Show that
$$
4\int_0^1 \int_0^1 \sqrt{1+x^2+y^2} \, dy \, dx = \frac{8}{3} \int_0^{\pi /4} (1+ \sec^2 \theta )^{1.5} \, d\theta – \frac{2\pi}{3}
$$
I have tried letting $1+x^2 = \lambda^2$ , $y=\lambda \tan \theta$ , $\frac{dy}{d\theta} = \lambda \sec^2 \theta$. After which
$$\begin{aligned}
4\int_0^1 \int_{\arctan (0/ \lambda)}^{\arctan (1/ \lambda)} \lambda^2 \sec^3 \theta \, d\theta \, dx &= 4\int_0^1 \lambda^2 \left[ 0.5 \sec \theta \tan \theta + 0.5 \ln (\sec \theta + \tan \theta) \right]_{\arctan (0/ \lambda)}^{\arctan (1/ \lambda)}\, dx
\end{aligned} $$
And then I was convinced this was not the way to go.
Best Answer
The function and the region are symmetric about the line $y=x$, so you can re-write your integral as
$$8\int_0^1 \int_0^x f(x,y) \; dy \; dx.$$
So you're integrating over the triangle with vertices $(0,0),$ $(1,0),$ and $(1,1).$
The vertical line $x=1$ has polar equation $r\cos \theta =1$ or $r=\sec \theta.$ So in polar, the integral is
$$8\int_0^{\pi/4} \int_0^{\sec \theta} \sqrt{1+r^2}\; r \; dr \; d\theta.$$
Then things are straightforward.