EXERCISE 6.2 – Genesismath

$1-8$ Find the volume of the solid that results when the shaded region is revolved about the indicated axis.

(1)

$\begin{aligned} & f(x)=\sqrt{3-x} ; \quad \text { revolving about } x \text {-axis} \quad \\ \\ & a=-1, \quad b=3 \\ \\ & V=\int_{a}^{b} \pi[f(x)]^{2} d x \\ \\ & V=\int_{-1}^{3} \pi(\sqrt{3-x})^{2} d x=\pi \int_{-1}^{3}(3-x) d x \\ \\ & V=\pi\left[3 x-\frac{x^{2}}{2}\right]_{-1}^{3} \\ \\ & V=\pi\left[\left(3(3)-\frac{3^{2}}{2}\right)-\left(3(-1)-\frac{(-1)^{2}}{2}\right)\right] \\ \\ & V=\pi\left[9-\frac{9}{2}+3+\frac{1}{2}\right] \\ \\ & V=\pi\left[\frac{9}{2}+\frac{1}{2}+3\right]=\pi\left[\frac{9+1+6}{2}\right]=\pi\left(\frac{16}{2}\right)=8 \pi . \end{aligned}$

(2)

$\begin{aligned} & f(x)=2-x^{2}, \quad g(x)=x ; \quad \text { revolving about } x \text {-axis } \\ \\ & a=0, \quad b=1 \\ \\ & V=\int_{a}^{b} \pi\left[(f(x))^{2}-(g(x))^{2}\right] d x \\ \\ & V=\int_{0}^{1} \pi\left[\left(2-x^{2}\right)^{2}-x^{2}\right] d x \end{aligned}$

$\begin{aligned} & V=\int_{0}^{1} \pi\left[4+x^{4}-4 x^{2}-x^{2}\right] d x \\ \\ & V=\pi \int_{0}^{1}\left[4+x^{4}-5 x^{2}\right] d x \\ \\ & V=\pi\left[4 x+\frac{x^{5}}{5}-\frac{5 x^{3}}{3}\right]_{0}^{1} \\ \\ & V=\pi\left(4(1)+\frac{1}{5}-\frac{5(1)}{3}\right)-0 \\ \\ & V=\pi\left(4+\frac{1}{5}-\frac{5}{3}\right)=\pi\left(\frac{60+3-25)}{15}\right. \\ \\ & V=\pi\left(\frac{38}{15}\right)=\frac{38}{15} \pi . \end{aligned}$

(3)

$y=3-2 x ; \quad\quad \text { revolving about } y \text {-axis }$

Solving for $x$

$\begin{aligned} & 2 x=3-y \\ \\ & x=\frac{1}{2}(3-y) \\ \\ & u(y)=\frac{1}{2}(3-y) \\ \\ & c=0, \quad d=2 \\ \\ & V=\int_{0}^{2} \pi[u(y)]^{2} d y \\ \\ & V=\int_{0}^{2} \pi\left[\frac{1}{2}(3-y)\right]^{2} d y=\pi \int_{0}^{2} \frac{1}{4}\left(9+y^{2}-6 y\right) d y \\ \\ & V=\frac{\pi}{4} \int_{0}^{2}\left(9+y^{2}-6 y\right) d y \end{aligned}$

$\begin{aligned} & V=\frac{\pi}{4}\left[9 y+\frac{y^{3}}{3}-6 \frac{y^{2}}{2}\right]_{0}^{2}=\frac{\pi}{4}\left[9 y+\frac{y^{3}}{3}-3 y^{2}\right]_{0}^{2} \\ \\ & V=\frac{\pi}{4}\left[9(2)+\frac{2^{3}}{3}-3(2)^{2}\right]-0 \\ \\ & V=\frac{\pi}{4}\left[18+\frac{8}{3}-12\right]=\frac{\pi}{4}\left[6+\frac{8}{3}\right]=\frac{\pi}{4}\left[\frac{18+8}{3}\right] \\ \\ & V=\frac{\pi}{4}\left[\frac{26}{3}\right]=\frac{13 \pi}{6} \end{aligned}$

(4)

$y=\frac{1}{x}, \quad x=2 ; \quad \quad \text { revolving about } y \text {-axis }$

Solving for $x$

$x=\frac{1}{y}, \quad x=2$

$\Rightarrow u(y)=\frac{1}{y}, \quad v(y)=2$

when $x=2$ , then $y=\frac{1}{2}$ :

$\begin{aligned} & c=\frac{1}{2}, \quad d=2 \\ \\ & V=\int_{e}^{d} \pi\left[(v(y))^{2}-(u(y))^{2}\right] d y \\ \\ & V=\int_{\frac{1}{2}}^{2} \pi\left[2^{2}-\left(\frac{1}{y}\right)^{2}\right] d y \\ \\ & V=\int_{\frac{1}{2}}^{2} \pi\left(4-\frac{1}{y^{2}}\right) d y=\pi \int_{\frac{1}{2}}^{2}\left(4-y^{-2}\right) d y \\ \\ & V=\pi\left[4 y-\frac{y^{-2+1}}{-2+1}\right]_{1}^{2} \end{aligned}$

$\begin{aligned} & V=\pi\left[4 y-\frac{y^{-1}}{-1}\right]_{\frac{1}{2}}^{2}=\pi\left[4 y+\frac{1}{y}\right]_{\frac{1}{2}}^{2} \\ \\ & V=\pi\left[\left(4(2)+\frac{1}{2}\right)-\left(4\left(\frac{1}{2}\right)+\frac{1}{1 / 2}\right)\right] \\ \\ & V=\pi\left[8+\frac{1}{2}-2-2\right]=\pi\left[4+\frac{1}{2}\right]=\pi\left(\frac{9}{2}\right)=\frac{9}{2} \pi \end{aligned}$

(5)

$\begin{aligned} & f(x)=\sqrt{\cos x}; \quad \text { revolving about } x \text {-axis } \\ \\ & a=\frac{\pi}{4}, \quad b=\frac{\pi}{2} \\ \\ & V=\int_{\pi / 4}^{\pi / 2} \pi \left(\sqrt{\cos x}\right)^{2} d x \\ \\ & V=\pi \int_{\pi / 2}^{\pi / 4} \cos x d x=\pi \left[\sin x \right]_{\pi / 2}^{\pi / 4} \\ \\ & V=\pi\left[\sin \frac{\pi}{2}-\sin \pi / 4\right]=\pi\left[1-\frac{1}{\sqrt{2}}\right]=\frac{\pi}{\sqrt{2}}[\sqrt{2}-1] . \end{aligned}$

(6)

$\begin{aligned} & f(x)=x^{2}, \quad g(x)=x^{3}; \quad \text { revolving about } x \text {-axis } \\ \\ & a=0, \quad b=1 \\ \\ & V=\int_{a}^{b} \pi\left[(f(x))^{2}-(g(x))^{2}\right] d x \\ \\ & V=\pi \int_{0}^{1}\left[\left(x^{2}\right)^{2}-\left(x^{3}\right)^{2}\right] d x \\ \\ & V=\pi \int_{0}^{1}\left(x^{4}-x^{6}\right) d x \\ \\ & V=\pi\left[\frac{x^{5}}{5}-\frac{x^{7}}{7}\right]_{0}^{1}=\pi\left[\frac{1}{5}-\frac{1}{7}-0\right]=\frac{2 \pi}{35} \end{aligned}$

(7)

$\begin{aligned} & f(y)=\sqrt{1+y} ; \quad \text { revolving about } y \text {-axis} \quad \\ \\ & c=-1, \quad d=3 \\ \\ & V=\int_{c}^{d} \pi[f(y)]^{2} d y\\ \\ & V=\int_{-1}^{3} \pi(\sqrt{1+y})^{2} d y=\pi \int_{-1}^{3}(1+y) d y \\ \\ & V=\pi\left[y+\frac{y^{2}}{2}\right]_{-1}^{3} \\ \\ & V=\pi\left[\left((3)+\frac{3^{2}}{2}\right)-\left((-1)+\frac{(-1)^{2}}{2}\right)\right] \\ \\ & V=\pi\left[3+\frac{9}{2}+1-\frac{1}{2}\right] \\ \\ & V=\pi\left[\frac{9}{2}-\frac{1}{2}+4\right] \\ \\ & V=\pi\left[\frac{9-1+8}{2}\right]=\pi\left(\frac{16}{2}\right)=8 \pi . \end{aligned}$

(8)

$y=x^2-1, \quad x=2; \quad \quad \text { revolving about } y \text {-axis }$

Solving for $x$

$x^2=1+y$

$\Rightarrow x=\sqrt{1+y}, \quad \text {and} \quad x=2$

$\Rightarrow u(y)=\sqrt{1+y}, \quad v(y)=2$

when $x=2$ , then $y=2^2-1=4-1=3$

$\begin{aligned} \text { Thus we have} \quad & c=0, \quad d=3 \\ \\ & V=\int_{c}^{d} \pi\left[(v(y))^{2}-(u(y))^{2}\right] d y \\ \\ & V=\int_{0}^{3} \pi\left[2^{2}-(\sqrt{1+y})^{2}\right] d y \\ \\ & V=\int_{0}^{3} \pi\left(4-(1+y\right)) d y \\ \\ & V=\pi \int_{0}^{3}\left(4-1-y)\right) d y \\ \\ & V=\pi \int_{0}^{3}\left(3-y\right) d y \\ \\ & V=\pi\left[3 y-\frac{y^{2}}{2}\right]_{0}^{3} \\ \\ & V=\pi\left[\left 3(3)-\frac{3^{2}}{2}\right] -0 \\ \\ & V=\pi\left[9-\frac{9}{2}\right]=\pi\left[\frac{9}{2}\right]=\frac{9}{2} \pi \end{aligned}$