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The curve $y=\sqrt{4-x^{2}}$ , $-1 \leq X \leq 1$ is an arc of the circle $x^{2}+y^{2}=4$ Find the area of the surface obtained by rotating this arc about the x-axis

area $=2 \pi \int y d s=2 \pi \int_{-1}^{1} y \sqrt{1+\left(\frac{dy}{d x}\right)^{2}} d x$

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

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

(3) $1+\left(y^{\prime}\right)^{2}=1+\left(\frac{-x}{\sqrt{4-x^{2}}}\right)^{2}=1+\frac{x^{2}}{4-x^{2}}=\frac{4-x^{2}}{4-x^{2}}+\frac{x^{2}}{4-x^{2}}=\frac{4}{4-x^{2}}$

area $=2 \pi \int_{-1}^{1} y \cdot \sqrt{1+y^{\prime 2}} d x=2 \pi \int_{-1}^{1} \sqrt{4-x^{2}} \cdot \sqrt{\frac{4}{4-x^{2}}} d x$

=$2 \pi \int_{-1}^{1} \sqrt{4-x^{2}} \cdot \frac{2}{\sqrt{4-x^{2}}} d x=2 \pi \int_{-1}^{1}2 d x=4 \pi \int_{-1}^{1} d x=4 \pi x |_{-1}^{1}$

$=4 \pi ({1} )-4 \pi(-1)=4 \pi+4 \pi=8 \pi$

Find the area of the surface obtained by the rotating the arc $4 x=y^{2}$ between (0,0) and (1,2) around the x-axis

area $=2 \pi \int y d s=2 \pi \int_{0}^{2} y \cdot \sqrt{1+\left(\frac{{dx}}{d y}\right)^{2}} d y$

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

area $=S=2 \pi \int_{0}^{2} y \cdot \sqrt{1+\frac{y^{2}}{4}} d y$

let $t=1+\frac{y^{2}}{4} \Rightarrow d t=\frac{y}{2} d y$

when $y= 0\rightarrow t=1$

when $y=2 \rightarrow t=2$

area $=s=2 \pi \int_{1}^{2} \sqrt{t}(2) \frac{1}{2} y \cdot d y$

$=2 \pi \cdot 2 \int_{1}^{2} t^{\frac{1}{2}} \cdot \frac{1}{2} y d y=4 \pi \int_{1}^{2} t^\frac{1}{2} dt$

$S=\left.\frac{4 \pi t^{\frac{3}{2}}}{\frac{3}{2}}\right|_{1} ^{2}=4 \pi \times \frac{2}{3} t^{3 / 2} |_{1}^{2}=\frac{8 \pi}{3}\left[2^{\frac{3}{2}}-1^{3 / 2}\right]$

$=\frac{8 \pi}{3}[\sqrt{8}-1]$