Answer
(a) $\int_0^{\pi/4} \int_{\tan x}^{1} f(x,y) dy dx$
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(b) $\int_{0}^{1} \int_{0}^{\tan^{-1} y} f(x,y) dx dy$
Work Step by Step
(a) The region $R$ for vertical cross-sections can be written as follows:
$R=$ { $( x,y) | \tan x \leq y \leq 1 , 0 \leq x \leq \dfrac{\pi}{4}$}
$\iint_{R} dA=\int_0^{\pi/4} \int_{\tan x}^{1} f(x,y) dy dx$
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(b) The region $R$ for horizontal cross-sections can be written as follows:
$R=$ { $( x,y) | 0 \leq x \leq \tan^{-1} y , 0 \leq y \leq 1$}
So, $\iint_{R} dA=\int_{0}^{1} \int_{0}^{\tan^{-1} y} f(x,y) dx dy$
