Answer
$\int_0^3 \int_{x^2}^{3x} f(x,y) dy dx$ and $\int_0^{9} \int_{(1/3) y}^{\sqrt y} f(x,y) dx dy$
Work Step by Step
We have $y=x^2; y=3x \implies 3x-x^2$
Solve for $x$, we get $x=0, 3$ , that is, $x^2 \leq y \leq 3x$
(a) For vertical cross-sections: 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 $\int_0^3 \int_{x^2}^{3x} f(x,y) dy dx$
(b) For horizontal cross-sections: 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 $\int_0^{9} \int_{(1/3) y}^{\sqrt y} f(x,y) dx dy$
