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) \space dx \space dy+ \int_{1}^{4} \int_{y-2}^{\sqrt y} f(x,y) \space dx \space dy$
Work Step by Step
a) $y= x^2; y=x+2; \implies x^2 -x-2=0$
or, $x=-1, 2$
The region $R$ for vertical cross-sections can be written as follows:
$\iint_{R} dA= \int_{-1}^{2} \int_{x^2}^{x+2} f(x,y) dy dx $
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b) The region $R$ for horizontal cross-sections can be written as follows:
$\iint_{R} dA=\int_{R_1} dA+\iint_{R_2} dA = \int_{0}^{1} \int_{-\sqrt y}^{\sqrt y} f(x,y) \space dx \space dy+ \int_{1}^{4} \int_{y-2}^{\sqrt y} f(x,y) \space dx \space dy$
