Exercise 8.1 · Q2: Area of the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$

Basics you need

Solution

1. Read off $a$ and $b$⁠, sketch, and use symmetry

   Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$⁠: $a^2 = 4$ and $b^2 = 9$⁠, so $a = 2$ and $b = 3$⁠.

   Putting $y = 0$⁠: $x^2 = 4$⁠, $x = \pm 2$⁠. Putting $x = 0$⁠: $y^2 = 9$⁠, $y = \pm 3$⁠. So the ellipse meets the axes at $(\pm 2, 0)$ and $(0, \pm 3)$⁠. Only $x^2$ and $y^2$ appear, so it is symmetrical about both axes:

   $$\begin{aligned} \text{Area} &= 4 \times (\text{area of the} \\ &\qquad \text{first-quadrant quarter}). \end{aligned}$$
   In the book's notation $a$ is always the number that goes with $x$⁠, whether or not it is the larger one. Here $b > a$⁠, so the long axis is vertical; this changes the picture but not the method.
   One quarter, times four. Basics: one quarter, times four

2. Find the top edge of the quarter

   $$\begin{aligned} \frac{y^2}{9} &= 1 - \frac{x^2}{4} = \frac{4 - x^2}{4} \\ y^2 &= \frac{9}{4}\left(4 - x^2\right) \\ y &= \pm\frac32\sqrt{4 - x^2}. \end{aligned}$$

   In the first quadrant $y \ge 0$⁠, so the top edge is $y = \frac32\sqrt{4 - x^2}$ for $0 \le x \le 2$⁠.

   $\sqrt{\frac94} = \frac32$⁠. The $+$ root is the upper half of the ellipse. For $0 \le x \le 2$⁠, $4 - x^2 \ge 0$⁠, so the root is real.
   Take the positive root above the $x$-axis. Basics: solve for y to get the top edge

3. Write the area as an integral

   $$\begin{aligned} \text{Area} &= 4\int_0^2 y\,dx \\ &= 4\int_0^2\frac32\sqrt{4 - x^2}\,dx \\ &= 6\int_0^2\sqrt{4 - x^2}\,dx. \end{aligned}$$
   The quarter is bounded by the curve, the $x$-axis and the ordinates $x = 0$ and $x = 2$⁠. The constants combine as $4 \cdot \frac32 = 6$⁠.
   Section 8.2: $A = \int_a^b y\,dx$⁠. Lesson: the area under a curve

4. Integrate with the square-root formula

   With $a = 2$⁠, $\frac{a^2}{2} = 2$⁠:

   $$\begin{aligned} &\int_0^2\sqrt{4 - x^2}\,dx \\ &= \Big[\frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\frac{x}{2}\Big]_0^2 \\ &= \big(1 \times 0 + 2\sin^{-1}1\big) - \big(0 + 2\sin^{-1}0\big) \\ &= 2\cdot\frac{\pi}{2} = \pi. \end{aligned}$$
   At $x = 2$ the root is 0 and $\sin^{-1}1 = \frac{\pi}{2}$⁠; at $x = 0$ both terms are 0. The value $\pi$ is a quarter of the area $4\pi$ of the circle of radius 2, whose top half is $\sqrt{4 - x^2}$⁠.
   Section 7.6.2 (iii); Theorem 2, Section 7.8.3. Basics: the quarter-circle integral

5. Multiply

   $$\text{Area} = 6 \times \pi = 6\pi \text{ sq. units}.$$
   The factor 6 already includes the 4 quarters and the $\frac32$ of the top edge.
   Area of the whole ellipse $= 4 \times$ the quarter. Lesson: areas of ellipses

6. Check with $\pi ab$

   By Example 2, the area is $\pi ab = \pi \cdot 2 \cdot 3 = 6\pi$⁠, the same.

   $\pi ab$ is symmetric in $a$ and $b$⁠, so it does not matter which one is larger. In an exam the integral must still be shown.
   Area of an ellipse $= \pi ab$⁠. Lesson: circle and ellipse, Examples 1–2

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