Exercise 8.1 · Q1: Area of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$

Basics you need

Solution

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

   Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$⁠: $a^2 = 16$ and $b^2 = 9$⁠, so $a = 4$ and $b = 3$⁠.

   Putting $y = 0$⁠: $x^2 = 16$⁠, $x = \pm 4$⁠. Putting $x = 0$⁠: $y^2 = 9$⁠, $y = \pm 3$⁠. So the ellipse meets the axes at $(\pm 4, 0)$ and $(0, \pm 3)$⁠. Replacing $x$ by $-x$ or $y$ by $-y$ does not change the equation, so it is symmetrical about both axes:

   $$\begin{aligned} \text{Area} &= 4 \times (\text{area of the} \\ &\qquad \text{first-quadrant quarter}). \end{aligned}$$
   The number under $x^2$ fixes how far the ellipse reaches along the $x$-axis, and the number under $y^2$ how far along the $y$-axis. The four quarters are mirror images, so one integral over the first quadrant is enough.
   One quarter, times four. Basics: one quarter, times four

2. Find the top edge of the quarter

   $$\begin{aligned} \frac{y^2}{9} &= 1 - \frac{x^2}{16} = \frac{16 - x^2}{16} \\ y^2 &= \frac{9}{16}\left(16 - x^2\right) \\ y &= \pm\frac34\sqrt{16 - x^2}. \end{aligned}$$

   In the first quadrant $y \ge 0$⁠, so the top edge is $y = \frac34\sqrt{16 - x^2}$ for $0 \le x \le 4$⁠.

   $\sqrt{\frac{9}{16}} = \frac34$⁠. The $+$ root is the upper half of the ellipse, which is the top of every vertical strip in the first quadrant. For $0 \le x \le 4$⁠, $16 - x^2 \ge 0$⁠, 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

   $$\begin{aligned} \text{Area} &= 4\int_0^4 y\,dx \\ &= 4\int_0^4\frac34\sqrt{16 - x^2}\,dx \\ &= 3\int_0^4\sqrt{16 - x^2}\,dx. \end{aligned}$$
   The quarter is bounded by the curve, the $x$-axis and the ordinates $x = 0$ and $x = 4$ (where the ellipse meets the $x$-axis). Vertical strips of height $y$ and width $dx$ add up to $\int_0^4 y\,dx$⁠; the constants combine as $4 \cdot \frac34 = 3$⁠.
   Section 8.2: $A = \int_a^b y\,dx$⁠. Lesson: the area under a curve

4. Integrate with the square-root formula

   With $a = 4$⁠, $\frac{a^2}{2} = 8$⁠:

   $$\begin{aligned} &\int_0^4\sqrt{16 - x^2}\,dx \\ &= \Big[\frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\frac{x}{4}\Big]_0^4 \\ &= \big(2 \times 0 + 8\sin^{-1}1\big) - \big(0 + 8\sin^{-1}0\big) \\ &= 8\cdot\frac{\pi}{2} = 4\pi. \end{aligned}$$
   At $x = 4$ the root is $\sqrt{0} = 0$ and $\sin^{-1}1 = \frac{\pi}{2}$ (principal value); at $x = 0$ both terms are 0. The value $4\pi$ is a quarter of the area $16\pi$ of the circle of radius 4, as it should be: $\sqrt{16 - x^2}$ is that circle's top half.
   Section 7.6.2 (iii); Theorem 2, Section 7.8.3. Basics: the quarter-circle integral

5. Multiply

   $$\text{Area} = 3 \times 4\pi = 12\pi \text{ sq. units}.$$
   The factor 3 already includes the 4 quarters and the $\frac34$ of the top edge.
   Area of the whole ellipse $= 4 \times$ the quarter. Lesson: areas of ellipses

6. Check with $\pi ab$

   For $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ the area is $\pi ab$ (Example 2). Here $\pi \cdot 4 \cdot 3 = 12\pi$⁠, the same.

   The general result is a quick check. In an exam answer the integral must still be shown; the formula alone does not earn the marks.
   Area of an ellipse $= \pi ab$⁠. Lesson: circle and ellipse, Examples 1–2

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