Science:Math Exam Resources/Courses/MATH103/April 2013/Question 04 (b)

MATH103 April 2013

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[hide] Question 04 (b)
Torricelli's Trumpet The surface area of a solid of revolution obtained from revolving $\displaystyle y=f(x)$ about the x-axis for $\displaystyle a\leq x\leq b$ is given by $\displaystyle A=2\pi \int _{a}^{b}y{\sqrt {1+{\big (}{\frac {dy}{dx}}{\big )}^{2}}}\,dx$ Consider the surface area of the surface obtained by revolving $\displaystyle y=1/x$ about the x-axis, where x has domain $\displaystyle [1,\infty )$ . Show whether this surface area exists (integral converges) or diverges. If it exists, provide an upper bound.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
First remind yourself of the comparison test for integrals (its similar to that for series). If $\displaystyle 0\leq f(x)\leq g(x)$ and if $\displaystyle \int _{a}^{b}g(x)\,dx$ converges then $\displaystyle \int _{a}^{b}f(x)\,dx$ also converges. If instead $\displaystyle \int _{a}^{b}f(x)\,dx$ diverges, then $\displaystyle \int _{a}^{b}g(x)\,dx$ diverges

[show] Hint 2
Plug in $\displaystyle y=1/x$ into the surface area formula, we know that the square root function has to be at least one on the domain in question. Use this to your advantage.

[show] Hint 3
Show this integral diverges.

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[show] Solution
In our scenario, the surface area is $\displaystyle A=2\pi \int _{1}^{\infty }{\frac {1}{x}}{\sqrt {1+{\big (}{\frac {dy}{dx}}{\big )}^{2}}}\,dx$ As mentioned in the hint, the square root must always be at least 1 on the domain $\displaystyle [1,\infty )$ since $\displaystyle {\frac {1}{x}}{\sqrt {1+{\big (}{\frac {dy}{dx}}{\big )}^{2}}}={\frac {1}{x}}{\sqrt {1+{\big (}-{\frac {1}{x^{2}}}{\big )}^{2}}}={\frac {1}{x}}{\sqrt {1+{\frac {1}{x^{4}}}}}>{\frac {1}{x}}$ Now, since ${\begin{aligned}\int _{1}^{\infty }{\frac {1}{x}}\,dx=\lim _{b\to \infty }\int _{1}^{b}{\frac {1}{x}}\,dx=\lim _{b\to \infty }\ln \|x\|{\big \|}_{1}^{b}=\lim _{b\to \infty }\ln \|b\|-\ln(1)\end{aligned}}$ diverges, we have that our integral diverges by the integral comparison test. Thus the surface area diverges. NOTE: Even though you can fill the horn with a finite amount of paint, you cannot paint the inside with a finite amount of paint.