Area inside a circle or ellipse

The circle and the ellipse have known total areas: $\pi r^2$ for a circle of radius $r$, $\pi a b$ for an ellipse with semi-axes $a$ and $b$. But the problems you will face involve parts of these regions , a quadrant, a strip between two vertical chords, a region between an ellipse and an inscribed line. Definite integrals handle all of these.

The circle $x^2 + y^2 = r^2$

Solving for the upper half: $y = \sqrt{r^2 - x^2}$. Area of the upper half: $A_{\text{half}} = \int_{-r}^r \sqrt{r^2 - x^2}\,dx = \left[\dfrac{x}{2}\sqrt{r^2 - x^2} + \dfrac{r^2}{2}\sin^{-1}(x/r)\right]_{-r}^r.$ At $x = r$: first term zero, second $= \dfrac{r^2}{2} \cdot \dfrac{\pi}{2} = \dfrac{\pi r^2}{4}$. At $x = -r$: $0 + \dfrac{r^2}{2}\cdot (-\dfrac{\pi}{2}) = -\dfrac{\pi r^2}{4}$. Difference: $\dfrac{\pi r^2}{2}$. So the full circle has area $\pi r^2$. The integral confirms the elementary formula.

The ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

Upper half: $y = b\sqrt{1 - x^2/a^2} = \dfrac{b}{a}\sqrt{a^2 - x^2}$. $A_{\text{half}} = \int_{-a}^a \dfrac{b}{a}\sqrt{a^2 - x^2}\,dx = \dfrac{b}{a} \cdot \dfrac{\pi a^2}{2} = \dfrac{\pi a b}{2}.$ Total area: $\pi a b$.

Strip cut from a circle by vertical lines

For a circle $x^2 + y^2 = r^2$ and vertical lines $x = c$ and $x = d$ with $-r \le c < d \le r$, the area inside the circle between the lines is $A = 2\int_c^d \sqrt{r^2 - x^2}\,dx.$ Use the standard integral $\int \sqrt{r^2 - x^2}\,dx = \dfrac{x}{2}\sqrt{r^2 - x^2} + \dfrac{r^2}{2}\sin^{-1}(x/r)$.

Quadrant or sector

For a sector of a circle, the trig-substitution form integrates beautifully because $\sin^{-1}$ delivers the central angle. For a quadrant the answer is $\dfrac{\pi r^2}{4}$.

Worked examples

Example 1. Area of the circle $x^2 + y^2 = 16$.

By formula: $\pi r^2 = 16\pi$.

Example 2. Area inside the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$.

$a = 3$, $b = 2$. Area $= \pi \cdot 3 \cdot 2 = 6\pi$.

Example 3. Area inside $x^2 + y^2 = 4$ between $x = 0$ and $x = 1$ (upper and lower halves).

Total: $A = 2\int_0^1 \sqrt{4 - x^2}\,dx = 2\left[\dfrac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}(x/2)\right]_0^1 = 2\left[\dfrac{1}{2}\sqrt 3 + 2 \cdot \dfrac{\pi}{6}\right] = \sqrt 3 + \dfrac{2\pi}{3}$.

Example 4. Area inside the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ in the first quadrant.

By symmetry: $\dfrac{1}{4}$ of total $= \dfrac{1}{4} \cdot \pi \cdot 4 \cdot 3 = 3\pi$.

Example 5. Area enclosed by $x^2 + y^2 = 4$ above the line $y = 1$.

Region: circle of radius $2$ above $y = 1$. By horizontal slices (since the line is horizontal): at height $y \in [1, 2]$, the strip extends from $x = -\sqrt{4 - y^2}$ to $x = \sqrt{4 - y^2}$. Area $= \int_1^2 2\sqrt{4 - y^2}\,dy = \left[y\sqrt{4 - y^2} + 4\sin^{-1}(y/2)\right]_1^2 = (0 + 4 \cdot \pi/2) - (\sqrt 3 + 4 \cdot \pi/6) = 2\pi - \sqrt 3 - \dfrac{2\pi}{3} = \dfrac{4\pi}{3} - \sqrt 3$.

Example 6. Area of the region $\{(x, y) : x^2 + y^2 \le 1,\ x + y \ge 1\}$.

This is the smaller circular segment cut off by the chord $x + y = 1$. The chord meets the circle at $(1, 0)$ and $(0, 1)$. Central angle subtended: $\pi/2$. Area of the sector: $\dfrac{1}{4}\pi r^2 = \pi/4$. Area of the triangle with vertices $(0, 0)$, $(1, 0)$, $(0, 1)$: $1/2$. Segment area = sector $-$ triangle $= \dfrac{\pi}{4} - \dfrac{1}{2}$.

Try it yourself

1. Area inside $x^2 + y^2 = 25$.
2. Area inside $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$.
3. Area inside the circle $x^2 + y^2 = 9$ in the first quadrant.
4. Area inside $x^2 + y^2 = 4$ between $x = -1$ and $x = 1$.
5. Area inside $\dfrac{x^2}{4} + \dfrac{y^2}{16} = 1$ above the $x$-axis.
6. Area enclosed by $x^2 + y^2 = 4$ and $y = x$ (region above the line, inside the circle).
7. Area of the smaller segment cut from $x^2 + y^2 = 4$ by $y = 1$.
8. Area enclosed by $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ between the chord $\dfrac{x}{3} + \dfrac{y}{2} = 1$ and the arc.
9. Area enclosed by $x^2 + y^2 \le 4$ and $|y| \le 1$.
10. Area enclosed by the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ in the first quadrant.
11. Area inside the circle $(x - 1)^2 + y^2 = 1$. (translation does not change area)
12. Area enclosed by $x^2 + y^2 \le r^2$ and $x \ge h$, where $0 \le h \le r$.
13. Area enclosed by $x^2 + y^2 = a^2$ and $|x| + |y| = a$ , find the region inside the circle but outside the square.
14. Area common to two circles $x^2 + y^2 = 1$ and $(x - 1)^2 + y^2 = 1$.

Pitfalls and tricks

• Symmetry is your friend. Compute a quarter or half and multiply.
• Sketch carefully. It is easy to integrate the wrong region , drawing reveals which is which.
• For circular segments, the sector-minus-triangle decomposition is faster than integration.
• Horizontal vs vertical slices. A circle is symmetric in both, but for problems cut by horizontal lines, horizontal slices give cleaner integrals.
• Use the standard integral $\int \sqrt{r^2 - x^2}\,dx$ from your formula sheet , don't re-derive.