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$$\int_0^{2\pi} \sqrt{a^2 +b^2+2ab \cos\varphi}\,\mathrm{d}\varphi$$

Where $a$ and $b$ are constants. I had to find the distance travelled by a point at a distance of $b$ from the centre of a rolling disc with radius $a$ in one full rotation. I got the final answer as this integral, but I was not able to solve this integral.

Jean-Claude Arbaut
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avz2611
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1 Answers1

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First consider the integrand \begin{align} \sqrt{a^{2} + b^{2} + 2 a b \cos(\phi)} \end{align} which can be seen as \begin{align} \sqrt{a^{2} + b^{2} + 2 a b \cos(\phi)} &= \sqrt{(a+b)^{2} - 4ab \sin^{2}\left(\frac{\phi}{2}\right) } \\ &= (a+b) \sqrt{ 1 - \left( \frac{4ab}{(a+b)^{2}} \right) \, \sin^{2}\left( \frac{\phi}{2}\right) }. \end{align} Now, \begin{align} I &= \int_{0}^{2\pi} \sqrt{a^{2} + b^{2} + 2 a b \cos(\phi)} \, d\phi \\ &= (a+b) \, \int_{0}^{2\pi} \sqrt{ 1 - \left( \frac{4ab}{(a+b)^{2}} \right) \, \sin^{2}\left( \frac{\phi}{2}\right) } \, d\phi \\ &= 2(a+b) \int_{0}^{\pi} \sqrt{1- k^{2} \sin^{2}(x) } \, dx \end{align} where $(a+b)^{2} k^{2} = 4ab$. The integral remaining if the complete elliptical integral of the second kind. The "solution" of the integral is \begin{align} I = 4(a+b) E\left(\frac{4ab}{(a+b)^{2}}\right). \end{align}
Hence, \begin{align} \int_{0}^{2\pi} \sqrt{a^{2} + b^{2} + 2 a b \cos(\phi)} \, d\phi = 4(a+b) E\left(\frac{4ab}{(a+b)^{2}}\right). \end{align}

Leucippus
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    While it's implicit here, it's worth pointing out that you're taking $k^2$ as the argument of your elliptic integral rather than $k$ itself. – Semiclassical Sep 04 '14 at 18:55
  • sorry i have not studied elliptical integrals , but i have read somewhere that their perimeter cannot be evaluated ,is the solution related to that ? – avz2611 Sep 04 '14 at 18:59
  • @user142634: The perimeter of an ellipse can be expressed in terms of the elliptic integral of the first kind $K(k^2)$. The elliptic integral used above is that of the second kind. Both are outside the realm of elementary functions. – Semiclassical Sep 04 '14 at 19:29