Example 1: Area enclosed by the circle $x^2 + y^2 = a^2$

Basics you need

Solution

1. Sketch the circle and use its symmetry

   The circle has centre $O(0, 0)$ and radius $a$⁠. Putting $y = 0$ gives $x = \pm a$⁠, and putting $x = 0$ gives $y = \pm a$⁠, so it meets the axes at $(\pm a, 0)$ and $(0, \pm a)$⁠. Call $A(a, 0)$ and $B(0, a)$⁠.

   If $(x, y)$ is on the circle, so are $(x, -y)$ and $(-x, y)$⁠, because only $x^2$ and $y^2$ appear in the equation. So the circle is symmetrical about both axes, and

   $$\begin{aligned} \text{Area} &= 4 \times (\text{area of} \\ &\qquad \text{the region } AOBA). \end{aligned}$$
   The axes cut the circle into four congruent quarters. Working with one quarter keeps the curve as a single, simple top edge, $y \ge 0$⁠, between $x = 0$ and $x = a$⁠.
   Replacing $y$ by $-y$ (or $x$ by $-x$⁠) leaves $x^2 + y^2$ unchanged. Basics: one quarter, times four

2. Find the top edge of the quarter

   $$\begin{aligned} x^2 + y^2 &= a^2 \\ \Rightarrow\; y^2 &= a^2 - x^2 \\ \Rightarrow\; y &= \pm\sqrt{a^2 - x^2}. \end{aligned}$$

   The region $AOBA$ lies in the first quadrant, where $y \ge 0$⁠, so its top edge is

   $$y = \sqrt{a^2 - x^2}, \quad 0 \le x \le a.$$
   The $+$ root is the upper semicircle, the $-$ root the lower one. A strip in the first quadrant reaches up to the upper semicircle, so its height is the positive root. For $0 \le x \le a$⁠, $a^2 - x^2 \ge 0$⁠, so the root is real.
   $\sqrt{\;}$ always means the non-negative root. Basics: solve for y to get the top edge

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

   A vertical strip at $x$ has height $y = \sqrt{a^2 - x^2}$ and width $dx$⁠. Adding the strips from $x = 0$ to $x = a$⁠:

   $$\begin{aligned} \text{area of } AOBA &= \int_0^a y\,dx \\ &= \int_0^a\sqrt{a^2 - x^2}\,dx. \end{aligned}$$

   So the whole area is $4\displaystyle\int_0^a\sqrt{a^2 - x^2}\,dx$⁠.

   The quarter is bounded by the curve, the $x$-axis and the ordinates $x = 0$ and $x = a$⁠, which is exactly the situation of the formula $A = \int_a^b y\,dx$⁠. The integrand is continuous and $\ge 0$ on $[0, a]$⁠, so the integral is the area itself (no part is below the axis).
   Section 8.2: $A = \int_a^b y\,dx$⁠. Lesson: the area under a curve is a sum of strips

4. Integrate with the square-root formula

   By the formula of Section 7.6.2 (iii), an antiderivative of $\sqrt{a^2 - x^2}$ is

   $$F(x) = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}.$$

   Check: differentiate. For $-a < x < a$⁠, by the product and chain rules,

   $$\begin{aligned} F'(x) &= \frac12\sqrt{a^2 - x^2} - \frac{x^2}{2\sqrt{a^2 - x^2}} \\ &\quad + \frac{a^2}{2}\cdot\frac{1}{\sqrt{a^2 - x^2}} \\ &= \frac12\sqrt{a^2 - x^2} + \frac{a^2 - x^2}{2\sqrt{a^2 - x^2}} \\ &= \sqrt{a^2 - x^2}. \end{aligned}$$
   The derivative of $\sin^{-1}\frac{x}{a}$ is $\frac{1}{\sqrt{1 - x^2/a^2}}\cdot\frac{1}{a} = \frac{1}{\sqrt{a^2 - x^2}}$ for $a > 0$⁠. The last two terms combine because $\frac{a^2 - x^2}{\sqrt{a^2 - x^2}} = \sqrt{a^2 - x^2}$⁠. So $F$ really is an antiderivative of the integrand.
   Section 7.6.2 (iii), which comes from integration by parts. Basics: the quarter-circle integral

5. Put in the limits

   $$\begin{aligned} &\int_0^a\sqrt{a^2 - x^2}\,dx \\ &= \Big[F(x)\Big]_0^a \\ &= \Big(\frac{a}{2}\cdot 0 + \frac{a^2}{2}\sin^{-1}1\Big) \\ &\quad - \Big(0 + \frac{a^2}{2}\sin^{-1}0\Big) \\ &= \frac{a^2}{2}\cdot\frac{\pi}{2} = \frac{\pi a^2}{4}. \end{aligned}$$
   At $x = a$ the root is $\sqrt{0} = 0$ and $\frac{x}{a} = 1$⁠; at $x = 0$ both terms are 0. The principal value $\sin^{-1}1 = \frac{\pi}{2}$ because $\sin\frac{\pi}{2} = 1$ and $\frac{\pi}{2}$ lies in the principal range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠; similarly $\sin^{-1}0 = 0$⁠. One fine point: $F'(x) = \sqrt{a^2 - x^2}$ was checked for $x < a$ only, since $\sin^{-1}$ has no derivative at 1. That is enough, because $F$ is continuous at $x = a$⁠: $\int_0^a = \lim_{t \to a^-}\int_0^t = \lim_{t \to a^-}\big(F(t) - F(0)\big) = F(a) - F(0)$⁠.
   Theorem 2, Section 7.8.3; principal value branch of $\sin^{-1}$ (Chapter 2). Basics: a definite integral is a number

6. Multiply by four

   $$\text{Area} = 4\cdot\frac{\pi a^2}{4} = \pi a^2 \text{ sq. units}.$$
   The integral gave one quarter, $AOBA$⁠; the symmetry of Step 1 makes the whole circle four such quarters. This is the familiar formula $\pi r^2$ with $r = a$⁠, now proved by integration.
   Area of a region symmetrical about both axes $= 4 \times$ the first-quadrant part. Lesson: circles and ellipses, one quarter times four

7. The same area with horizontal strips

   The book also works the quarter with horizontal strips. A horizontal strip at height $y$ has length $x$ and width $dy$⁠. Solving for $x$⁠, the right half of the circle is $x = \sqrt{a^2 - y^2}$⁠, and $y$ runs from 0 to $a$⁠:

   $$\begin{aligned} \text{Area} &= 4\int_0^a x\,dy = 4\int_0^a\sqrt{a^2 - y^2}\,dy \\ &= 4\Big[\frac{y}{2}\sqrt{a^2 - y^2} + \frac{a^2}{2}\sin^{-1}\frac{y}{a}\Big]_0^a \\ &= 4\cdot\frac{a^2}{2}\cdot\frac{\pi}{2} = \pi a^2. \end{aligned}$$
   This answers the book's "(Why?)": the quarter is bounded by the curve $x = \sqrt{a^2 - y^2}$⁠, the $y$-axis and the lines $y = 0$⁠, $y = a$⁠, so the formula $A = \int_c^d x\,dy$ applies. The integral is the one of Steps 4 and 5 with $y$ in place of $x$⁠, so it has the same value, $\frac{\pi a^2}{4}$⁠.
   Section 8.2: $A = \int_c^d x\,dy$ for horizontal strips. Lesson: horizontal strips

Every fact used, and where it comes from