The point $P(a,b)$ lies on the curve $y=\mathrm{arsinh}\,x$. $R$ is the region bounded by the curve, the $x$- and $y$-axes and the line $x=a$. When $R$ is rotated $2\pi$ radians about the $x$-axes the solid formed has volume $\pi(b-\sqrt2)^2$ cubic units. Find the coordinates for point $P$, give your answer in exact form.
I've integrated $\pi\,\mathrm{arsinh}^2\,x$ and I got $$\pi\,\mathrm{arsinh}\, x(x\,\mathrm{arsinh}\, x-2\sqrt{1+x^2}+2x)$$ but I don't know what to do from there. Please help.
To find the volume, plug in $a$ into the antiderivative that you computed, and take away the result of plugging in $0$. Since $b=\text{arcsinh}\,a$ we get that the volume is $\pi(b^2-2b\sqrt{1+a^2}+2a)$. But we know the volume, it is $\pi(b^2-2b\sqrt{2}+2)$. Set the computed volume equal to the known volume, and do the obvious cancellations. We get $$-2b\sqrt{1+a^2}+2a=-2b\sqrt{2}+2.$$ Note that $a=1$ is a solution of this equation. I think there is no other solution, but have not verified it.