Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 468: 21

$$ \ln (5+\sqrt{26})$$

Since \begin{align*} s&=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x\\ &=\int_{-a}^{a} \sqrt{1+(\sinh x)^{2}} d x\\ &=\int_{-a}^{a} \sqrt{1+\sinh ^{2} x} d x\\ &=\int_{-a}^{a} \sqrt{\cosh ^{2} x} d x\\ &=\int_{-a}^{a} \cosh x d x\\ &=[\sinh x]_{-a}^{a}\\ &=2 \sinh a \end{align*} Setting this expression equal to 10 and solving for $a$ yields $$a=\sinh ^{-1}(5)=\ln (5+\sqrt{26})$$