Calculate $\iint\limits_D {\sqrt {{x^2} - {y^2}} }\,dA$ ... 
Calculate $$\iint\limits_D {\sqrt {{x^2} - {y^2}} }\,dA$$ where $D$ is
  the triangle with vertices $(0,0), (1,1)$ and $(1,-1)$.

I get the following integral $$I = 2\int\limits_0^1 {\int\limits_0^x {\sqrt {{x^2} - {y^2}} } \,dydx} $$ I would appreciate some hints or help in solving this integral. 
The answer $\frac{\pi }{6}$ somehow implies that the solution may be obtained by introducing polar coordinates. I tried solving it with polar coordinates setting the integration limits $0 \ldots \frac{\pi }{4}$ for $\theta $ and $0 \ldots \tan \theta $ for $r$ and multiplying the result by 2. However, i could not get the right answer...

For the polar approach:
$$D=\left\{(r,\theta)\mid 0\le r\le\sec\theta,\,0\le\theta\le\frac{\pi}{4}\right\}$$
and so you have
$$2\iint_D dA=2\int_0^{\pi/4}\int_0^{\sec\theta}r\,dr\,d\theta$$
