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

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

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

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

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