Let $a, b$ be positive real numbers. Let $E$ be the region defined as:
$$E=\left \{ (x, y): 0\leq x \leq \frac{\pi}{2}, \;\;\; 0\leq y\leq \min \left \{\frac{a}{\cos x}, \; \frac{b}{\sin x} \right \} \right \}$$
Prove that $\displaystyle \iint \limits_{E} y \, {\rm d}x \, {\rm d}y =ab$.
Solution
Let us sketch the graph:
The region $E$ can be represented as a union of the following disjoint regions:
$$A=\left\{\left(x,y\right)\in\mathbb{R}^2: 0\leq x\leq \arctan \left(\frac{b}{a}\right)\,,0\leq y\leq \frac{a}{\cos x}\right\}$$
$$B=\left\{\left(x,y\right)\in\mathbb{R}^2: \arctan \left(\frac{b}{a}\right)< x\leq \frac{\pi}{2}\,,0\leq y\leq \frac{b}{\sin x}\right\}$$
These two sets are Jordan countable and the function $f(x, y)= y, \; (x, y) \in E$ is continuous hence integrable on either of the two sets. Therefore:
$$\iint \limits_{E} f(x, y)\, {\rm d}(x, y) = \iint \limits_{A} f(x, y) \, {\rm d}(x, y) + \iint \limits_{B} f(x, y) \, {\rm d}(x, y)$$
So we have to evaluate either of the two integrals individually. With guide the sketch of the region we have that:
${\color{gray} \blacksquare}$
$$\begin{aligned} \iint \limits_{A}f\left(x,y\right)\,{\rm d} \left(x,y\right)&=\int_{0}^{\arctan \left(\frac{b}{a}\right)}\int_{0}^{a/\cos x}y\,{\rm d} y\,{\rm d} x\\&=\int_{0}^{\arctan \left(\frac{b}{a}\right)}\left[\frac{y^2}{2}\right]_{0}^{a/\cos x}\,{\rm d} x\\&=\int_{0}^{\arctan \left(\frac{b}{a}\right)}\frac{a^2}{2\,\cos^2\,x}\,{\rm d} x\\&=\left[\frac{a^2\,\tan x}{2}\right]_{0}^{\arctan \left(\frac{b}{a}\right)}\\&=\frac{a^2}{2}\,\tan \left(\arctan \left[\frac{b}{a}\right)\right]=\frac{a\,b}{2}\end{aligned}$$
${\color{gray} \blacksquare}$
$$\begin{aligned}\iint \limits_{B}f\left(x,y\right)\,{\rm d}\left(x,y\right)&=\int_{t}^{\pi/2}\int_{0}^{b/\sin x}y\,{\rm d}y\,{\rm d}x\\&=\int_{t}^{\pi/2}\left[\frac{y^2}{2}\right]_{0}^{b/\sin x}\,{\rm d}x\\&=\int_{t}^{\pi/2}\frac{b^2}{2\,\sin^2 x}\,{\rm d}x\\&=\lim_{x\to t^{+}}\int_{x}^{\pi/2}\frac{b^2}{2\,\sin^2\,z}\,{\rm d}z\\&=\lim_{x\to t^{+}}\left[-\frac{b^2}{2}\,\cot z\right]_{x}^{\pi/2}\\&=\lim_{x\to t^{+}}\left(\frac{b^2}{2}\,\cot x\right)\\&=\frac{a\,b}{2}\end{aligned} $$
where $t=\arctan \frac{b}{a}$. Finally we deduce the wanted result:
$$\iint \limits_{E}f\left(x,y\right)\,{\rm d}\left(x,y\right)=\iint \limits_{E}y\,{\rm dy}\,{\rm d}x=ab$$
as wanted.
$$E=\left \{ (x, y): 0\leq x \leq \frac{\pi}{2}, \;\;\; 0\leq y\leq \min \left \{\frac{a}{\cos x}, \; \frac{b}{\sin x} \right \} \right \}$$
Prove that $\displaystyle \iint \limits_{E} y \, {\rm d}x \, {\rm d}y =ab$.
Solution
Let us sketch the graph:
$$A=\left\{\left(x,y\right)\in\mathbb{R}^2: 0\leq x\leq \arctan \left(\frac{b}{a}\right)\,,0\leq y\leq \frac{a}{\cos x}\right\}$$
$$B=\left\{\left(x,y\right)\in\mathbb{R}^2: \arctan \left(\frac{b}{a}\right)< x\leq \frac{\pi}{2}\,,0\leq y\leq \frac{b}{\sin x}\right\}$$
These two sets are Jordan countable and the function $f(x, y)= y, \; (x, y) \in E$ is continuous hence integrable on either of the two sets. Therefore:
$$\iint \limits_{E} f(x, y)\, {\rm d}(x, y) = \iint \limits_{A} f(x, y) \, {\rm d}(x, y) + \iint \limits_{B} f(x, y) \, {\rm d}(x, y)$$
So we have to evaluate either of the two integrals individually. With guide the sketch of the region we have that:
${\color{gray} \blacksquare}$
$$\begin{aligned} \iint \limits_{A}f\left(x,y\right)\,{\rm d} \left(x,y\right)&=\int_{0}^{\arctan \left(\frac{b}{a}\right)}\int_{0}^{a/\cos x}y\,{\rm d} y\,{\rm d} x\\&=\int_{0}^{\arctan \left(\frac{b}{a}\right)}\left[\frac{y^2}{2}\right]_{0}^{a/\cos x}\,{\rm d} x\\&=\int_{0}^{\arctan \left(\frac{b}{a}\right)}\frac{a^2}{2\,\cos^2\,x}\,{\rm d} x\\&=\left[\frac{a^2\,\tan x}{2}\right]_{0}^{\arctan \left(\frac{b}{a}\right)}\\&=\frac{a^2}{2}\,\tan \left(\arctan \left[\frac{b}{a}\right)\right]=\frac{a\,b}{2}\end{aligned}$$
${\color{gray} \blacksquare}$
$$\begin{aligned}\iint \limits_{B}f\left(x,y\right)\,{\rm d}\left(x,y\right)&=\int_{t}^{\pi/2}\int_{0}^{b/\sin x}y\,{\rm d}y\,{\rm d}x\\&=\int_{t}^{\pi/2}\left[\frac{y^2}{2}\right]_{0}^{b/\sin x}\,{\rm d}x\\&=\int_{t}^{\pi/2}\frac{b^2}{2\,\sin^2 x}\,{\rm d}x\\&=\lim_{x\to t^{+}}\int_{x}^{\pi/2}\frac{b^2}{2\,\sin^2\,z}\,{\rm d}z\\&=\lim_{x\to t^{+}}\left[-\frac{b^2}{2}\,\cot z\right]_{x}^{\pi/2}\\&=\lim_{x\to t^{+}}\left(\frac{b^2}{2}\,\cot x\right)\\&=\frac{a\,b}{2}\end{aligned} $$
where $t=\arctan \frac{b}{a}$. Finally we deduce the wanted result:
$$\iint \limits_{E}f\left(x,y\right)\,{\rm d}\left(x,y\right)=\iint \limits_{E}y\,{\rm dy}\,{\rm d}x=ab$$
as wanted.
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