Area under a curve and along the y-axis

What you will be able to do

• find the area between a curve, the $x$-axis and two vertical lines;
• find the area between a curve, the $y$-axis and two horizontal lines;
• use the areas of a line, a parabola, a circle and an ellipse.

Area under a curve between two ordinates

Take a curve $y = f(x)$ that does not go below the $x$-axis between $x = a$ and $x = b$. The region bounded by the curve, the $x$-axis and the lines $x = a$ and $x = b$ is cut into vertical strips. A strip at $x$ has height $y$ and a thin width $dx$. The area is

A word on words. Many Indian books call the vertical lines $x = a$ and $x = b$ the ordinates. Strictly, an ordinate is the $y$-coordinate of a point, but "the ordinates $x = a$ and $x = b$" always means the vertical lines. A-Level, IB and AP questions simply say "the lines $x = a$ and $x = b$", or "between $x = a$ and $x = b$".

Area along the y-axis

Some regions are better cut into horizontal strips. This happens when the curve is given as $x = g(y)$. The region is bounded by the curve, the $y$-axis and the lines $y = c$ and $y = d$. A strip at height $y$ runs from the $y$-axis to the curve, so its length is $x = g(y)$ and its thickness is $dy$. The area is

The rule is the same as before with the two axes swapped. Always match the strip to the variable: vertical strips go with $dx$ and limits in $x$, horizontal strips go with $dy$ and limits in $y$.

Four standard shapes

A line. The region under $y = 2x + 1$ from $x = 1$ to $x = 4$ is a trapezium with parallel sides $3$ and $9$ and width $3$. By the trapezium formula its area is $\tfrac{1}{2}(3 + 9) \times 3 = 18$. The integral $\left[x^2 + x\right]_1^4 = 20 - 2 = 18$ agrees.

A parabola. The curve $y^2 = 4x$ with $y \ge 0$ is $y = 2\sqrt{x}$. The area under it from $x = 0$ to $x = 4$ is $\displaystyle\int_0^4 2\sqrt{x}\,dx = \dfrac{4}{3} \times 8 = \dfrac{32}{3}$. Cut along the $y$-axis instead and the region between the curve and the $y$-axis, for $0 \le y \le 4$, has area $\dfrac{16}{3}$. The two pieces fill the $4$ by $4$ square: $\dfrac{32}{3} + \dfrac{16}{3} = 16$.

A circle. For $x^2 + y^2 = a^2$, the upper half is $y = \sqrt{a^2 - x^2}$. The standard integral from the previous chapter is

From $0$ to $a$ the first term gives $0$ at both ends, and the second gives $\dfrac{a^2}{2} \times \dfrac{\pi}{2}$. So one quarter of the circle has area $\dfrac{\pi a^2}{4}$, and the whole circle has area $4 \times \dfrac{\pi a^2}{4} = \pi a^2$.

An ellipse. The ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ has $y = \dfrac{b}{a}\sqrt{a^2 - x^2}$ in its upper half. Every height of the circle of radius $a$ is multiplied by $\dfrac{b}{a}$, so every area is too. The whole ellipse has area $\dfrac{b}{a} \times \pi a^2 = \pi ab$.

A worked example

The ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ has $a = 5$ and $b = 3$.

1. The whole ellipse. The area is $\pi ab = 15\pi \approx 47.12$ square units.
2. The first-quadrant part. It is a quarter of the ellipse: $\dfrac{15\pi}{4} \approx 11.78$ square units.
3. A part of a circle. For the circle $x^2 + y^2 = 25$, find the area under the arc from $x = 0$ to $x = 4$. The standard antiderivative is $\dfrac{x}{2}\sqrt{25 - x^2} + \dfrac{25}{2}\sin^{-1}\dfrac{x}{5}$. At $x = 4$ it equals $6 + \dfrac{25}{2}\sin^{-1}\dfrac{4}{5}$, and at $x = 0$ it equals $0$. The area is $6 + \dfrac{25}{2}\sin^{-1}\dfrac{4}{5} \approx 17.59$ square units.

Practise with new numbers

Try it in Python

The program cuts a quarter of a circle of radius $1$ into thin strips, and multiplies by $4$. The result should be close to $\pi$. It then does the same for the ellipse of the worked example.

Common mistakes

• Mixing the strips. A horizontal strip needs $dy$ and limits in $y$. Writing $\int x\,dx$ between $y$-limits gives a wrong answer.
• Forgetting the sign of the side. If the curve lies to the left of the $y$-axis, $x$ is negative and the integral is negative. The area is its absolute value.
• Forgetting $\pi$. The area of a circle or an ellipse has a factor $\pi$.
• Taking one semi-axis only. The area of an ellipse is $\pi ab$, not $\pi a^2$.

Remember these

• Under a curve, between $x = a$ and $x = b$: $A = \displaystyle\int_a^b y\,dx$.
• Beside the $y$-axis, between $y = c$ and $y = d$: $A = \displaystyle\int_c^d x\,dy$.
• A circle of radius $a$ has area $\pi a^2$, and an ellipse with semi-axes $a$ and $b$ has area $\pi ab$.
• Check an answer with a shape you know, such as a trapezium.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the area bounded by $y = x^3$, the $x$-axis and the lines $x = 1$ and $x = 3$.

2. Find the area between the curve $x = y^2$, the $y$-axis and the lines $y = 1$ and $y = 3$.

3. Find the area under $y = 5 - x$ from $x = 1$ to $x = 4$, and check it with a trapezium.

4. Find the area of the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{4} = 1$.

5. Find the area of the quarter of the circle $x^2 + y^2 = 36$ in the first quadrant.

References

• OpenStax, Calculus Volume 1, Chapter 6, section on areas between curves, and Chapter 5, section on the definite integral.
• NCERT, Mathematics Part II for Class XII, Chapter 8, section on the area under simple curves.