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Consider the integral

$$4\pi\int_{0}^{1}\cosh(t)\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}{\mathrm{d}t}$$

This definite integral arose while computing the surface area of a hyperboloid. The hyperboloid is parameterized by $$ \begin{align*} {x}&=\cosh({t})\cos(\theta)\text{;}\\ {y}&=\cosh({t})\sin(\theta)\text{;}\\ {z}&=\sinh({t})\\ \end{align*} $$ for $$ \begin{align*} 0&\leq\theta\leq{2}\pi\text{;}\\ -1&\leq{t}\leq{1}\text{.}\\ \end{align*} $$ Recall that $\cosh^{2}({t})-\sinh^{2}({t})=1$. The position vector $\vec{r}$ is given by $$ \vec{r}({x}({t},\theta),{y}({t},\theta),{z}({t},\theta))=\cosh({t})\cos(\theta)\hat{\imath}+\cosh({t})\sin(\theta)\hat{\jmath}+\sinh({t})\hat{k}\text{,} $$ and its partial derivatives are $$ \begin{align*} \frac{\partial\vec{r}}{\partial{t}}&=\sinh({t})\cos(\theta)\hat{\imath}+\sinh({t})\sin(\theta)\hat{\jmath}+\cosh({t})\hat{k}\text{;}\\ \frac{\partial\vec{r}}{\partial\theta}&=\cosh({t})(-\sin(\theta))\hat{\imath}+\cosh({t})\cos(\theta)\hat{\jmath}+0\hat{k}\\ \end{align*} $$ with respect to $t$ and $\theta$. The area of the parallelogram formed by the two partial derivatives is given by the magnitude of their cross product. $$ \begin{align*} \frac{\partial\vec{r}}{\partial{t}}\times\frac{\partial\vec{r}}{\partial\theta}&=\begin{vmatrix} \hat{\imath}&\hat{\jmath}&\hat{k}\\\sinh({t})\cos(\theta)&\sinh({t})\sin(\theta)&\cosh({t})\\\cosh({t})(-\sin(\theta))&\cosh({t})\cos(\theta)&0\end{vmatrix}\\ &=-\cosh^{2}({t})\cos(\theta)\hat{\imath}-\cosh^{2}({t})\sin(\theta)\hat{\jmath}+\sinh({t})\cosh({t})\hat{k}\text{;}\\ \left\|\frac{\partial\vec{r}}{\partial{t}}\times\frac{\partial\vec{r}}{\partial\theta} \right\|&=\cosh({t})\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}\text{.}\\ \end{align*} $$ Integrating over the surface yields $$\int_{0}^{2\pi}\int_{-1}^{1}\cosh(t)\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}{dt}{d}\theta=2\pi\int_{-1}^{1}\cosh({t})\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}{dt}\text{.}$$ By symmetry, the integral is equivalent to $$4\pi\int_{0}^{1}\cosh({t})\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}{dt}\text{.}$$ According to Wolfram Alpha, the surface area of a hyperboloid is $$4\pi\int_{0}^{1}\cosh({t})\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}=\pi\left(\sqrt{2}\sinh^{-1}(\sqrt{2}\sinh(1))+2\sinh(1)\sqrt{\cosh(2)}\right)\text{.}$$ How does one solve the integral $4\pi\int_{0}^{1}\cosh(t)\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}{dt}$.

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    $\begingroup$ When in doubt, replace (hyperbolic) trig functions by exponentials. It can help you wade through the mess much more easily. $\endgroup$ Jun 1, 2019 at 17:38

3 Answers 3

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Here comes the first steps.

You use the hyperbolic one, $$ \cosh^2t-\sinh^2t=1, $$ which gives you $$ 4\pi\int_0^1\cosh t\sqrt{1+2\sinh^2t}\,dt $$ Now, let $u=\sinh t$, and you will get $du=\cosh t\,dt$, and so $$ 4\pi\int_0^{\sinh 1}\sqrt{1+2u^2}\,du. $$ Can you proceed from here, finding a primitive of $\sqrt{1+2u^2}$?

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As mickep suggests, $$ 4\pi\int_{0}^{1}\cosh(t)\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}\,{dt}$$ can be converted to $$ 4\pi\int_0^{\sinh 1}\sqrt{1+2u^2}\,du$$ The next step is to substitute again with $$ u = \tfrac{\sqrt{2}}{2}\sinh v $$ and $$ du = \tfrac{\sqrt{2}}{2}\cosh v\,dv $$ so that the integral becomes $$ 4\pi\int_0^{\sinh^{-1}\left(\sqrt{2}\sinh 1\right)}\sqrt{\sinh^2 v + 1}\cdot\tfrac{\sqrt{2}}{2}\cosh v\,dv$$ and $$2\pi\sqrt{2}\int_0^{\sinh^{-1}\left(\sqrt{2}\sinh 1\right)}\cosh^2 v\,dv $$ which evaluates to $$2\pi\sqrt{2}\left(\tfrac{1}{2}\sinh v \cosh v + \tfrac{1}{2} v\right|_0^{\sinh^{-1}\left(\sqrt{2}\sinh 1\right)}$$ $$ \pi\sqrt{2}\left(\sinh v \cosh v + v\right|_0^{\sinh^{-1}\left(\sqrt{2}\sinh 1\right)}$$ $$ \pi\sqrt{2}\left(\sinh v \sqrt{\sinh^2 v + 1} + v\right|_0^{\sinh^{-1}\left(\sqrt{2}\sinh 1\right)}$$ $$ \pi\sqrt{2}\left(\sqrt{2}\sinh 1 \sqrt{2\sinh^2 1 + 1} + \sinh^{-1}\left(\sqrt{2}\sinh 1\right)\right) $$ and finally $$\pi\left(2\sinh 1 \sqrt{\cosh 2} + \sqrt{2}\sinh^{-1}\left(\sqrt{2}\sinh 1\right)\right)$$

That last can be changed to $$\pi\sqrt{2}\left(\sqrt{2}\sinh 1 \sqrt{\cosh 2} + \ln\left(\sqrt{2}\sinh 1+\sqrt{\cosh 2}\right)\right)$$ which follows a common pattern of hyperbolic integrals where the result is a product of two values plus the natural logarithm of the sum of those values.

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Another way (with the help of Wolfy):

$\begin{array}\\ I &=\int_{0}^{1}\cosh(t)\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}{dt}\\ &=\int_{0}^{1}\frac12(e^t+e^{-t})\sqrt{\frac14(e^t+e^{-t})^{2}+\frac14(e^t-e^{-t})^{2}}{dt}\\ &=\frac14\int_{0}^{1}(e^t+e^{-t})\sqrt{(e^{2t}+2+e^{-2t})+(e^{2t}-2+e^{-2t})}{dt}\\ &=\frac14\int_{0}^{1}(e^t+e^{-t})\sqrt{2(e^{2t}+e^{-2t})}{dt}\\ &=\frac{\sqrt{2}}{4}\int_{0}^{1}(e^t+e^{-t})\sqrt{e^{2t}+e^{-2t}}{dt}\\ &=\frac{\sqrt{2}}{4}\int_{0}^{1}(1+e^{-2t})\sqrt{e^{4t}+1}{dt}\\ &=\frac{\sqrt{2}}{4}\int_{1+e^{-2}}^{2}x\sqrt{\frac1{(x-1)^2}+1}\dfrac{dx}{2(x-1)}\\ &=\frac{\sqrt{2}}{8}\int_{1+e^{-2}}^{2}\dfrac{x\sqrt{(x-1)^2+1}}{(x-1)^2}dx\\ &=\frac{\sqrt{2}}{8}\int_{e^{-2}}^{1}\dfrac{(x+1)\sqrt{x^2+1}}{x^2}dx\\ &=\frac{\sqrt{2}}{8}\left(\int_{e^{-2}}^{1}\dfrac{\sqrt{x^2+1}}{x}dx+\int_{e^{-2}}^{1}\dfrac{\sqrt{x^2+1}}{x^2}dx\right)\\ &=\frac{\sqrt{2}}{8}\left( \sqrt{x^2 + 1} - \ln(\sqrt{x^2 + 1} + 1) + \ln(x)+\sinh^{-1}(x) - \dfrac{\sqrt{x^2 + 1}}{x}\right)\big|_{e^{-2}}^{1} \qquad\text{Wolfy did this}\\ &=\frac{\sqrt{2}}{8}\left( (1-\dfrac1{x})\sqrt{x^2 + 1} - \ln(\sqrt{x^2 + 1} + 1) + \ln(x)+\sinh^{-1}(x) \right)\big|_{e^{-2}}^{1}\\ &=\frac{\sqrt{2}}{8} (f(1)-f(e^{-2}) )\\ f(x) &= (1-\dfrac1{x})\sqrt{x^2 + 1} - \ln(\sqrt{x^2 + 1} + 1) + \ln(x)+\sinh^{-1}(x) \\ f(1) &= - \ln(\sqrt{2} + 1) +\sinh^{-1}(1) \\ f(e^{-2}) &= (1-e^2)\sqrt{e^{-4} + 1} - \ln(\sqrt{e^{-4} + 1} + 1) -2+\sinh^{-1}(e^{-2})\\ I &=\frac{\sqrt{2}}{8} ( - \ln(\sqrt{2} + 1) +\operatorname{arcsinh}(1)\\ &\quad - ((1-e^2)\sqrt{e^{-4} + 1} - \ln(\sqrt{e^{-4} + 1} + 1) -2+\operatorname{arcsinh}(e^{-2})))\\ \end{array} $

$x = e^{-2t}+1, dx = -2e^{-2t}dt =-2(x-1)dt, dt =\dfrac{-dx}{2(x-1)} $

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