I'm having troubles evaluating this double integral. Can somebody help me? I've gone to the part that I need to use trigonometric substitution, but performing the said sub, I think I'm kind of unsure of the next steps.
$$\int \limits_{0}^{\sqrt{2}}\int \limits_{x}^{\sqrt{4-x^2}}{\sqrt{x^2+y^2}} \, dy \, dx $$
On
Have you sketched the region of integration? It is a sector of a circle of radius $2$, one-eighth of the circle.
The function $z=\sqrt{z^2+y^2}$ forms a cone with its tip at the origin. Out to radius $2$, the volume of that cone would be $$V=\frac13\pi(2)^2(2)=\frac83\pi$$ using basic geometry formulas. The integral with the full circle as the region of integration would be the complement of that cone within a cylinder of radius $2$ and height $2$: $$\pi(2)^2(2)-\frac83\pi=\frac{16}{3}\pi$$
And since the region of integration is really a one-eight sector of that circle, the given integral is $$\frac{1}{8}\left(\frac{16}{3}\pi\right)=\frac23\pi$$
Note that the integration region is in the first quadrant, between $y=x $ and $x^2+y^2=4$, which, in polar coordinates, leads to $$\int \limits_{0}^{\sqrt{2}}\int \limits_{x}^{\sqrt{4-x^2}}{\sqrt{x^2+y^2}} \ dy \ dx =\int_{\pi/4}^{\pi/2} \int_0^2 r \>rdrd\theta=\frac{2\pi}3 $$