2\pi \times 6\\ =12\pi \end{array}$

The complete circle has circumference as $12\pi \mathrm{cm}$.

Circumference of the opening of the conical cup is $C=2\pi R-s$.

Substitute $12\pi$ for circumference of circle and $2\pi$ for $s$ in $C=2\pi R-s$:

$\begin{array}{c}C=12\pi -2\pi \\ =10\pi \\ \approx 31.4\end{array}$

Thus, the circumference $C$ of the opening of the cup is $\underset{¯}{31.4\mathrm{cm}}$.

To determine

The radius $r$ of the opening of the cup.

Answer to Problem 87E

The radius $r$ of the opening of the cup is $\underset{¯}{5\mathrm{cm}}$.

Explanation of Solution

From part (a), $C=10\pi$.

Substitute $C=10\pi$in $C=2\pi r$:

$\begin{array}{l}10\pi =2\pi \times r\\ r=5\end{array}$

Thus, the radius $r$ of the opening of the cup is $\underset{¯}{5\mathrm{cm}}$.

To determine

The height $h$ of the cup.

Answer to Problem 87E

The height $h$ of the cup is $\underset{¯}{3.32\mathrm{cm}}$.

Explanation of Solution

Given:

Radius of the circular piece of paper is $6\mathrm{cm}$.

Angle of the remaining circle is $\frac{5\pi }{3}$.

Slant height of the cone is $6\mathrm{cm}$.

Calculation:

The Pythagorean Theorem is $l^2=r^2+h^2$, where $l$ is the slant height of the cone, $r$ is the radius of the cone and $h$ is the height of the cone.

The formula for the Pythagorean Theorem is $l^2=r^2+h^2$.

From part (b), $r=5$.

Substitute $6\mathrm{cm}$ for $l$ and $5\mathrm{cm}$ for $r$ in $l^2=r^2+h^2$:

$\begin{array}{l}{\left(6\right)}^2={\left(5\right)}^2+h^2\\ h^2=36-25\\ h^2=11\\ h=\sqrt{11}\\ h\approx 3.32\end{array}$

Thus, the height $h$ of the cup is $\underset{¯}{3.32\mathrm{cm}}$.

To determine

The volume of the cup.

Answer to Problem 87E

The volume of the cup is $\underset{¯}{86.8{\mathrm{cm}}^3}$.

Explanation of Solution

Formula used:

Volume of a cone is $\mathrm{V}=\frac{1}{3}\pi r^{2}h$, were $r$ is the radius of a cone and $h$ is the height of a cone.

Calculation:

From part (b) and (c), $r=5$ and $h=3.32$.

Substitute $r=5$ and $h=3.32$ in $\mathrm{V}=\frac{1}{3}\pi r^{2}h$.

$\begin{array}{c}\mathrm{V}=\frac{1}{3}\times 3.14\times {\left(5\right)}^2\times 3.32\\ =\frac{3.14\times 25\times 3.32}{3}\\ \approx 86.8\end{array}$

Thus, the volume of the cup is $\underset{¯}{86.8{\mathrm{cm}}^3}$.