Chapter 6, Problem 3P

(a)

To determine

To show: The volume of a segment of a sphere $V=\frac{1}{3}\pi h^2\left(3r-h\right)$.

Answer to Problem 3P

The volume of the segment of a sphere is $\underset{\_}{V=\frac{1}{3}\pi h^2\left(3r-h\right)}$.

Explanation of Solution

Given information:

The segment of a sphere with radius r and height h.

Consider that the Equation of circle as $x^2+y^2=r^2$ (1)

Rearrange Equation (1).

$\begin{array}{l}{x}^2=r^2-y^2\\ x=\sqrt{r^2-y^2}\end{array}$

The dimensions of the sphere as shown in Figure 1.

Refer to Figure 1.

Consider that the sphere is obtained by rotating the circle $x^2+y^2=r^2$ about y-axis.

The region lies between $a=r-h$ and $b=r$.

Calculation:

The expression to find the volume of the segment of a sphere as shown below:

$V={\displaystyle {\int }_a^{b}A\left(y\right)dy}$ (2)

Find the area of the segment of a sphere as shown below.

$\begin{array}{c}A\left(y\right)=\pi r^2\\ =\pi {\left(\sqrt{r^2-y^2}\right)}^2\\ =\pi \left(r^2-y^2\right)\end{array}$

Substitute $\left(r-h\right)$ for a, r for b, and $\pi \left(r^2-y^2\right)$ for $A\left(y\right)$ in Equation (2).

$\begin{array}{c}V={\displaystyle {\int }_{r-h}^r\pi \left(r^2-y^2\right)dy}\\ =\pi {\left(r^{2}y-\frac{y^3}{3}\right)}_{r-h}^r\\ =\pi \left(r^2\times r-\frac{r^3}{3}-\left(r^2\left(r-h\right)-\frac{{\left(r-h\right)}^3}{3}\right)\right)\\ =\pi \left(r^3-\frac{r^3}{3}-\left(r-h\right)\left(r^2-\frac{{\left(r-h\right)}^2}{3}\right)\right)\end{array}$

$\begin{array}{c}V=\pi \left(\frac{2r^3}{3}-\frac{\left(r-h\right)}{3}\left(3r^2-\left(r^2+h^2-2rh\right)\right)\right)\\ =\frac{1}{3}\pi \left(2r^3-\left(r-h\right)\left(2r^2+2rh-h^2\right)\right)\\ =\frac{1}{3}\pi \left(2r^3-2r^3-2r^{2}h+r{h}^2+2r^{2}h+2r{h}^2-h^3\right)\\ =\frac{1}{3}\pi \left(3r{h}^2-h^3\right)\end{array}$

$=\frac{1}{3}\pi h^2\left(3r-h\right)$

Therefore, the volume of the segment of a sphere is $\underset{\_}{V=\frac{1}{3}\pi h^2\left(3r-h\right)}$.

(b)

To determine

To calculate: The value of x using Newton’s method.

Answer to Problem 3P

The value of x by using Newton’s method is 0.2235.

Explanation of Solution

Given information:

A sphere of radius 1 is sliced by a plane at a distance x from the center.

The volume of one segment is twice the volume of the other.

The Answer of the equation $3x^3-9x+2=0$, $0<x<1$ is x.

Calculation:

Find the volume of the sphere as shown below.

$V=\frac{4}{3}\pi r^3$ (3)

Substitute 1 for r in Equation (3).

$\begin{array}{c}V=\frac{4}{3}\pi {\left(1\right)}^3\\ =\frac{4}{3}\pi \end{array}$

Consider the smaller segment has height $h=1-x$.

Refer to part (a).

Volume of the segment of a sphere as shown below:

$V=\frac{1}{3}\pi h^2\left(3r-h\right)$ (4)

Substitute 1 for r and $1-x$ for h in Equation (4).

$\begin{array}{c}V=\frac{1}{3}\pi {\left(1-x\right)}^2\left(3\times 1-\left(1-x\right)\right)\\ =\frac{1}{3}\pi {\left(1-x\right)}^2\left(x+2\right)\end{array}$

This volume must be $\frac{1}{3}$ of the total volume of the sphere.

Substitute $\frac{4}{3}\pi$ for V in the above Equation.

$\begin{array}{l}\frac{1}{3}\left(\frac{4}{3}\pi \right)=\frac{1}{3}\pi {\left(1-x\right)}^2\left(x+2\right)\\ \frac{4}{3}=\left(1+x^2-2x\right)\left(x+2\right)\\ x+x^3-2x^2+2+2x^2-4x=\frac{4}{3}\\ x^3-3x+2=\frac{4}{3}\end{array}$

$\begin{array}{l}3x^3-9x+6=4\\ 3x^3-9x+2=0\end{array}$

Apply Newton’s method as shown below.

$x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$ (5)

Consider $f\left(x\right)=3x^3-9x+2$

Differentiate both sides of the Equation.

$f^{\prime }\left(x\right)=9x^2-9$

Substitute $3x^3-9x+2$ for $f\left(x\right)$ and $9x^2-9$ for $f^{\prime }\left(x\right)$ in Equation (5).

$x_{n+1}=x_n-\frac{x_n^3-9x_n+2}{3x_n^2-9}$ (6)

Consider $x_1=0$.

Substitute 1 for n in Equation (6).

$\begin{array}{c}{x}_2=x_1-\frac{x_1^3-9x_1+2}{3x_1^2-9}\\ =0-\frac{0+2}{0-9}\\ =0.2222\end{array}$

Substitute 2 for n in Equation (6).

$\begin{array}{c}{x}_3=x_2-\frac{x_2^3-9x_2+2}{3x_2^2-9}\\ =0.2222-\frac{{0.2222}^3-9\times 0.2222+2}{3\times {0.2222}^2-9}\\ =0.2222+0.0013\\ \approx 0.2235\end{array}$

Substitute 2 for n in Equation (6).

$\begin{array}{c}{x}_4=x_3-\frac{x_3^3-9x_3+2}{3x_3^2-9}\\ =0.2235-\frac{{0.2235}^3-9\times 0.2235+2}{3\times {0.2235}^2-9}\\ =0.2222+0.0013\\ \approx 0.2235\end{array}$

$\begin{array}{c}{x}_3\approx 0.2235\\ \approx x_4\end{array}$

Therefore, the value of x by using Newton’s method is 0.2235.

(c)

To determine

To calculate: sinking depth of the sphere.

Answer to Problem 3P

The sinking depth of the sphere is 0.6736 m.

Explanation of Solution

Given information:

Wooden sphere of radius as 0.5 m.

Specific gravity of the wooden sphere is 0.75.

The depth x to which a floating sphere of radius r sinks in water is a root of the equation as follows:

$x^3-3r{x}^2+4r^{3}s=0$

Calculation:

Show the root of the equation as shown below.

$x^3-3r{x}^2+4r^{3}s=0$ (7)

Substitute 0.5 m for r and 0.75 for s in Equation (7).

$\begin{array}{l}{x}^3-3\left(0.5\right)x^2+4{\left(0.5\right)}^3\times 0.75=0\\ x^3-\frac{3}{2}{x}^2+4\times \frac{1}{8}\times \frac{3}{4}=0\\ 8x^3-12x^2+3=0\end{array}$

Consider $f\left(x\right)=8x^3-12x^2+3$

Differentiate both sides of the Equation.

$f^{\prime }\left(x\right)=24x^2-24x$

Apply Newton’s method.

Substitute $8x^3-12x^2+3$ for $f\left(x\right)$ and $24x^2-24x$ for $f^{\prime }\left(x\right)$ in Equation (5).

$x_{n+1}=x_n-\frac{8x_n^3-12x_n^2+3}{24x_n^2-24}$ (8)

Consider $x_1=0.5$.

Substitute 1 for n in Equation (8).

$\begin{array}{c}{x}_2=x_1-\frac{8x_1^3-12x_1^2+3}{24x_1^2-24}\\ =0.5-\frac{8\times {0.5}^3-12\times {0.5}^2+3}{24\times {0.5}^2-24}\\ =0.5+0.0556\\ =0.5556\end{array}$

Substitute 2 for n in Equation (8).

$\begin{array}{c}{x}_3=x_2-\frac{8x_2^3-12x_2^2+3}{24x_2^2-24}\\ =0.5556-\frac{8\times {0.5556}^3-12\times {0.5556}^2+3}{24\times {0.5556}^2-24}\\ =0.5556+0.0402\\ =0.5958\end{array}$

Substitute 3 for n in Equation (8).

$\begin{array}{c}{x}_4=x_3-\frac{8x_3^3-12x_3^2+3}{24x_3^2-24}\\ =0.5958-\frac{8\times {0.5958}^3-12\times {0.5958}^2+3}{24\times {0.5958}^2-24}\\ =0.5958+0.0279\\ =0.6237\end{array}$

Approximately the depth should be $x=0.6736$.

Therefore, the sinking depth of the sphere is 0.6736 m.

(d) i)

To determine

To calculate: The speed of rising the water level in the bowl from instant of water level at 3 inches deep.

Answer to Problem 3P

The speed of rising the water level in the bowl from instant of water level at 3 inches deep is $\underset{\_}{0.003\text{in}/\text{s}}$

Explanation of Solution

Given information:

The shape of the bowl is hemisphere.

The radius of hemispherical bowl is 5 inches.

Water is running into the bowl at the rate of $0.2{\text{in}}^{\text{3}}/\text{s}$.

Depth of water is 3 inches.

Calculation:

Show the volume of the segment of a sphere as follows:

$\begin{array}{c}V=\frac{1}{3}\pi h^2\left(3r-h\right)\\ =\pi r{h}^2-\frac{1}{3}\pi h^3\end{array}$

Differentiate both sides of the Equation with respect to time t.

$\frac{dV}{dt}=2\pi rh\frac{dh}{dt}-\pi h^2\frac{dh}{dt}$ (9)

Find the speed of the water level in the bowl rising at the instant as shown below.

Substitute 5 inches for r, $0.2{\text{in}}^{\text{3}}/\text{s}$ for $\frac{dV}{dt}$, and 3 inches for h in Equation (9).

$\begin{array}{l}0.2=2\pi \times 5\times 3\frac{dh}{dt}-\pi \times 3^2\frac{dh}{dt}\\ 0.2=\left(94.248-28.274\right)\frac{dh}{dt}\\ \frac{dh}{dt}=0.003\text{in/s}\end{array}$

Therefore, the speed of rising the water level in the bowl from instant of water level at 3 inches deep is $\underset{\_}{0.003\text{in}/\text{s}}$

ii)

To determine

To calculate: The time taken to fill the bowl from 4 inch level of water.

Answer to Problem 3P

The time taken to fill the bowl from 4 inch level of water is $\underset{\_}{6.5\min }$.

Explanation of Solution

Given information:

The radius of hemispherical bowl is 5 inches.

Water is running into the bowl at the rate of $0.2{\text{in}}^{\text{3}}/\text{s}$.

Depth of water is 4 inches.

Calculation:

Find the volume of the water required to fill the bowl from the instant that the water is 4 in. depth as follows:

$\begin{array}{c}V=\frac{1}{2}\times \frac{4}{3}\pi r^3-\frac{1}{3}\pi h^2\left(3r-h\right)\\ =\frac{2}{3}\pi r^3-\pi r{h}^2+\frac{1}{3}\pi h^3\end{array}$

Substitute 5 inches for r and 4 inches for h in the Equation.

$\begin{array}{c}V=\frac{2}{3}\pi \times 5^3-\pi \times 5\times 4^2+\frac{1}{3}\pi \times 4^3\\ =\frac{2}{3}\times 125\pi -80\pi +\frac{64}{3}\pi \\ =\frac{\pi \left(250-240+64\right)}{3}\\ =\frac{74}{3}\pi \end{array}$

Find the time required to fill the bowl as shown below.

$\begin{array}{c}T=\frac{V}{\frac{dV}{dt}}\\ =\frac{\frac{74}{3}\pi }{0.2}\\ =387.5\text{sec}\times \frac{1\min }{60\sec }\\ =6.5\text{min}\end{array}$

Therefore, The time taken to fill the bowl from 4 inch level of water is $\underset{\_}{6.5\min }$.