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...

Write $D=\{(x,y)\in \mathbb R^2\colon 0\leq x\leq 1\land -x\leq y\leq x\}$.
The integral then comes equal to  $\displaystyle \int \limits_0^1\int \limits _{-x}^x\sqrt{x^2-y^2}\mathrm dy\mathrm dx$.
You can get away with the one dimensional substitution $y=x\sin(\theta)$.
