a $$\int_{0}^{\frac {4}{3}}\int_{0}^{\sqrt{4x-x^{2}}}\sqrt{x^{2}+y^{2}}\;dx\;dy$$ b $$\int_{0}^{1}\int_{0}^{\sqrt{4x-x^2}}\sqrt{x^{2}+y^{2}}\;dx\;dy$$
for (b) $x=1$ so, $y=\sqrt{4x-x^2}= \sqrt{4*1-1}= \sqrt3$
$r^2=x^2+y^2$
$r^2= (1)^1+(\sqrt3)^2$
$r^2=\sqrt4$ so $r=2$
now $\theta=\arctan (y/x)=\arctan(\sqrt3)=\pi/3$
so the double integral will be: $4\pi/3$ for ($0<\theta<\pi/3$ , and $0<r<2$ )
[I do not know if my answer correct or not and I'm not sure how I can solve (a) anyone can help me!]