Computing Areas of Plane Figures

1. The area under a curve

Curvilinear trapezoid If \( y = f(x) \ge 0 \) for \( x \in [a, b] \), then the area of the curvilinear trapezoid bounded by the arc of the curve, the lines \(x = a\), \(x = b\) and the segment \([a, b]\) of the \(x\)-axis is \[ P = \int_a^b f(x)\,dx. \]

Why. Cut the figure by vertical lines into thin strips of width \(dx\). Each strip is almost a rectangle of height \(f(x)\), so its area is about \(f(x)\,dx\). Adding all strips gives a Riemann sum, and as the strips get thinner, the sum tends to the integral. The error disappears in the limit because the "curved tops" of the strips become negligible compared to the strips themselves.

Figure 1. The area under \(y = f(x)\) between \(a\) and \(b\), approximated by rectangles. A typical strip (highlighted) has width \(dx\) and height \(f(x)\); the integral adds up infinitely many such strips.

This "element of area" way of thinking, \( dP = (\text{height})\cdot(\text{width}) \), is the key to every formula below. Instead of memorizing formulas, ask: what does a thin strip look like, and what is its area?

Example 1 · One arch of the sine \[ P = \int_0^\pi\sin x\,dx = \big[-\cos x\big]_0^\pi = 1 - (-1) = 2. \] A curved arch of height 1 and base \(\pi \approx 3.14\) has area exactly \(2\), a bit less than \( \frac23\cdot\pi\cdot1 \approx 2.09 \) of a parabolic arch of the same size.

2. Signed area: when \(f\) goes below the axis

The condition \( f(x) \ge 0 \) in the formula matters. Where \( f(x) < 0 \), the "height" \(f(x)\) of a strip is negative, so the integral counts that part of the area with a minus sign. The integral is a signed area: parts above the axis count positively, parts below negatively.

Split at the zeros \( \int_0^{2\pi}\sin x\,dx = 0 \), but the area between the sine and the \(x\)-axis on \([0, 2\pi]\) is \(4\): the two arches cancel in the integral. To get the true (geometric) area, split the interval at the zeros of \(f\) and add the absolute values: \[ P = \int_a^b|f(x)|\,dx = \int_0^\pi\sin x\,dx + \Big|\int_\pi^{2\pi}\sin x\,dx\Big| = 2 + 2 = 4. \]

3. Area between two curves

If \( f(x) \ge g(x) \) on \([a, b]\), a vertical strip between the curves has height \( f(x) - g(x) \), whatever their position relative to the \(x\)-axis:

\[ P = \int_a^b\big[f(x) - g(x)\big]\,dx. \]

The formula of Section 1 is the special case \(g = 0\). The limits are usually the \(x\)-coordinates of the intersection points, found by solving \( f(x) = g(x) \).

Example 2 The area between \(y = x\) and \(y = x^2\). They meet where \(x = x^2\): \(x = 0\) and \(x = 1\). On \([0, 1]\), \(x \ge x^2\), so \[ P = \int_0^1(x - x^2)\,dx = \frac12 - \frac13 = \frac16. \]

If the curves cross inside the interval, the "upper" and "lower" curves swap. Then split at the crossing points, or integrate \( |f(x) - g(x)| \).

4. Slicing horizontally

Nothing forces the strips to be vertical. If the region is bounded on the left and right by curves \( x = g(y) \) and \( x = h(y) \), with \( h(y) \ge g(y) \) for \( c \le y \le d \), horizontal strips of thickness \(dy\) and length \( h(y) - g(y) \) give

\[ P = \int_c^d\big[h(y) - g(y)\big]\,dy. \]

Example 3 · Choosing the better direction The region bounded by the parabola \( y^2 = 2x \) and the line \( y = x - 4 \). Intersections: \( y^2 = 2(y + 4) \), so \( y^2 - 2y - 8 = 0 \), \(y = -2\) or \(y = 4\): points \((2, -2)\) and \((8, 4)\).
With vertical strips, the lower boundary changes at \(x = 2\) (first the lower half of the parabola, then the line), and two integrals are needed. With horizontal strips, the left boundary is always the parabola \( x = \frac{y^2}{2} \) and the right boundary is always the line \( x = y + 4 \): \[ P = \int_{-2}^4\Big(y + 4 - \frac{y^2}{2}\Big)dy = \Big[\frac{y^2}{2} + 4y - \frac{y^3}{6}\Big]_{-2}^{4} = \frac{40}{3} - \Big(-\frac{14}{3}\Big) = 18. \]

Figure 2. The region of Example 3. A horizontal strip (green) always runs from the parabola to the line, so one integral in \(y\) suffices.

5. Curves given parametrically

If the curve is given by \( x = x(t) \), \( y = y(t) \), substitute in \( P = \int y\,dx \): \( dx = x'(t)\,dt \), so

\[ P = \int_{t_1}^{t_2}y(t)\,x'(t)\,dt, \]

where \(t_1\) and \(t_2\) correspond to \(x = a\) and \(x = b\). (If the parameter runs "backwards", from right to left, the result comes out negative; take the absolute value.)

Example 4 · Area of an ellipse \( x = a\cos t \), \( y = b\sin t \). The upper half is traced as \(t\) goes from \(\pi\) to \(0\) (\(x\) from \(-a\) to \(a\)): \[ P = 2\int_\pi^0 b\sin t\cdot(-a\sin t)\,dt = 2ab\int_0^\pi\sin^2 t\,dt = 2ab\cdot\frac\pi2 = \pi ab. \] For \(a = b = r\), the circle, this is \(\pi r^2\).

6. Areas in polar coordinates

For a curve given in polar coordinates, \( \rho = f(\varphi) \), vertical strips are awkward. The natural pieces are thin sectors from the pole \(O\).

Curvilinear triangle If the curve is given in polar coordinates by \( \rho = f(\varphi) \), the area of the curvilinear triangle \(OAB\), bounded by the arc \(AB\) of the curve and the rays \(\varphi = \alpha\) and \(\varphi = \beta\), is \[ P = \frac12\int_\alpha^\beta\big[f(\varphi)\big]^2\,d\varphi. \]

Why the \(\frac12\) and the square? A circular sector of radius \(\rho\) and angle \(\theta\) has area \( \frac12\rho^2\theta \) (a fraction \( \frac{\theta}{2\pi} \) of the disk \(\pi\rho^2\)). Divide the region into thin sectors of angle \(d\varphi\). Each is almost a circular sector of radius \( \rho = f(\varphi) \), with area

\[ dP = \frac12\rho^2\,d\varphi = \frac12\big[f(\varphi)\big]^2\,d\varphi, \]

and adding them up gives the formula. The \(\frac12\rho^2\) plays the role that "height" \(f(x)\) played for vertical strips.

Figure 3. The curvilinear triangle \(OAB\) is filled by thin sectors from the pole. A typical sector (green) is almost a circular sector of radius \(\rho = f(\varphi)\) and angle \(d\varphi\), with area \(\frac12\rho^2\,d\varphi\).

Example 5 · Cardioid \( \rho = a(1 + \cos\varphi) \), \( 0 \le \varphi \le 2\pi \): \[ P = \frac12\int_0^{2\pi}a^2(1 + \cos\varphi)^2\,d\varphi = \frac{a^2}{2}\int_0^{2\pi}\big(1 + 2\cos\varphi + \cos^2\varphi\big)\,d\varphi = \frac{a^2}{2}(2\pi + 0 + \pi) = \frac{3\pi a^2}{2}. \] That is \(1.5\) times the area of the circle of radius \(a\).

Figure 4. The cardioid \(\rho = a(1 + \cos\varphi)\), with area \(\frac32\pi a^2\).

Example 6 · Lemniscate of Bernoulli \( \rho^2 = a^2\cos 2\varphi \). The curve exists where \( \cos 2\varphi \ge 0 \) and has two symmetric loops; a quarter of it corresponds to \( 0 \le \varphi \le \frac\pi4 \). Here \(\rho^2\) is given directly, which is convenient: \[ P = 4\cdot\frac12\int_0^{\pi/4}a^2\cos 2\varphi\,d\varphi = 2a^2\Big[\frac{\sin 2\varphi}{2}\Big]_0^{\pi/4} = a^2. \]

Example 7 · Spiral of Archimedes The first turn of \( \rho = a\varphi \), \( 0 \le \varphi \le 2\pi \), together with the segment of the polar axis, bounds the area \[ P = \frac12\int_0^{2\pi}a^2\varphi^2\,d\varphi = \frac{a^2}{2}\cdot\frac{(2\pi)^3}{3} = \frac{4\pi^3a^2}{3}. \] This is exactly \(\frac13\) of the area of the disk of radius \(2\pi a\) (the circle through the endpoint), a result already known to Archimedes.

Use symmetry, and watch the range of \(\varphi\) Polar curves often trace parts of themselves more than once, or exist only for some angles (like the lemniscate). Before integrating, sketch the curve, find the range of \(\varphi\) that traces the region exactly once, and use symmetry to integrate over a smaller part.

7. Summary

Region	Element of area	Formula
Under \(y = f(x) \ge 0\)	vertical strip \(f(x)\,dx\)	\( \int_a^b f(x)\,dx \)
Between \(g(x) \le f(x)\)	\( [f(x) - g(x)]\,dx \)	\( \int_a^b[f(x) - g(x)]\,dx \)
Between \(g(y) \le h(y)\)	horizontal strip \( [h(y) - g(y)]\,dy \)	\( \int_c^d[h(y) - g(y)]\,dy \)
Parametric \(x(t), y(t)\)	\( y\,dx = y(t)x'(t)\,dt \)	\( \Big\|\int_{t_1}^{t_2}y(t)x'(t)\,dt\Big\| \)
Polar \(\rho = f(\varphi)\)	thin sector \( \frac12\rho^2\,d\varphi \)	\( \frac12\int_\alpha^\beta f(\varphi)^2\,d\varphi \)

8. Practice problems

Try each one before opening the solution. More solved problems from this topic are in Applications of Definite Integrals.

Problem 1: The area bounded by \(y = \ln x\), the \(x\)-axis and the line \(x = e\).

\( \ln x \ge 0 \) for \(x \ge 1\): \( \int_1^e\ln x\,dx = \big[x\ln x - x\big]_1^e = 0 - (-1) = 1 \).

Problem 2: The area enclosed by \(y = x^3\) and \(y = x\).

They meet at \(x = -1, 0, 1\). By symmetry, \( P = 2\int_0^1(x - x^3)\,dx = 2\Big(\frac12 - \frac14\Big) = \frac12 \). (Note that \( \int_{-1}^1(x - x^3)\,dx = 0 \): the signs cancel.)

Problem 3: The area under \( y = e^{-x} \) for \( x \ge 0 \).

An improper integral: \( \int_0^\infty e^{-x}\,dx = 1 \). An infinitely long region can have a finite area.

Problem 4: Check the area of a circle using the polar curve \( \rho = 2a\cos\varphi \).

This is the circle of radius \(a\) through the pole, traced once for \( -\frac\pi2 \le \varphi \le \frac\pi2 \): \( P = \frac12\int_{-\pi/2}^{\pi/2}4a^2\cos^2\varphi\,d\varphi = 2a^2\cdot\frac\pi2 = \pi a^2 \).

Problem 5: The area of the three-petal rose \( \rho = a\sin 3\varphi \).

One petal: \( 0 \le \varphi \le \frac\pi3 \), area \( \frac12\int_0^{\pi/3}a^2\sin^2 3\varphi\,d\varphi = \frac{a^2}{2}\cdot\frac\pi6 = \frac{\pi a^2}{12} \). Three petals: \( \frac{\pi a^2}{4} \).

Problem 6: The area of the region bounded by the parabola \( y = 4 - x^2 \) and the \(x\)-axis.

\( \int_{-2}^2(4 - x^2)\,dx = 16 - \frac{16}{3} = \frac{32}{3} \). Check with Archimedes' rule: \( \frac23 \) of the enclosing rectangle \(4\times4\), i.e. \( \frac{32}{3} \).

Final thoughts

All the area formulas are one idea in different coordinates: cut the figure into thin pieces whose area you know, and add them up. In Cartesian coordinates the pieces are rectangles, \(f(x)\,dx\); in polar coordinates they are circular sectors, \(\frac12\rho^2\,d\varphi\). The practical skill is choosing the slicing that makes the boundaries simplest, keeping an eye on signs, and using symmetry. The same idea will later give volumes, arc lengths and surface areas.

Related posts: integrating trigonometric functions (used in the polar examples) and improper integrals.