EXERCISE 6.5 – Genesismath

1-4 Find the area of the surface generated by revolving the given curve about the $x$ -axis.

$y=7 x, \quad 0 \leq x \leq 1--------(1)$

differentiating w.r.t $x$

$\frac{d y}{d x}=7 -------(2)$

Formula for the area of the Surface of revolution ( about $x$ -axis):

$S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x--------(3)$

Substituting (1) and (2) into (3) and integrating over the interval $[0,1]$

$\begin{aligned} S & =\int_{0}^{1} 2 \pi(7 x) \sqrt{1+(7)^{2}} d x=\int_{0}^{1} 14 \pi x \sqrt{1+49} d x \\ \\ S & =\int_{0}^{1} 14 \pi x \sqrt{50} d x=14 \sqrt{50} \pi \int_{0}^{1} x d x \\ \\ & =14 \sqrt{25 \times 2} \pi\left[\frac{x^{2}}{2}\right]_{0}^{1}=14(5 \sqrt{2}) \pi\left[\frac{1}{2}-0\right] \\ \\ & =\frac{70 \sqrt{2} \pi}{2}=35 \pi \sqrt{2} . \end{aligned}$

$y =\sqrt{x} \quad, \quad 1 \leq x \leq 4 \quad -------(1)$

differentiating w.r. t $x$

$\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}--------(2)$

Formula for the area of the surface of revolution( about $x$ -axis):

$S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x------(3)$

Substituting (1) and (2) into (3) and integrating over the interval $[1,4]$

$\begin{aligned} & S=\int_{1}^{4} 2 \pi \sqrt{x} \sqrt{1+\left(\frac{1}{2 \sqrt{x}}\right)^{2}} d x \\ \\ & =\int_{1}^{4} 2 \pi \sqrt{x} \sqrt{1+\frac{1}{4 x}} d x \\ \\ & =\int_{1}^{4} 2 \pi \sqrt{x} \sqrt{\frac{4 x+1}{4 x}} d x=\int_{1}^{4}2 \pi \frac{\sqrt{x}}{\sqrt{x}} \sqrt{\frac{4 x+1}{4}} d x \\ \\ & =\int_{1}^{4} 2 \pi\left(x+\frac{1}{4}\right)^{1 / 2} d x=2 \pi\left[\frac{\left(x+\frac{1}{4}\right)^{3 / 2}}{3 / 2}\right]_{1}^{4} \\ \\ & =2 \pi \cdot \frac{2}{3}\left[\left(x+\frac{1}{4}\right)^{3 / 2}\right]_{1}^{4}=\frac{4 \pi}{3}\left[\left(4+\frac{1}{4}\right)^{3 / 2}-\left(1+\frac{1}{4}\right)^{3 / 2}\right] \\ \\ & =\frac{4 \pi}{3}\left[\left(\frac{16+1}{4}\right)^{3 / 2}-\left(\frac{4+1}{4}\right)^{3 / 2}\right] \\ \\ & =\frac{4 \pi}{3}\left(\frac{(17)^{3 / 2}}{4^{3 / 2}}-\frac{(5)^{3 / 2}}{4^{3 / 2}}\right] \\ \\ & S=\frac{4 \pi}{3}\left[\frac{17 \sqrt{17}}{8}-\frac{5 \sqrt{5}}{8}\right]=\frac{\pi}{6}[17 \sqrt{17}-5 \sqrt{5}] \end{aligned}$

$y=\sqrt{4-x^{2}}, \quad-1 \leq x \leq 1---------(1)$

differentiating w.r.t $x$

$\frac{d y}{d x}=\frac{1}{2}\left(4-x^{2}\right)^{-1 / 2} 2 x=\frac{x}{\sqrt{4-x^{2}}}------(2)$

Formula for the area of the surface of resolution (about $x$ -axis):

$S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x--------(3)$

Substituting (1) and (2) into (3) and integrating over the interval $[-1,1]$ .

$S=\int_{-1}^{1} 2 \pi \sqrt{4-x^{2}} \sqrt{1+\left(\frac{x}{\sqrt{4-x^{2}}}\right)^{2}} d x$

$S=\int_{-1}^{1} 2 \pi \sqrt{4-x^{2}} \sqrt{1+\frac{x^{2}}{4-x^{2}}} d x$

$S=\int_{-1}^{1} 2 \pi \sqrt{4-x^{2}} \sqrt{\frac{\left(4-x^{2}\right)+x^{2}}{\left(4-x^{2}\right)}} d x$

$\begin{aligned} & S=\int_{-1}^{1} 2 \pi \sqrt{4-x^{2}+x^{2}} d x=\int_{-1}^{1} 2 \pi \sqrt{4} d x \\ \\ & S=\int_{-1}^{1} 4 \pi d x=4 \pi[x]_{-1}^{1}=4 \pi(1-(-1))=4 \pi(1+1)=8 \pi . \end{aligned}$

$y=x^{3}, \quad 1 \leq x \leq 2--------(1)$

differentiating w.r.t $x$

$\frac{d y}{d x}=3 x^{2}---------(2)$

Formula for the area of the surface of revolution (about $x$ -axis):

$S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x------(3)$

substituting (1) and (2) into (3) and integrating over the interval $[1,2]$

$S=\int_{1}^{2} 2 \pi x^{3} \sqrt{1+\left(3 x^{2}\right)^{2}} d x=2 \pi \int_{1}^{2} x^{3} \sqrt{1+9 x^{4}} d x$

The derivative of $\sqrt {1+9 x^{4}}$ is $36 x^3$ . Therefore multiplying and dividing by 36 to complete the derivative of $\sqrt{ 1+9 x^{4}}$ .

$S=\frac{2}{36} \pi \int_{1}^{2} 36x^{3} (1+9 x^{4}} )^{1/2}d x$

$S=\frac{2\pi}{36} \left[\frac{\left(1+9 x^{4}\right)^{3/2}}{3/2}\right]_{1}^{2}$

$S=\frac{\pi}{18} \cdot \frac{2}{3}\left[\left(1+9 x^{4}\right)^{3 / 2}\right]_{1}^{2}$

$S=\frac{\pi}{27}\left[(1+9(2)^{4})^{3 / 2}-( 1+9(1))^{3 / 2}\right]$

$\begin{aligned} & S=\frac{\pi}{27}\left[(1+9(16))^{3 / 2}-(10)^{3 / 2}\right] \\ \\ & S=\frac{\pi}{27}\left[(1+144)^{3 / 2}-(10)^{3 / 2}\right]=\frac{\pi}{27}\left[(145)^{3 / 2}-(10)^{3 / 2}\right] \end{aligned}$

5-8. Find the area of the surface generated by revolving the given curve about the $y$ -axis

5 .

$x=9 y+1, \quad 0 \leq y \leq 2-----------(1)$

differentiating w.r.t $y$

$\frac{d x}{d y}=9--------(2)$

Formula for the area of the surface of revolution (about $y$ -axis)

$S=\int_{c}^{d} 2 \pi x \sqrt{1+\left(\frac{d x}{d y}\right)^{2}} d y------(3)$

substituting (1) and (2) into (3) and integrating over the interval $[0,2]$

$\begin{aligned} & S=\int_{0}^{2} 2 \pi(9 y+1) \sqrt{1+(9)^{2}} d y=2 \pi \int_{0}^{2}(9 y+1) \sqrt{1+81} d y \\ \\ & S=2 \pi \sqrt{82} \int_{0}^{2}(9 y+1) d y=2 \pi \sqrt{82}\left[\frac{9 y^{2}}{2}+y\right]_{0}^{2} \\ \\ & S=2 \pi \sqrt{82}\left[\frac{9}{2}(2)^{2}+2-0\right] \\ \\ & S=2 \pi \sqrt{82}\left[18+2\right] \\ \\ & S=2 \pi \sqrt{82}(20)=40 \pi \sqrt{82} \end{aligned}$