W04 Regular

Due date: Sunday 2/8, 11:59pm

Arc length

01

03

Arc length - tricky integration

Find the arc length of the curve $y=e^x$ for $x\in [\mathrm{0,1}/2]$.

(Hint: the integral can be done using either: (i) $u$-sub then trig sub, or (ii) ‘rationalization’ then partial fractions.)

Link to original

Solution

Solutions - 0080-03

(1) Integral formula for arclength:

$$S={\int }_a^b\sqrt{1+{(f^{\prime })}^2}dx\gg \gg {\int }_0^{1/2}\sqrt{1+e^{2x}}dx$$

---

(2) Perform $u$-sub with $u=\sqrt{1+e^{2x}}$ and so $e^{2x}=u^2-1$ and also:

$$\begin{array}{ccc}& u=\sqrt{1+e^{2x}}& \text{(Note A)}\\ & & \\ & \gg \gg e^{2x}=u^2-1,du=\frac{e^{2x}}{\sqrt{1+e^{2x}}}dx& \\ & & \\ & \gg \gg dx=\frac{u}{u^2-1}du& \end{array}$$

Now transform the integral to $u$:

\int_0^{1/2} \sqrt{1+e^{2x}}\,dx \quad \gg\gg \quad \int_\sqrt{2}^\sqrt{1+e} \frac{u^2}{u^2-1}\,du ParseError: Got function '\sqrt' with no arguments as subscript at position 58: …\gg \quad \int_\̲s̲q̲r̲t̲{2}^\sqrt{1+e} …

Note A: Instead of this $u$-sub and partial fractions, one can set $e^x=u$ and obtain \int_1^\sqrt{e}\frac{\sqrt{1+u^2}}{u}\,du ParseError: Got function '\sqrt' with no arguments as superscript at position 8: \int_1^\̲s̲q̲r̲t̲{e}\frac{\sqrt{…. Then trig sub with $u=\tan \theta$ leads to (eventually) the same final answer.

---

(3) Integrate: partial fraction decomposition:

Number degree not lower → long division first:

$$\frac{u^2}{u^2-1}\gg \gg 1+\frac{1}{u^2-1}\gg \gg 1+\frac{1}{(u+1)(u-1)}$$

Write general PFD formula:

$$\frac{1}{(u+1)(u-1)}=\frac{A}{u+1}+\frac{B}{u-1}$$

Solve for $A$ and $B$. Cross multiply:

$$1=A(u-1)+B(u+1)$$

• $x=-1\gg \gg A=-1/2$
• $x=1\gg \gg B=1/2$

$${\int }_{\sqrt{2}}^{\sqrt{1+e}}\frac{u^2}{u^2-1}du\gg \gg {\int }_{\sqrt{2}}^{\sqrt{1+e}}1+\frac{-1/2}{u+1}+\frac{1/2}{u-1}du$$

---

(4) Evaluate integral:

$$\begin{array}{ccc}& {\int }_{\sqrt{2}}^{\sqrt{1+e}}1-\frac{1}{2(u+1)}+\frac{1}{2(u-1)}du& \\ & & \\ & \gg \gg {u-\frac{1}{2}\ln |u+1|+\frac{1}{2}\ln |u-1||}_{\sqrt{2}}^{\sqrt{1+e}}& \\ & & \\ & \gg \gg [\sqrt{1+e}-\frac{1}{2}\ln (\sqrt{1+e}+1)+\frac{1}{2}\ln (\sqrt{1+e}-1)]& \\ & & \\ & -[\sqrt{2}-\frac{1}{2}\ln (\sqrt{2}+1)+\frac{1}{2}\ln (\sqrt{2}-1)]& \text{(Note B)}\\ & & \\ & \gg \gg \sqrt{1+e}-\sqrt{2}+\frac{1}{2}(\ln \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)+\ln \left(\frac{\sqrt{1+e}-1}{\sqrt{1+e}+1}\right))& \end{array}$$

Note B: This answer is sufficient. It is not necessary to simplify as in the last step.

---

Alternative:

(1) Sub $u=e^x$, $du=e^{x}dx$, $dx=\frac{1}{u}du$:

$${\int }_0^{1/2}\sqrt{1+e^{2x}}dx\gg \gg {\int }_1^{\sqrt{e}}\frac{\sqrt{1+u^2}}{u}du$$

---

(2) Trig sub $u=\tan \theta$, $du={\sec }^2\theta d\theta$ and simplify the integrand:

$$\begin{array}{c}\frac{\sqrt{1+{\tan }^2\theta }}{\tan \theta }\cdot {\sec }^2\theta d\theta \gg \gg \frac{\sec \theta }{\tan \theta }(1+{\tan }^2\theta )d\theta \\ \\ \gg \gg (\frac{\sec \theta }{\tan \theta }+\sec \theta \tan \theta )d\theta \gg \gg (\csc \theta +\tan \theta \sec \theta )d\theta \end{array}$$

---

(3) Integrate:

$$\int (\csc \theta +\tan \theta \sec \theta )d\theta \gg \gg \ln |\csc \theta -\cot \theta |+\sec \theta +C$$

---

(4) Back-substitute $\csc \theta =\frac{\sqrt{u^2+1}}{u}$, $\cot \theta =\frac{1}{u}$, $\sec \theta =\sqrt{u^2+1}$:

$$\gg \gg \ln \left|\frac{\sqrt{u^2+1}-1}{u}\right|+\sqrt{u^2+1}+C$$

---

(5) Evaluate at the $u$-bounds:

$$\begin{array}{c}\gg \gg \ln \left|\frac{\sqrt{u^2+1}-1}{u}\right|+\sqrt{u^2+1}{|}_1^{\sqrt{e}}\\ \\ \gg \gg \ln \left(\frac{\sqrt{e+1}-1}{\sqrt{e}}\right)+\sqrt{e+1}-\ln (\sqrt{2}-1)-\sqrt{2}\\ \\ \gg \gg \ln (\sqrt{e+1}-1)-\frac{1}{2}+\sqrt{e+1}-\ln (\sqrt{2}-1)-\sqrt{2}\\ \\ \gg \gg \approx 0.8210\end{array}$$Link to original

Surface areas of revolutions - thin bands

02

02

Surface area: cone

A cone may be described as the surface of revolution of a ray emanating from the origin, revolved around the $x$-axis.

Let $f(x)=mx$ for some $m>0$. Find the surface area of the cone given by revolving the graph of $f$ around the $x$-axis over $x\in [0,h]$.

Now calculate this area using geometry, and verify that the two methods give the same formula. (Hint: ‘unroll’ the cone into a sector.)

Link to original

Solution

Solutions - 0090-02

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

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

---

(2) Work out integrand:

$$\begin{array}{c}f(x)=mx\gg \gg f^{\prime }(x)=m\\ \\ S\gg \gg {\int }_0^{h}2\pi x\sqrt{1+m^2}dx\end{array}$$

---

(4) Evaluate integral:

$$\begin{array}{c}{\int }_0^{h}2\pi mx\sqrt{1+m^2}dx\\ \gg \gg 2\pi m\sqrt{1+m^2}{\int }_0^{h}xdx\\ \gg \gg 2\pi m\sqrt{1+m^2}{\left(\frac{x^2}{2}\right)|}_0^h\\ \gg \gg \pi h^{2}m\sqrt{1+m^2}\end{array}$$

---

(5) Verify with geometry:

Note that unrolling the cone forms a sector with radius $h\sqrt{1+m^2}$ and arc length $2\pi mh$. The total circumference is $2\pi h\sqrt{1+m^2}$. So the area of the sector is:

$$\begin{array}{c}A=\frac{\text{sector arc}}{\text{total arc}}\cdot \text{area circle}\\ \\ \gg \gg \frac{2\pi mh}{2\pi h\sqrt{1+m^2}}\cdot \pi {\left(h\sqrt{1+m^2}\right)}^2\\ \\ \gg \gg \pi h^{2}m\sqrt{1+m^2}\end{array}$$

Notes:

• Sector radius is the lateral length of the cone: hypotenuse of right triangle with legs $h$ (on $x$-axis) and $mh$ (on $y$-axis).
• Sector arc is $2\pi mh$ because $mh$ is radius of the base.

Link to original

03

03

Surface area: parabolic reflector

A parabolic reflector is given by rotating the curve $y=x^2$ around the $y$-axis for $x\in [\mathrm{0,2}]$.

What is the surface area of this reflector?

Link to original

Solution

Solutions - 0090-03

Method 1: integrate in $x$

(1) Integral formula for surface area:

$$S={\int }_a^{b}2\pi f(x)\sqrt{1+{(f^{\prime })}^2}dx\gg \gg 2\pi {\int }_0^{2}x\sqrt{1+4x^2}dx$$

---

(2) Integrate: perform $u$-sub with $u=1+4x^2$ and so $du=8xdx$:

$$\begin{array}{c}S\gg \gg \frac{\pi }{4}{\int }_1^{17}\sqrt{u}du\\ \\ \gg \gg \frac{\pi }{4}{\frac{2}{3}{u}^{\frac{3}{2}}|}_1^{17}\\ \\ \gg \gg \frac{\pi }{6}({17}^{3/2}-1)\end{array}$$

---

Method 2: integrate in $y$

(1) Integral formula for surface area using $g(y)=\sqrt{y}$:

$$\begin{array}{c}S={\int }_c^{d}2\pi g(y)\sqrt{1+{(g^{\prime })}^2}dy\gg \gg {\int }_0^{4}2\pi \sqrt{y}\sqrt{1+\frac{1}{4}{y}^{-1}}dy\\ \\ \gg \gg \pi {\int }_0^4\sqrt{4y+1}dy\end{array}$$

---

(2) Integrate: perform $u$-sub with $u=4y+1$ and so $du=4dy$:

$$\begin{array}{c}\pi {\int }_0^4\sqrt{4y+1}dy\gg \gg \frac{\pi }{4}{\int }_1^{17}\sqrt{u}du\\ \\ \gg \gg \frac{\pi }{4}{\frac{2}{3}{u}^{\frac{3}{2}}|}_1^{17}\\ \\ \gg \gg \frac{\pi }{6}({17}^{3/2}-1)\end{array}$$Link to original