정적분의 뜻과 성질

(a) $$\begin{aligned} (ax)^2+(by)^2&=1\\ by&=\sqrt{1-(ax)^2}\\ y &= \frac{1}{b}\sqrt{1-(ax)^2} \end{aligned}$$ (b) The values of $x$ in the domain of the function above are those that satisfy $1-(ax)^2 \geq 0$. That is, $-\frac{1}{a}\leq x \leq \frac{1}{a}$. Therefore, the upper half of the ellipse has area

$$\begin{aligned} \displaystyle\frac{1}{b}&\displaystyle\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{1-(ax)^2}\ \mathrm{d}x \\ \text{The upper half of a circle has equation $y=\sqrt{r^2-x^2}$. } \\ &=\frac{1}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{a^2\left(\frac{1}{a^2}-x^2\right)}\ \mathrm{d}x\\ &=\frac{1}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}a\sqrt{\frac{1}{a^2}-x^2}\ \mathrm{d}x\\ &=\frac{a}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{\frac{1}{a^2}-x^2}\ \mathrm{d}x \end{aligned}$$ (c) The function $y=\sqrt{\dfrac{1}{a^2}-x^2}$ is the upper-half of the circle centred at the origin with radius $\dfrac{1}{a}$. So, the expression from (b) evaluates to $\left(\dfrac{a}{b}\right)\dfrac{\pi}{2a^2} = \dfrac{\pi}{2ab}$.

The expression from (b) was half of the ellipse, so the area of the ellipse is $\dfrac{\pi}{ab}$.

Remark: this was a slightly long-winded way of getting the result. The reasoning is basically this:

• The area of the unit circle $x^2+y^2=1$ is $\pi$ .
• The ellipse $(ax)^2+y^2=1$ is obtained by shrinking the unit circle horizontally by a factor of $a$. So, its area is $\dfrac{\pi}{a}$ .
• Further, the ellipse $(ax)^2+(by)^2=1$ is obtained from the previous ellipse by shrinking it vertically by a factor of $b$. So, its area is $\dfrac{\pi}{ab}$ .