Example 2: Area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Basics you need

Solution

1. Sketch the ellipse and use its symmetry

   Putting $y = 0$⁠: $\frac{x^2}{a^2} = 1$⁠, so $x = \pm a$⁠. Putting $x = 0$⁠: $y = \pm b$⁠. The ellipse meets the axes at $A(a, 0)$⁠, $A'(-a, 0)$⁠, $B(0, b)$ and $B'(0, -b)$⁠.

   Replacing $x$ by $-x$⁠, or $y$ by $-y$⁠, does not change the equation, so the ellipse is symmetrical about both axes:

   $$\text{Area} = 4 \times (\text{area of } AOBA).$$
   Each quarter is the mirror image of the first-quadrant quarter $AOBA$⁠, so all four have the same area. In $AOBA$ the top edge is a single curve over $0 \le x \le a$⁠.
   A region symmetrical about both axes is four congruent quarters. Basics: one quarter, times four

2. Find the top edge of the quarter

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

   In the first quadrant $y \ge 0$⁠, so the top edge of $AOBA$ is $y = \frac{b}{a}\sqrt{a^2 - x^2}$⁠, $0 \le x \le a$⁠.

   $\sqrt{b^2/a^2} = \frac{b}{a}$ because $a, b > 0$⁠. The $+$ sign gives the upper half of the ellipse, the part the vertical strips of $AOBA$ reach up to.
   Take the positive root above the $x$-axis. Basics: solve for y to get the top edge

3. Write the area as an integral (vertical strips)

   $$\begin{aligned} \text{Area} &= 4\int_0^a y\,dx \\ &= 4\int_0^a\frac{b}{a}\sqrt{a^2 - x^2}\,dx \\ &= \frac{4b}{a}\int_0^a\sqrt{a^2 - x^2}\,dx. \end{aligned}$$
   $AOBA$ is bounded by the curve, the $x$-axis and the ordinates $x = 0$⁠, $x = a$⁠, so its area is $\int_0^a y\,dx$⁠. The constant $\frac{b}{a}$ comes out of the integral.
   Section 8.2: $A = \int_a^b y\,dx$⁠; a constant factor can be taken outside an integral. Lesson: the area under a curve

4. Integrate with the square-root formula

   $$\begin{aligned} &\int_0^a\sqrt{a^2 - x^2}\,dx \\ &= \Big[\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\Big]_0^a \\ &= \Big(\frac{a}{2}\times 0 + \frac{a^2}{2}\sin^{-1}1\Big) - 0 \\ &= \frac{a^2}{2}\cdot\frac{\pi}{2} = \frac{\pi a^2}{4}. \end{aligned}$$
   This is the book's first "(Why?)": the formula of Section 7.6.2 (iii). Its derivative is $\sqrt{a^2 - x^2}$ for $-a < x < a$ (checked in Example 1, Step 4), and the antiderivative is continuous at $x = a$⁠, so Theorem 2 gives the value. At $x = a$ the root is 0 and $\sin^{-1}1 = \frac{\pi}{2}$⁠; at $x = 0$ both terms are 0, since $\sin^{-1}0 = 0$⁠.
   Section 7.6.2 (iii); principal values $\sin^{-1}1 = \frac{\pi}{2}$⁠, $\sin^{-1}0 = 0$⁠. Basics: the quarter-circle integral

5. Multiply out

   $$\text{Area} = \frac{4b}{a}\cdot\frac{\pi a^2}{4} = \pi ab \text{ sq. units}.$$
   The factor 4 (four quarters) and the $\frac{1}{4}$ of the quarter-circle value cancel, and $\frac{a^2}{a} = a$⁠.
   One quarter, times four. Lesson: circles and ellipses, one quarter times four

6. The same area with horizontal strips

   Solving the equation for $x$ instead: $x = \pm\frac{a}{b}\sqrt{b^2 - y^2}$⁠, and in the first quadrant $x = \frac{a}{b}\sqrt{b^2 - y^2}$ for $0 \le y \le b$⁠. A horizontal strip has length $x$ and width $dy$⁠:

   $$\begin{aligned} \text{Area} &= 4\int_0^b x\,dy \\ &= \frac{4a}{b}\int_0^b\sqrt{b^2 - y^2}\,dy \\ &= \frac{4a}{b}\Big[\frac{y}{2}\sqrt{b^2 - y^2} \\ &\quad + \frac{b^2}{2}\sin^{-1}\frac{y}{b}\Big]_0^b \\ &= \frac{4a}{b}\Big[\Big(\frac{b}{2}\times 0 \\ &\quad + \frac{b^2}{2}\sin^{-1}1\Big) - 0\Big] \\ &= \frac{4a}{b}\cdot\frac{b^2}{2}\cdot\frac{\pi}{2} = \pi ab. \end{aligned}$$
   This is the book's second "(Why?)": $AOBA$ is bounded by the curve, the $y$-axis and the lines $y = 0$⁠, $y = b$⁠, so $A = \int_0^b x\,dy$⁠. The roles of $a$ and $b$ swap, and the square-root formula is used with $b$ in place of $a$⁠. Both methods give $\pi ab$⁠, as they must.
   Section 8.2: $A = \int_c^d x\,dy$⁠. Lesson: horizontal strips

7. Compare with the circle

   When $b = a$⁠, the ellipse is the circle $x^2 + y^2 = a^2$⁠, and $\pi ab$ becomes $\pi a^2$⁠, the area found in Example 1.

   A special case that agrees with a known answer is a quick test of a general formula. The ellipse is the circle of radius $a$ scaled vertically by the factor $\frac{b}{a}$⁠, and every strip's height, so the area, is multiplied by that factor: $\pi a^2\cdot\frac{b}{a} = \pi ab$⁠.
   The top edges differ only by the constant factor $\frac{b}{a}$⁠. Lesson: circle and ellipse, Examples 1–2

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