W04 Stepwise

Due date: Thursday 2/5, 11:59pm

Arc length

01

01

Arc length - reversed $x$ and $y$ roles

Find the arc length of the curve that satisfies the equation $x=\frac{1}{12}{y}^3+y^{-1}$ over $y\in [\mathrm{1,2}]$.

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Solution

Solutions - 0080-01

(1) Integral formula for arclength:

$$S={\int }_c^d\sqrt{1+{(g^{\prime })}^2}dx$$

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(2) Work out integrand:

$$\begin{array}{c}g(x)=\frac{1}{12}{y}^3+y^{-1}\gg \gg g^{\prime }(x)=\frac{1}{4}{y}^2-y^{-2}\\ \\ {(g^{\prime })}^2\gg \gg \frac{1}{16}{y}^4-\frac{1}{2}+y^{-4}\\ \\ 1+{(g^{\prime })}^2\gg \gg \frac{1}{16}{y}^4+\frac{1}{2}+y^{-4}\gg \gg {(\frac{1}{4}{y}^2+y^{-2})}^2\\ \\ \sqrt{1+{(g^{\prime })}^2}\gg \gg \frac{1}{4}{y}^2+y^{-2}\end{array}$$

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(3) Integrate:

$$\begin{array}{c}S\gg \gg {\int }_1^2\frac{1}{4}{y}^2+y^{-2}dy\\ \\ \gg \gg {\frac{1}{12}{y}^3-y^{-1}|}_1^2\gg \gg \frac{13}{12}\end{array}$$Link to original

Surface areas of revolutions - thin bands

02

01

Surface area: revolved cubic

The curve $y=x^3$ over $x\in [\mathrm{0,2}]$ is revolved around the $x$-axis.

Find the area of the resulting surface.

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Solution

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:

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