Surface Area of Curve about y-axis

Question

I'm trying to rotate the curve $$ \frac{1}{4} x^{2}-\frac{1}{2} \ln x $$ with $$ 1 \leq x \leq 2 $$ about the y-axis. I know the formula for this is: $$ \int_{a}^{b}\left(2 \pi f(y) \sqrt{1+\left(f^{\prime}(y)\right)^{2}}\right) d y $$ The only difficulty I'm having is getting the formula in terms of x. How should I go about this?

Answer 1

There is solution to the problem but it is quite complex.

Since $x>1$ rewrite $$y=\frac 14 x^2-\frac 12 \log(x)=\frac 14 x^2-\frac 14 \log(x^2)$$ The inverse is given in terms of Lambert function $$x=\sqrt{-W_{-1}\left(-e^{-4 y}\right)}$$ $$\frac {dx}{dy}=-\frac{2 \sqrt{-W_{-1}\left(-e^{-4 y}\right)}}{W_{-1}\left(-e^{-4 y}\right)+1}$$ All of that makes the integrand to be $$2\pi\sqrt{-W_{-1}\left(-e^{-4 y}\right)} \sqrt{1-\frac{4 W_{-1}\left(-e^{-4 y}\right)}{\left(W_{-1}\left(-e^{-4 y}\right)+1\right){}^2}}$$ which (thanks to a CAS) can be integrated (have a look here).