Answer
a) $ \int_{0}^1 \int_{x}^{3-2x} f(x,y) dy dx $
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(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) $y= 3-2x; y=x; x=0 \implies x=1; y=1$
The region $R$ for vertical cross-sections can be written as follows:
$\iint_{R} \space dA= \int_{0}^1 \int_{x}^{3-2x} f(x,y) dy dx $
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(b) $y= 3-2x; y=x; x=0 \implies x=1; y=1$
The region $R$ for horizontal cross-sections can be written as follows:
$\iint_{R} \space dA=\iint_{R_1} dA+\iint_{R_2} dA = \int_{0}^{1} \int_{0}^{y} f(x,y) dx dy+ \int_{1}^{3} \int_{0}^{(3-y)/2} f(x,y) \space dx \space dy$
