Answer
a) $ \int_{0}^1 \int_{x}^{3-2x} f(x,y) dy dx $
(b) $ \int_{0}^{1} \int_{0}^{y} f(x,y) dx dy+ \int_{1}^{3} \int_{0}^{(3-y)/2} f(x,y) dx dy$
Work Step by Step
(a) For vertical cross-sections, the region $R$ can be defined as:
$y= 3-2x; y=x; x=0 \implies x=1; y=1$
$\iint_{R} dA= \int_{0}^1 \int_{x}^{3-2x} f(x,y) dy dx $
(b) For horizontal cross-sections, the region $R$ can be defined as:
$\iint_{R} dA=iint_{R_1} dA+\iint_{R_1} dA$
or, $= \int_{0}^{1} \int_{0}^{y} f(x,y) dx dy+ \int_{1}^{3} \int_{0}^{(3-y)/2} f(x,y) dx dy$
