How to evaluate this definite double integral? $$\int\int_D 6x\sqrt{y^2-x^2}dA, D=\{(x,y)|0\leq y\leq 2, 0 \leq x \leq y\}$$
I tried: $$\int_0^2 \int_0^y 6x\sqrt{y^2-x^2}dxdy$$
But that is incorrect. 

Notice, $$\int_{0}^2\int_0^y 6x\sqrt{y^2-x^2} \ dxdy=\int_{0}^2\left(\int_0^y 6x\sqrt{y^2-x^2} \ dx\right)dy $$
$$=\int_{0}^2\left(-3\int_0^y (y^2-x^2)^{1/2} \ d(y^2-x^2)\right)dy $$
$$=\int_{0}^2dy\left(-3\frac{2(y^2-x^2)^{3/2}}{3}\right)_{0}^{y}$$
$$=\int_{0}^2\left(2y^3\right)\ dy$$
$$=\left(2\frac{y^4}{4}\right)_{0}^{2}=\left(\frac{2^4}{2}-0\right)=\color{red}{8}$$
