Surface Area of the solid by revolution of a logarithm function around $y$-axis?

Question

I am trying to get the surface area of this equation in $y$ axis but I can't integrate it.

$$ \begin{align*} x=f(y)&= \frac{7\ln \left(4^\frac{5}{7}y\right)}{\ln(4)}\\ \frac{dx}{dy}=f'(y) &= \frac{7}{y\ln4} \end{align*}$$ so to get the surface area $$ \begin{align*}S_x&=\int_a^b2\pi x\,\sqrt{1+\Big(\frac{dx}{dy}\Big)^2}\,dy\\ &=\int_1^42\pi \left(\frac{7\ln(4^\frac{5}{7}y)}{\ln4}\right)\,\sqrt{1+\Big(\frac{7}{y\ln4}\Big)^2}\,dy\\ &=\frac{7\cdot 2\pi}{\ln4} \int_1^4\ln \left(4^\frac{5}{7}y\right)\,\sqrt{1+\Big(\frac{7}{y\ln4}\Big)^2}\,dy \end{align*}$$ how can I integrate this? I have spent nearly 3 hours but I can't integrate it.

Answer 1

I suppose this integral is very hard to be evaluated (or cannot be evaluated) by analytical methods. You need to approximate it using the Simpson's rule.

For your reference, here is a link (Check question no. 17) to a similar question I found that uses the rule to evaluate the surface area obtained from the revolution of a logarithm function. Below is the quote from the same link.

$y=ln(x), 1\leq x \leq 3.$

$\begin{align*}S=\int_1^32\pi\ln(x)\sqrt{1+\frac{1}{x^2}}\,dx.\end{align*}$

Let $f(x)=\ln(x)\sqrt{1+\frac{1}{x^2}}$ , Since $n=10$, $\Delta x=\frac{3-1}{10}=\frac{1}{5}.$

Then $S=S_{10}=2\pi\frac{1/5}{3}[f(1)+4f(1.2)+2f(1.4)+..+2f(2.6)+4f(2.8)+f(3)]\approx9.023754.$

The quoted result is correct to six decimal places. I hope it helps a little.