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I'm doing some calculus and don't get what they did in the solution, maybe someone can tell me :

$$\int_0^Rdr \int_0^{2\pi}\sqrt{r^2+a^2} \ d\phi =2\pi^2a \int_0^{\text{arcsinh}(R/a)} \cosh^2x \ dx \quad $$

where $f(r,\phi)=(r\cos\phi,r \sin\phi,a\phi)$

How did they got from the left hand-side to the right-handside ?

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The right-hand side is not quite correct. It should read $2\pi a^2\int_0^{\text{arsinh}(R/a)}\cosh^2(x)\,dx$.

To obtain this expression, use the classical hyperbolic substitution $r=a\sinh(x)$ so that $dr=a\cosh(x)$ and $\sqrt{r^2+a^2}=a\cosh(x)$.

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    Thanks, did they used that $cosh(arsinh(x))=\sqrt(x^2+1)$ ? – 2017-02-15
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    You're welcome. It is true that $\cosh(\text{arsinh}(x))=\sqrt{1+x^2}$. But we don't need this here. We only need $\cosh^2(x)-\sinh^2(x)=1$ and $\frac{d\sinh(x)}{dx}=\cosh(x)$. – 2017-02-15