Answer
a) $ \int_{-1}^{2} \int_{x^2}^{x+2} f(x,y) dy dx $
(b) $ \int_{0}^{1} \int_{-\sqrt y}^{\sqrt y} f(x,y) dx dy+ \int_{1}^{4} \int_{y-2}^{\sqrt y} f(x,y) dx dy$
Work Step by Step
(a) For vertical cross-sections, the region $R$ can be defined as:
$y= x^2; y=x+2; \implies x^2 -x-2=0$
or, $x=-1, 2$
$\iint_{R} dA= \int_{-1}^{2} \int_{x^2}^{x+2} f(x,y) dy dx $
(b) For horizontal cross-sections, the region $R$ can be defined as:
$\iint_{R} dA=\int_{R_1} dA+\iint_{R_1} dA$
or, $= \int_{0}^{1} \int_{-\sqrt y}^{\sqrt y} f(x,y) dx dy+ \int_{1}^{4} \int_{y-2}^{\sqrt y} f(x,y) dx dy$
