Solutions - 0090-01

(1) Integral formula for surface area, revolution about $x$-axis:

$$S={\int }_a^{b}2\pi f(x)\sqrt{1+{\left(f^{\prime }(x)\right)}^2}dx$$

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(2) Work out integrand:

$$a=0,b=2,f(x)=x^3,f^{\prime }(x)=3x^2$$

Then:

$$\sqrt{1+{(f^{\prime })}^2}\gg \gg \sqrt{1+9x^4}$$

So:

$$S\gg \gg {\int }_0^{2}2\pi x^3\sqrt{1+9x^4}dx$$

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(3) Perform $u$-sub with $u=1+9x^4$ and so $du=36x^{3}dx$:

$${\int }_0^{2}2\pi x^3\sqrt{1+9x^4}dx\gg \gg \frac{\pi }{18}{\int }_1^{145}\sqrt{u}du$$

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(4) Integrate:

$$\begin{array}{cc}\frac{\pi }{18}{\int }_1^{145}\sqrt{u}du& \gg \gg \frac{\pi }{27}{\left[u^{3/2}\right]}_1^{145}\\ & \\ & \gg \gg \frac{\pi }{27}(145\sqrt{145}-1)\end{array}$$