Splitting a region and choosing the strips

What you will be able to do

• split a region at each crossing and add the areas of the pieces;
• use symmetry to shorten the working;
• choose vertical or horizontal strips, so that one integral is enough.

When the upper curve changes

Draw $y = x^3$ and $y = 4x$. They cross at $x = -2$, $x = 0$ and $x = 2$. For $x$ between $0$ and $2$ the line is above the cubic. For $x$ between $-2$ and $0$ the cubic is above the line. The curves enclose two lobes.

If you integrate $4x - x^3$ from $-2$ to $2$ in one go, the right lobe counts as positive and the left lobe as negative. They cancel and the answer is $0$. That is wrong. The area is always positive, so split at every crossing.

The rule is simple.

1. Solve $f(x) = g(x)$ and list all the meeting points inside the range.
2. On each piece, find which curve is above.
3. Integrate (upper $-$ lower) on each piece and add the results.

In symbols, the area is $\displaystyle\int_a^b |f(x) - g(x)|\,dx$, with the absolute value removed piece by piece.

A worked example: two lobes

Find the area enclosed by $y = x^3$ and $y = 4x$.

1. Meeting points. $x^3 = 4x$ gives $x(x^2 - 4) = 0$, so $x = -2, 0, 2$.
2. Symmetry. Both curves are odd functions. The left lobe is the right lobe turned half a turn about the origin. The two lobes have equal areas.
3. One lobe. On $[0, 2]$ the line is above, so the area is $\displaystyle\int_0^2 (4x - x^3)\,dx = 8 - 4 = 4$.
4. Both lobes. The total area is $2 \times 4 = 8$.

The same idea works for $\sin x$ and $\cos x$ between $0$ and $\dfrac{\pi}{2}$. They cross at $x = \dfrac{\pi}{4}$, and the upper curve changes there.

Circles and ellipses with lines

A line can cut a piece off a circle or an ellipse. Use the area of the whole shape, or a quarter of it, and subtract the triangle.

Take the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ and the chord $\dfrac{x}{5} + \dfrac{y}{3} = 1$, which joins $(5, 0)$ to $(0, 3)$. The quarter of the ellipse in the first quadrant has area $\dfrac{\pi \times 5 \times 3}{4}$. The triangle under the chord has area $\dfrac{5 \times 3}{2}$. The small piece between the chord and the ellipse has area $\dfrac{15\pi}{4} - \dfrac{15}{2} \approx 4.28$.

Choosing vertical or horizontal strips

Look at the region bounded by the parabola $y^2 = 2x$ and the line $y = x - 4$. They meet at $(2, -2)$ and $(8, 4)$.

With vertical strips, the bottom of a strip is the lower half of the parabola for $x$ from $0$ to $2$. For $x$ from $2$ to $8$ the bottom is the line. The bottom changes at $x = 2$, so two integrals are needed.

With horizontal strips, every strip runs from the parabola on the left, $x = \dfrac{y^2}{2}$, to the line on the right, $x = y + 4$. The left and right ends never change, so one integral is enough:

The area is $18$. Vertical strips give the same number, with more work: $\displaystyle\int_0^2 2\sqrt{2x}\,dx + \int_2^8 \big(\sqrt{2x} - (x - 4)\big)\,dx = 18$.

A rule of thumb. Ask which edge of the strip changes along the way. Choose the direction in which the top and the bottom stay the same curve. Curves written as $x = g(y)$ usually want horizontal strips.

Practise with new numbers

Try it in Python

The program adds strips for $y = x^3$ and $y = 4x$ from $-2$ to $2$. It keeps two running totals: one with the sign, and one with each strip made positive.

Common mistakes

• One integral across a crossing. The lobes cancel. Split at every crossing, or take the absolute value of each strip.
• Missing a meeting point. A cubic can cross a line three times. Solve the equation fully.
• Keeping vertical strips when they need splitting. Try horizontal strips before you split.
• Mixing up the variable. With horizontal strips integrate $dy$, with limits that are $y$-values, not $x$-values.

Remember these

• Split the region at every crossing, and add the pieces. Each piece is (upper $-$ lower).
• Symmetry about the origin or an axis halves the work.
• Choose the strip direction in which the top and the bottom never change.
• For horizontal strips the strip width is (right $-$ left), and the limits are $y$-values.

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 enclosed by $y = x$ and $y = x^3$.

2. Find the area of the region between $y = |x|$ and $y = 2 - x^2$.

3. Find the area between the parabola $x = y^2$ and the line $x = 4$. Which strips are shorter to work with?

4. Find the area between $y = \sin x$ and $y = \cos x$ from $x = 0$ to $x = \dfrac{\pi}{2}$.

5. A student works out $\displaystyle\int_{-3}^{3} (9x - x^3)\,dx = 0$ for the region enclosed by $y = 9x$ and $y = x^3$, and says the area is $0$. What is wrong, and what is the area?

References

• OpenStax, Calculus Volume 1, Chapter 6, section 6.1 on areas between curves.
• NCERT, Mathematics Part II for Class XII, Chapter 8, the section on the area between two curves.