Question 1e14b

Dec 19, 2017

$A = \setminus {\int}_{x = 0}^{x = a} \setminus {\int}_{y = 0}^{y = \frac{b}{a} \setminus \sqrt{{a}^{2} - {x}^{2}}} \mathrm{dx} \mathrm{dy} = \setminus \pi a b .$
a=|\sqrt{9}|=3; \qquad b=|\sqrt{4}|=2; \qquad A=\piab=6\pi.

Explanation:

Let us consider the general case of an ellipse of semi-major axis $a$ and semi-minor axis $b$,

Equation of ellipse:$\setminus q \quad {x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$

Area of the ellipse: Because the ellipse is symmetrical about both the x and y axes, the total area can be written as $4$ times the area of the first quadrant,

$A = 4 \setminus {\int}_{x = 0}^{x = a} \setminus {\int}_{y = 0}^{y = \frac{b}{a} \setminus \sqrt{{a}^{2} - {x}^{2}}} \mathrm{dx} \mathrm{dy}$
$A = 4 \setminus {\int}_{x = 0}^{x = a} \left[\setminus {\int}_{y = 0}^{y = \frac{b}{a} \setminus \sqrt{{a}^{2} - {x}^{2}}} \mathrm{dy}\right] \mathrm{dx}$
$A = 4 \setminus {\int}_{x = 0}^{x = a} \frac{b}{a} \setminus \sqrt{{a}^{2} - {x}^{2}} . \mathrm{dx}$
$A = \frac{4 b}{a} \setminus {\int}_{x = 0}^{x = a} \setminus \sqrt{{a}^{2} - {x}^{2}} \mathrm{dx}$

x\equiv a\sin\theta; \qquad dx = a\cos\thetad\theta
Limits:
x=0 \qquad \rightarrow \qquad \theta=0; \qquad x=a \qquad \rightarrow \qquad \theta = \pi/2#

$A = \frac{4 b}{a} \setminus {\int}_{0}^{\setminus \frac{\pi}{2}} \left[\setminus \sqrt{{a}^{2} - {a}^{2} \setminus {\sin}^{2} \setminus \theta}\right] . a \setminus \cos \setminus \theta d \setminus \theta$
$A = 4 a b \setminus {\int}_{0}^{\setminus \frac{\pi}{2}} \setminus {\cos}^{2} \setminus \theta . d \setminus \theta = 4 a b \setminus {\int}_{0}^{\setminus \frac{\pi}{2}} \frac{1}{2} \left\{1 + \setminus \cos 2 \setminus \theta\right\} d \setminus \theta$
$A = 2 a b \setminus {\int}_{0}^{\setminus \frac{\pi}{2}} \left\{1 + \setminus \cos 2 \setminus \theta\right\} . d \setminus \theta$
$A = 2 a b \left[\setminus \frac{\pi}{2} + \setminus {\int}_{0}^{\setminus \frac{\pi}{2}} \setminus \cos 2 \setminus \theta . d \setminus \theta\right] = 2 a b \left[\setminus \frac{\pi}{2} + \left\{{\cancel{\frac{\setminus \sin 2 \setminus \theta}{2}}}_{0}^{\setminus \frac{\pi}{2}}\right\}\right]$
$A = \setminus \pi a b$