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Find the area of the surface obtained by rotating the arc \(y=x^{3}\) where \(0 \leq x \leq 2\) about the x-axis

area \(=S=2 \pi \int y d s=2 \pi \int y \sqrt{1+\left(y^{\prime}\right)^{2}} d x\)

(1) \(y=x^{3} \Rightarrow y^{\prime}=3 x^{2} \Rightarrow\left(y^{\prime}\right)^{2}=\left(3 x^{2}\right)^{2}=9 x^{4} \Rightarrow 1+(y)^{2}=1+9 x^{4}\)

\(\Rightarrow \sqrt{1+\left(y^{\prime}\right)^{2}}=\sqrt{1+9 x^{4}} \Rightarrow \text { area }=s=2 \pi \int_{0}^{2} x^{3} \sqrt{1+9 x^{4}} d x\)

let \(t=1+9 x^{4} \Rightarrow d t=36 x^{3} d x\)

when \(x=0 \Rightarrow t=1\)

when \(x=2 \Rightarrow t=1+9(2)^{4}=145\)

area \(=S=2 \pi \int_{1}^{145} x^{3} \sqrt{t} d x \frac{36}{36}=\frac{2 \pi}{36} \int_{1}^{145} \sqrt{t} \cdot {36 x^{3} d x}\)

\(=\frac{\pi}{18} \int_{1}^{145} t^{1 / 2} d t=\frac{\pi}{18}\left.\frac{t^{3 / 2}}{3 / 2}\right|_{1} ^{145}\)

\(=S=\frac{\pi}{18} \cdot \frac{2}{3} \cdot\left.t^{3 / 2}\right|_{1} ^{145}=\frac{\pi}{27} t^{3 / 2}|_{1} ^{145}=\)

\(=\frac{\pi}{27}\left[(145)^{3 / 2}-1\right]\)

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