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 are given that $y=x^2; y=3x$
$3x-x^2 \implies x=0, 3$
or, $x^2 \leq y \leq 3x$
(a) For vertical cross-sections, the region $R$ can be defined as:
$\int_0^3 \int_{x^2}^{3x} f(x,y) dy dx$
(b) For horizontal cross-sections, the region $R$ can be defined as:
$\int_0^{9} \int_{(1/3) y}^{\sqrt y} f(x,y) dx dy$
