## Thomas' Calculus 13th Edition

$\int_0^2 \int_{x^3}^8 dy dx; \int_0^{8} \int_{0}^{y^{1/3}} dx dy$
We have $y=x^3; y=8; x^3=8$ Solve for $x$; we get $x=2$ (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 $\int_0^2 \int_{x^3}^8 dy dx$ (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 $\int_0^{8} \int_{0}^{y^{1/3}} dx dy$