Answer
$\int_0^2 \int_{x^3}^8 dy dx; \int_0^{8} \int_{0}^{y^{1/3}} dx dy$
Work Step by Step
We are given that $y=x^3; y=8; x^3=8$
This gives $x=2$
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 $\int_0^2 \int_{x^3}^8 dy dx; \int_0^{8} \int_{0}^{y^{1/3}} dx dy$
