Composite regions and area between three or more curves

Real exam problems often involve three boundaries , say two curves and a line , that together enclose a region. To find such an area you must locate every intersection point, identify which curve forms the top (or right) boundary on each sub-interval, and sum the appropriate integrals.

The general procedure

1. Sketch. Plot all the bounding curves and shade the enclosed region.
2. Find all pairwise intersections of the boundary curves. These give the candidates for limits.
3. Identify the boundary functions on each sub-interval: which curve is on top, which is on the bottom (or right, left, for horizontal slices).
4. Set up one integral per sub-interval and add them.

The key insight: as you sweep across the region, the top (or right) boundary may change at intersection points. So may the bottom. Each change forces a new integral on a new sub-interval.

Three-curve example schematic

Suppose the region is bounded by $y = f(x)$, $y = g(x)$, and $y = h(x)$, with intersections:

• $f \cap g$ at $x = p$
• $g \cap h$ at $x = q$
• $f \cap h$ at $x = r$

A common configuration: from $x = p$ to $x = q$, the region is between $g$ (top) and $f$ (bottom); from $x = q$ to $x = r$, between $h$ (top) and $f$ (bottom). Area = $\int_p^q (g - f)\,dx + \int_q^r (h - f)\,dx$.

The exact partition depends on the geometry; always confirm by sketching.

Region bounded by a curve and two lines

A typical setup: the region bounded by a curve $y = f(x)$ and two lines $y = m_1 x$ and $y = m_2 x$ through the origin. Find where each line meets the curve, then either split into two integrals or use horizontal slices.

Worked examples

Example 1. Area of the triangle bounded by the lines $y = x$, $y = 2x$, and $x + y = 6$.

Step 1: find vertices. $y = x \cap y = 2x \Rightarrow x = 0$, vertex $(0,0)$. $y = x \cap x + y = 6 \Rightarrow x = 3$, vertex $(3, 3)$. $y = 2x \cap x + y = 6 \Rightarrow 3x = 6 \Rightarrow x = 2$, vertex $(2, 4)$.

Vertical sweep: on $[0, 2]$, top is $y = 2x$, bottom is $y = x$. On $[2, 3]$, top is $y = 6 - x$, bottom is $y = x$.

Area $= \int_0^2 (2x - x)\,dx + \int_2^3 (6 - x - x)\,dx = \int_0^2 x\,dx + \int_2^3 (6 - 2x)\,dx = 2 + (6x - x^2)\big|_2^3 = 2 + ((18-9)-(12-4)) = 2 + 1 = 3$.

Example 2. Area enclosed by $y^2 = 4x$ and $x^2 = 4y$.

These are two parabolas , one opening right, one opening up. Solve: from the first, $x = y^2/4$. Substitute into second: $(y^2/4)^2 = 4y \Rightarrow y^4 = 64 y \Rightarrow y(y^3 - 64) = 0 \Rightarrow y = 0$ or $y = 4$. So intersections at $(0, 0)$ and $(4, 4)$.

Vertical sweep on $[0, 4]$: top is $y = 2\sqrt x$ (from $y^2 = 4x$, upper branch), bottom is $y = x^2/4$. Area $= \int_0^4 \left(2\sqrt x - \dfrac{x^2}{4}\right)\,dx = \dfrac{2 \cdot 2}{3} x^{3/2}\big|_0^4 - \dfrac{x^3}{12}\big|_0^4 = \dfrac{4 \cdot 8}{3} - \dfrac{64}{12} = \dfrac{32}{3} - \dfrac{16}{3} = \dfrac{16}{3}$.

Example 3. Area enclosed by the parabola $y = x^2$, the line $y = x + 2$, and the $x$-axis.

Intersections: parabola meets $x$-axis at $(0, 0)$; line meets $x$-axis at $(-2, 0)$; parabola meets line at $x^2 = x + 2 \Rightarrow x = -1, 2$.

Sketch reveals: region is bounded on the left by the line $y = x + 2$ from $(-2, 0)$ to $(-1, 1)$, then by the parabola $y = x^2$ from $(-1, 1)$ to $(0, 0)$, then back along the $x$-axis from $(0, 0)$ to $(-2, 0)$. Use vertical sweep on $[-2, -1]$: top is $y = x + 2$, bottom is $y = 0$. On $[-1, 0]$: top is $y = x^2$, bottom is $y = 0$.

Area $= \int_{-2}^{-1}(x + 2)\,dx + \int_{-1}^0 x^2\,dx = \left[\dfrac{x^2}{2} + 2x\right]_{-2}^{-1} + \left[\dfrac{x^3}{3}\right]_{-1}^0 = (\dfrac{1}{2} - 2 - (2 - 4)) + (0 - (-\dfrac{1}{3})) = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$.

Example 4. Area enclosed by $y = |x|$ and $y = 1$.

By symmetry across the $y$-axis: area $= 2 \int_0^1 (1 - x)\,dx = 2 \cdot \dfrac{1}{2} = 1$.

Example 5. Area enclosed by $y = x^2$, $y = -x^2 + 4$, and $x = 1$ (in the first quadrant, right of the $y$-axis).

Parabolas meet at $x^2 = -x^2 + 4 \Rightarrow x^2 = 2 \Rightarrow x = \sqrt 2$ in the first quadrant.

But the line $x = 1$ is to the left of $\sqrt 2$. So the region runs from $x = 1$ to $x = \sqrt 2$, bounded above by $y = -x^2 + 4$ and below by $y = x^2$. Area $= \int_1^{\sqrt 2} ((-x^2 + 4) - x^2)\,dx = \int_1^{\sqrt 2}(4 - 2x^2)\,dx = [4x - \dfrac{2x^3}{3}]_1^{\sqrt 2}= (4\sqrt 2 - \dfrac{4\sqrt 2}{3}) - (4 - \dfrac{2}{3}) = \dfrac{8\sqrt 2}{3} - \dfrac{10}{3} = \dfrac{8\sqrt 2 - 10}{3}$.

Example 6. Region bounded by $y = \sin x$, $y = \cos x$, $x = 0$, $x = \pi/2$.

Curves cross at $x = \pi/4$. From $0$ to $\pi/4$, $\cos x \ge \sin x$. From $\pi/4$ to $\pi/2$, $\sin x \ge \cos x$. The region between the two curves on $[0, \pi/2]$ has area $\int_0^{\pi/2} |\sin x - \cos x|\,dx = 2 \int_0^{\pi/4}(\cos x - \sin x)\,dx = 2(\sqrt 2 - 1)$.

Try it yourself

1. Area enclosed by $y = x$, $y = 0$, $x = 1$, $x = 4$.
2. Area enclosed by $y = x^2$, $y = x + 6$, $x = 0$.
3. Triangle bounded by $y = x$, $y = 2x$, $x = 4$.
4. Area enclosed by $x = y^2$, $x = 4 - y^2$.
5. Area enclosed by $y = x^2$, $y = x$, and $x = 2$.
6. Area enclosed by $y = \sin x$, $y = \cos x$ on $[0, \pi]$.
7. Region enclosed by the triangle with vertices $(0, 0)$, $(4, 0)$, $(2, 3)$.
8. Area enclosed by $y = 2$, $y = x^2$, $x = 0$, $x = 2$.
9. Area enclosed by $y = e^x$, $y = e$, $x = 0$.
10. Area enclosed by $y^2 = x$, $y = x$, $y = 2$.
11. Area enclosed by $y = 2 - x^2$, $y = -x$, $x = 0$, $x = 1$.
12. Area enclosed by $|x| + |y| = 1$.
13. Area enclosed by $y = e^x$, $y = \ln x$, $x = 1$, $x = e$. (Note: these curves are far apart on $[1, e]$.)
14. Area enclosed by $y^2 = 4x$ and $4y = x^2$.

Pitfalls and tricks

• Always sketch the region first. Many composite problems are obvious once drawn.
• Find every intersection. Each one is a potential limit or split point.
• Read which curve is on top carefully on each piece , sign errors compound across multiple integrals.
• Symmetry across the $y$-axis halves the work for regions involving $|x|$.
• Watch for "extra" lines in the problem , the $x$-axis, $y$-axis, or a vertical line may be implicit boundaries.
• Box check: areas should be positive. If an integral comes out negative, you have the top and bottom reversed.