## 175 – Quiz 4

Here is quiz 4.

Problem 1 is Exercise 7.1.36 from the book. Here is a graph showing the curve ${y=x\sin x}$ for ${0\le x\le\pi.}$

We rotate about the line ${x=\pi}$ the region bounded by the ${x}$-axis and this curve.

To find its volume, a first natural attempt would be to use the washer method. We would then attempt to compute the volume as

$\displaystyle \int_0^{Max}\pi((\pi-x_1)^2-(\pi-x_2)^2)\,dy,$

where ${Max}$ is the maximum of ${y}$ for ${0\le x\le \pi}$ and, for any given value of ${y}$ with ${0\le y\le Max,}$ ${x_1}$ and ${x_2}$ are the values of ${x}$ with ${0\le x_1 such that ${x\sin x=y.}$

Unfortunately, this does not seem to be the best approach, as there does not seem to be a reasonable way of solving for ${x}$ the equation ${x\sin x=y.}$

(In fact, this equation cannot be solved in terms of elementary functions.

Similarly, there is no way of finding exactly what the value of ${Max}$ is in terms of elementary functions.)

Since this approach seems to lead us nowhere, we now try to compute the volume using the shell method. Now the volume is expressed as

$\displaystyle \int_0^\pi 2\pi(\pi-x)x\sin x\,dx.$

This expression looks approachable with the techniques we have studied. First, let’s rewrite the integral as

$\displaystyle 2\pi \int_0^\pi\pi x\sin x\,dx-2\pi\int_0^\pi x^2\sin x\,dx.$

We compute both expressions using integration by parts:

To find ${\displaystyle \int x\sin x\,dx,}$ we use ${u=x}$ and ${dv=\sin x\,dx,}$ so ${du=dx}$ and we can take ${v=-\cos x.}$ Hence

$\displaystyle \int_0^\pi\pi x\sin x=-\pi x\cos x\vert_0^\pi+\int_0^\pi \pi\cos x\,dx.$

We recognize from the graph of ${\cos x}$ that the second expression is zero, and we have:

$\displaystyle \int_0^\pi\pi x\sin x=-\pi (-\pi-0)=\pi^2.$

Similarly, for ${\displaystyle \int x^2\sin x\,dx}$ we have ${u=x^2}$ and ${dv=\sin x\,dx,}$ so ${du=2x\,dx}$ and ${v=-\cos x,}$ and

$\displaystyle \int_0^\pi x^2\sin x=-x^2\cos x\vert_0^\pi+\int_0^\pi 2x\cos x\,dx$ $\displaystyle=\pi^2+2\int_0^\pi x\cos x\,dx.$

The last expression ${\displaystyle \int x\cos x\,dx}$ is once more computed using parts, now with ${u=x}$ and ${dv=\cos x\,dx,}$ so ${du=dx}$ and ${v=\sin x.}$ This gives

$\displaystyle \int_0^\pi x\cos x\,dx=x\sin x\vert_0^\pi-\int_0^\pi\sin x\,dx=0+\cos x\vert_0^\pi=-2.$

Hence

$\displaystyle \int_0^\pi x^2\sin x\,dx=\pi^2-4.$

Finally, the required volume is

$\displaystyle 2\pi(\pi^2)-2\pi(\pi^2-4)=8\pi.$

Problem 2 asked to evaluate ${\displaystyle \int_0^3\sqrt{x^2+6x}\,dx.}$

A first attempt may go by using integration by parts, with ${u=\sqrt x}$ and ${dv=\sqrt{x+6}\,dx.}$ Unfortunately, this approach would not lead to simpler expressions, as both integrals and derivatives of ${\sqrt x}$ and ${\sqrt{x+6}}$ carry radicals.

If the expression inside the square root were of the form ${x^2+a^2}$ or ${x^2-a^2}$ we could use a trigonometric substitution. However, ${x^2+6x}$ is not of this form. On the other hand, in Chapter 9 we saw that it is sometimes useful to complete squares, so we may want to try that here. We have:

$\displaystyle x^2+6x=x^2+6x+9-9=(x+3)^2-9.$

This suggest trying the trigonometric substitution ${x+3=3\sec\theta}$ for ${0\le\theta<\pi/2.}$ We have ${dx=3\sec\theta\tan\theta\,d\theta}$ and${\sqrt{x^2+6x}=3\tan\theta.}$ Also, when ${x=0}$ we have ${\sec\theta=1,}$ or ${\theta=0,}$ and when ${x=3}$ we have ${\sec\theta=2,}$ or ${\theta=\pi/3.}$

In terms of ${\theta,}$ the integral becomes

$\displaystyle \int_0^{\pi/3}3\tan\theta\,3\sec\theta\tan\theta\,d\theta=9\int_0^{\pi/3}\sec\theta\tan^2\theta\,d\theta.$

To evaluate expressions of this form, we use the identity ${\tan^2\theta=\sec^2\theta-1,}$ and obtain

$\displaystyle \int\sec\theta\tan^2\theta\,d\theta=\int\sec^3\theta\,d\theta-\int\sec\theta\,d\theta.$

The second expression we recognize as ${-\ln|\sec\theta+\tan\theta|+ C_1.}$ For the first, we use either the reduction formula found in lecture, or integration by parts:

$\displaystyle \int\sec^3\theta\,d\theta=\int\sec\theta\sec^2\theta\,d\theta$ $\displaystyle=\sec\theta\tan\theta-\int\tan\theta\sec\theta\tan\theta\,d\theta,$

or ${\displaystyle \int\sec^3\theta\,d\theta=}$

$\displaystyle \sec\theta\tan\theta-\int(\sec^2\theta-1)\sec\theta\,d\theta$ $\displaystyle=\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|-\int\sec^3\theta\,d\theta,$

from which we get

$\displaystyle \int\sec^3\,d\theta=\frac{\sec\theta\tan\theta}2+\frac12\ln|\sec\theta+\tan\theta| + C_2.$

Finally,

$\displaystyle 9\int_0^{\pi/3}\sec\theta\tan^2\theta\,d\theta$ $\displaystyle=9\left(\left(\frac{2\sqrt3}2-\frac12\ln(2+\sqrt3)\right)-\left(\frac02-\frac12\ln(1+0)\right)\right),$

so the required integral equals ${\displaystyle 9\sqrt3-\frac{9}2\ln(2+\sqrt3).}$

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