EXERCISE 6.3

 

1-4 use cylindrical shells to find the volume of the solid generated when the shaded region is revolved about the indicated axis.

(1)

$$
\begin{aligned}
& f(x)=x^{2}, \quad \text { revolving about } y \text {-axis } \\ \\
& a=1 \quad b=2
\end{aligned}
$$

$$
\begin{aligned}
& V=\int_{a}^{b} 2 \pi x f(x) d x \\ \\
& V=\int_{1}^{2} 2 \pi x\left(x^{2}\right) d x=2 \pi \int_{1}^{2} x^{3} d x=2 \pi\left[\frac{x^{4}}{4}\right]_{1}^{2} \\ \\
& V=\frac{2 \pi}{4}\left[2^{4}-1\right]=\frac{\pi}{2}[16-1]=15 \frac{\pi}{2}
\end{aligned}
$$

(2)

$$
y=\sqrt{4-x^{2}}, \quad  y=x \quad \text { revolving about } y \text {-axis }
$$

solving $ y=\sqrt{4-x^{2}}$ and $y=x$ simultaneously

$$
\begin{aligned}
& x=\sqrt{4-x^{2}} \Rightarrow x^{2}=4-x^{2} \Rightarrow 2 x^{2}=4 \\ \\
& \Rightarrow x^{2}=2 \quad \Rightarrow x=\sqrt{2} \\ \\
& \therefore \quad a=0, \quad b=\sqrt{2}
\end{aligned}
$$

$$
\begin{aligned}
& V=\int_{a}^{b} 2 \pi x\left[\sqrt{4-x^{2}}-x\right] d x \\ \\
& V=\int_{0}^{\sqrt{2}} 2 \pi\left[ x \sqrt{4-x^{2}}-x^{2}\right] d x
\end{aligned}
$$

$$
\begin{aligned}
& V=2 \pi \int_{0}^{\sqrt{2}} x\left(4-x^{2}\right)^{1 / 2} d x-2 \pi \int_{0}^{\sqrt{2}} x^{2} d x \\ \\
& V=-\pi \int_{0}^{\sqrt{2}}-2 x\left(4-x^{2}\right)^{1 / 2} d x-2 \pi\left[\frac{x^{3}}{3}\right]_{0}^{\sqrt{2}} \\ \\
& V=-\pi\left[\frac{\left(4-x^{2}\right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{\sqrt{2}}-2 \frac{\pi}{3}\left[(\sqrt{2})^{3}-0\right] \\ \\
& V=-\pi\left[\frac{\left(4-x^{2}\right)^{3 / 2}}{3 / 2}\right]_{0}^{\sqrt{2}}-2 \frac{\pi}{3}(2 \sqrt{2}) \\ \\
& V=-\pi\left[\frac{2}{3}\left(4-(\sqrt{2})^{2}\right)^{3 / 2}-\frac{2}{3}(4-0)^{3 / 2}\right]-4 \sqrt{2} \pi / 3 \\ \\
& V=-\frac{2 \pi}{3}\left[(4-2)^{3 / 2}-(4)^{3 / 2}\right]-4 \sqrt{2} \pi / 3 \\ \\
& V=-\frac{2 \pi}{3}\left[2^{3 / 2}-2^{3}\right]-4 \sqrt{2} \pi / 3 \\ \\
& V=-\frac{2 \pi}{3}[2 \sqrt{2}-8]-4 \sqrt{2} \frac{\pi}{3} \\ \\
& V=-4 \sqrt{2} \frac{\pi}{3}+16 \frac{\pi}{3}-4 \sqrt{2} \pi / 3 \\ \\
& V=16 \frac{\pi}{3}-8 \sqrt{2} \frac{\pi}{3}=\frac{8 \pi}{3}(2-\sqrt{2})
\end{aligned}
$$

(3)

$$
\begin{aligned}
& x=f(y)=2 y-2 y^{2},  \quad \text { revolving about } x \text {-axis }\\ \\
& c=0, \quad d=1 \\ \\
& V=\int_{c}^{d} 2 \pi y f(y) d y
\end{aligned}
$$

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

(4) $$

\begin{aligned}

y=\sqrt{x+2}, \quad y=x \quad \text { revolving about } x \text {-axis }\\ \\

\end{aligned}

$$

Solving

$$y=\sqrt{y+2} \Rightarrow y^{2}=y+2 \Rightarrow y^{2}-y-2=0$$

$$y^{2}-2 y+y-2=0 $$ $$ y(y-2)+1(y-2)=0$$

$$(y-2)(y+1)=0 $$ $$ \Rightarrow y-2=0, y+1=0$$

$$\Rightarrow y=2, \quad y=-1$$   Neglecting   $y=-1 $ because curves lie in the first quadrant

$$\therefore \quad c=0, \quad d=2$$

 

$$
\begin{aligned}
\text { Solving } y & =\sqrt{x+2} \quad \text { for } x \\ \\
y^{2} & =x+2 \\ \\
x & =y^{2}-2
\end{aligned}
$$

Thus we have

$$
\begin{aligned}
& x=y^{2}-2, \quad x=y, \quad \text { revolving about } x \text {-axis } \\ \\
& c=0 \quad d=2 \\ \\
& V=\int_{c}^{d} 2 \pi y\left[y-\left(y^{2}-2\right)\right] d y \\ \\
& V=\int_{c} 2 \pi y\left[y-y^{2}+2\right] d y \\ \\
& V=\int_{0}^{2} 2 \pi\left[y^{2}-y^{3}+2 y\right] d y \\ \\
& V=2 \pi\left[\frac{y^{3}}{3}-\frac{y^{4}}{4}+2 \frac{y^{2}}{2}\right]_{0}^{2} \\ \\
& V=2 \pi\left[\frac{2^{3}}{3}-\frac{2^{4}}{4}+2 \frac{(2)^{2}}{2}\right]-0 \\ \\
& V=2 \pi\left[\frac{8}{3}-\frac{16}{4}+4\right] \\ \\
& V=2 \pi\left[\frac{8}{3}-4+4\right]=2 \pi(8 / 3)=\frac{16 \pi}{3}
\end{aligned}
$$

(5)

$$
\begin{aligned}

& y=x^{3}, x=1, y=0 \quad \text { revolving about } x \text {-axis } \\ \\
& \text { When } y=0, \text { then } x=0 \\ \\
& \therefore \quad a=0, \quad b=1 \\ \\
& V=\int_{0}^{1} 2 \pi x\left(x^{3}\right) d x=2 \pi \int_{0}^{4} x^{4} d x=2 \pi\left[\frac{x^{5}}{5}\right]_{0}^{1} \\ \\
& V=\frac{2 \pi}{5}(1-0)=\frac{2 \pi}{5}
\end{aligned}
$$

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