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<x_2\le\pi} 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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This entry was posted on Sunday, October 18th, 2009 at 1:27 pm and is filed under 175: Calculus II. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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