Simple way to evaluate the double integration of sqrt(x^2 + y^2 + 1) limits x=0 to 1, y=0 to 1
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?
$\endgroup$2 Answers
$\begingroup$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
$\endgroup$ 3 $\begingroup$$$\frac{1}{\sqrt{3}}-\frac{\pi}{18}+\frac{2}{3}\log(2+\sqrt{3})\approx 1.28.$$
hint :$$\int_0^1\int_0^1\sqrt{x^2+y^2+1}dxdy$$ turn to polar coordinate $$x^2+y^2=r^2\\dxdy=rdrd\theta\\w.r.t [0,1]\times[0,1]\\\to 0\leq r\leq 1 \\0\leq \theta\leq\frac{\pi}{2}$$ $$2\int_0^{\frac{\pi}{4}}\int_{9}^{\sec \theta}\sqrt{r^2+1}\times rdrd\theta$$
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