While tutoring a student in introductory calculus, I inquired as to how she would set up an integral to determine the volume of the solid created by revolving the region bounded by the functions $\sin(x)$, $\ln(x)$, and the $x$-axis about the line $y=5$. She set up the integral as follows: $$\pi\int_{1}^{a} 25-(\ln(x))^2 dx + \pi\int_{a}^{\pi} 25-\sin^2(x)dx $$ where $a$ is the solution to $\sin(x) = \ln(x)$. However, I couldn't give an explanation as to how to find this point. Is this something that could be solved given two years of introductory calculus, or does it require a larger toolkit? In searching for an answer to this problem online, I was bombarded by hits on $\ln(x)$ differentiation resources which obviously weren't helpful. Any light you could shed would be appreciated!

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    $\begingroup$ It might just so happen that it will disappear in the final calculations. Maybe it does this time $\endgroup$ – Jakobian Mar 21 at 23:43
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    $\begingroup$ It's a numerical matter, it's not likely that there's any sort of closed form solution. Newton's method should be fine, or even a simple binary search. WolframAlpha tells us the answer is about $2.21911$. $\endgroup$ – lulu Mar 21 at 23:44
  • $\begingroup$ $\displaystyle x \in \left[{1 \over \mathrm{e}},\mathrm{e}\right]$. $\endgroup$ – Felix Marin Mar 22 at 5:52

Numerical methods are the way to go. Clearly there would be only one solution which can be figured by considering the monotonicity of $\ln x$ and the boundedness of $\sin x$. It would help if you're allowed to use a calculator.

Define $f(x)=\sin x-\ln x$. By Newton Raphson method we have the following iterative formula. $$x_{n+1}=x_n-\dfrac{\sin x_n-\ln x_n}{\cos x_n-1/x_n}$$

I'm going to make a first guess say $x_0=2$. Now use the iteration formula to obtain an accurate value of the solution: $$\begin{array}{|p{3cm}||p{3cm}|p{3cm}|p{3cm}|}\hline n & x_n \\ \hline 0 & 2 \\ 1 & 2.23593406389 \\ 2& 2.21918552153 \\ 3& 2.21910715064\\\hline \end{array}$$

  • $\begingroup$ No, You can't be a class 10 student (from your profile) , I am sure about that... $\endgroup$ – Darkrai Mar 22 at 6:57
  • $\begingroup$ Well I am @Darkrai $\endgroup$ – Paras Khosla Mar 22 at 6:57
  • $\begingroup$ What!!!! Which school/University? $\endgroup$ – Darkrai Mar 22 at 6:59
  • $\begingroup$ I'm in Cambridge School in Punjab. Why do you ask? $\endgroup$ – Paras Khosla Mar 22 at 7:00
  • $\begingroup$ And where are you pursuing your Complex Analysis course? I am asking because most of the schools don't offer a "Complex Analysis" course to a class 10 student in India... $\endgroup$ – Darkrai Mar 22 at 7:00

If the student is allowed a graphing calculator, then you can set plot y=sin(x) and y=ln(x) on the graph and use the calculator to find the intersection (I'm a high-schooler and I did this to find intersections in Calc 2)


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