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To evaluate the integral, I performed the integration with respect to x and put the limits of x. I got this result:

$$\sqrt{y^2 + 2}/2 + (y^2 + 1) \ln(1+ \sqrt{y^2 + 2})/2 - (y^2 + 1)/2 \ln(\sqrt{y^2 + 1}) +c$$

Now I need to integrate the above expression with respect to y, which seems to be rather complex. Is there a simple way to evaluate the original integral?

Nosrati
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coolscitist
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2 Answers2

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First, by symmetry, your integral equals $$ 2\int_0^1\int_0^x\sqrt{x^2+y^2+1}\,dy\,dx. $$ Changing to polar coordinates, you will get new limits $$ 0\leq \phi\leq \pi/4\quad\text{and}\quad 0\leq r\leq 1/\cos\phi, $$ and so you end up with $$ 2\int_0^{\pi/4}\int_0^{1/\cos\phi}\sqrt{1+r^2}\,r\,dr\,d\phi. $$ The inner integral is easily calculated, $$ \int_0^{1/\cos\phi}\sqrt{1+r^2}r\,dr=\Bigl[\frac13\bigl(1+r^2\bigr)^{3/2}\Bigr]_0^{1/\cos\phi}=\frac{1}{3}\bigl(1+1/\cos^2\phi\bigr)^{3/2}-\frac{1}{3}. $$ For the outer one, set $u=\tan\phi$, and you will have to tackle $$ \int\frac{(2+u^2)^{3/2}}{1+u^2}\,du. $$ Next, set $$ s=\frac{u}{\sqrt{2+u^2}} $$ and you will end up with a rational integrand, $$ \int\frac{4}{(1-s^2)^2(1+s^2)}\,ds. $$ I suppose you are now on safe ground? Continue with partial fraction decomposition. If I do not do too many errors in my calculations, the result is

$$\frac{1}{\sqrt{3}}-\frac{\pi}{18}+\frac{2}{3}\log(2+\sqrt{3})\approx 1.28.$$

mickep
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  • Tried to do this on my own. How did you know we were supposed to put s = u/sqrt(2+u^2)? – coolscitist Sep 05 '17 at 17:33
  • It is close to the Euler substitution, and I have just found that it works very well in expressions like this. I'm sorry, but I do not have a better reference than that. – mickep Sep 05 '17 at 17:34
  • Hmm, ok I understand. It's been several years since I last did calculus so that intuitive nature of knowing exactly what technique to use by just looking at the problem is gone :-) Thanks anyways! – coolscitist Sep 05 '17 at 23:56
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hint :$$\int_0^1\int_0^1\sqrt{x^2+y^2+1}dxdy$$ turn to polar coordinate $$x^2+y^2=r^2\\dxdy=rdrd\theta\\w.r.t [0,1]\times[0,1]\\\to 0\leq r\leq 1 \\0\leq \theta\leq\frac{\pi}{2}$$ $$2\int_0^{\frac{\pi}{4}}\int_{9}^{\sec \theta}\sqrt{r^2+1}\times rdrd\theta$$

Khosrotash
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    Using it I got this result: 0.957362. But when I evaluated the original integral in python, I got: 1.28078927. Also, how is dx dy = r dr dTheta? – coolscitist Sep 04 '17 at 18:04
  • @coolscitist :see this page ,https://math.stackexchange.com/questions/1636021/rigorous-proof-that-dx-dy-r-dr-d-theta/1636057#1636057 – Khosrotash Sep 04 '17 at 18:08
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    Hmm got the r dr dTheta part. But still the two integrals are not giving same results when I did them in python. – coolscitist Sep 04 '17 at 18:26
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    @Khosrotash I think integral limits are incorrect! – Nosrati Sep 04 '17 at 18:30
  • $$\int _0^1\int _0^{\frac{\pi }{2}}r \sqrt{r^2+1}dtdr=\frac{1}{6} \left(2 \sqrt{2}-1\right) \pi$$ while $$\int _0^1\int _0^1\sqrt{x^2+y^2+1}dydx=\frac{1}{3} \left(\sqrt{3}+\log \left(7+4 \sqrt{3}\right)\right)-\frac{\pi }{18}$$ Why? – Raffaele Sep 04 '17 at 18:32
  • $$\int _{\frac{\pi }{4}}^{\frac{\pi }{2}}\int _0^{\frac{1}{\sin\theta}}r \sqrt{r^2+1.}drd\theta +\int _0^{\frac{\pi }{4}}\int _0^{\frac{1}{\cos\theta }}r \sqrt{r^2+1.}drd\theta$$ – Nosrati Sep 04 '17 at 18:34
  • You are right , I'll fix it . – Khosrotash Sep 04 '17 at 18:41