Answer
$\int_0^3 \int_{0}^{2x} dy dx; \int_0^{6} \int_{1/2y}^{3} dx dy$
Work Step by Step
We have $y=2x; y=0$
Solve for $x$, we get $x=0$
(a) Use the top curve for the top bound and the bottom curve for the bottom bound to find the $dy$ value and the maximum and minimum values for $x$ on that interval to find the $dy$ value.
Hence, we have $\int_0^3 \int_{0}^{2x} dy dx$
(b) Use the right curve for the top bound and the left curve for the bottom bound to find the $dx$ value and the maximum and minimum values for $y$ on that interval for $dy$ values.
So, we have $\int_0^{6} \int_{1/2y}^{3} dx dy$
