Answer
$\int_0^3 \int_{0}^{2x} dy dx; \int_0^{6} \int_{1/2y}^{3} dx dy$
Work Step by Step
As we are given that $y=2x; y=0$
a) For the $dy$ value, we will have to use the top curve for the top bound and the bottom curve for the bottom bound. For $dx$, we need to use the maximum and minimum values for $x$ on that interval.
b) For the $dx$ value, we will have to use the right curve for the top bound and the left curve for the bottom bound. For $dy$, we need to use the maximum and minimum values for $y$ on that interval.
Hence, we have $\int_0^3 \int_{0}^{2x} dy dx; \int_0^{6} \int_{1/2y}^{3} dx dy$
