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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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