Area bounded by a Curve and a Line

Introduction to the Concept

Finding the area bounded by a curve and a straight line is a natural extension of finding the area under a simple curve. In this case, the region is enclosed not by a curve and a coordinate axis, but by two different graphs.

The key idea is simple:

• at each point in the interval, take the upper function minus lower function if integrating with respect to $x$,
• or take the right function minus left function if integrating with respect to $y$.

Before writing the integral, we must first identify the shape of the enclosed region. This requires finding the points of intersection of the given curve and line, because these points provide the limits of integration.

So every problem of this type begins with geometry and algebra together:

1. understand the region,
2. find intersection points,
3. decide whether to integrate with respect to $x$ or $y$,
4. subtract the lower boundary from the upper boundary (or left from right).

Steps to Find the Area

To solve such problems systematically, follow these steps:

Step 1: Sketch the Region Draw a rough graph of the given curve and line. This helps you see:

• where they intersect,
• which graph lies above the other,
• whether symmetry can be used,
• whether one integral or split integrals are needed.

Step 2: Find the Points of Intersection Solve the equations simultaneously. These points usually provide the lower and upper limits of the integral.

Step 3: Choose the Variable of Integration

• If the graphs are naturally written as $y=f(x)$, integrate with respect to $x$.
• If the graphs are naturally written as $x=f(y)$, integrate with respect to $y$.
• Choose the variable that avoids splitting the region unnecessarily.

Step 4: Set up the Correct Difference

• Integrating with respect to $x$: If the top curve is $y=f(x)$ and the bottom curve is $y=g(x)$, then $$\text{Area} = \int_a^b [f(x)-g(x)]\,dx$$
• Integrating with respect to $y$: If the right boundary is $x=f(y)$ and the left boundary is $x=g(y)$, then $$\text{Area} = \int_c^d [f(y)-g(y)]\,dy$$

Step 5: Evaluate the Integral Carefully Always simplify first if possible, then integrate and apply the limits.

The most common mistake in these problems is reversing the order and writing lower minus upper, which gives a negative result. Since area must be positive, always check the geometry before writing the integrand.

Symmetry and Standard Results

Symmetry can reduce calculation substantially. For example, if the bounded region is symmetric about the x-axis or y-axis, it is often easier to find half the area and multiply by 2.

For the parabola $y^2 = 4ax$, the line $x=a$ is its latus rectum. Since the parabola is symmetric about the x-axis, the area cut off by this line can be found using one half and doubling it.

A useful standard result is:

Area bounded by the parabola $y^2 = 4ax$ and the line $y = mx$ $$\text{Area} = \frac{8a^2}{3m^3}$$

This formula is obtained by finding the points of intersection and integrating the difference between the upper branch of the parabola and the line. It is useful for checking your answer quickly.

Example 1: Parabola and a Horizontal Line

Find the area of the region bounded by the parabola $y = x^2$ and the line $y = 4$.

Solution: Step 1: Understand the geometry. The parabola $y=x^2$ opens upward, and the horizontal line $y=4$ cuts it in two points. The enclosed region lies between the line and the parabola and is symmetric about the y-axis.

Step 2: Find the points of intersection. Set $$x^2 = 4$$ So, $$x = -2,\ 2$$ Thus the intersection points are $(-2,4)$ and $(2,4)$.

Step 3: Identify upper and lower curves. Between $x=-2$ and $x=2$:

• upper curve: $y=4$
• lower curve: $y=x^2$

Hence, $$\text{Area} = \int_{-2}^{2} (4-x^2)\,dx$$ Using symmetry, $$\text{Area} = 2\int_0^2 (4-x^2)\,dx$$

Step 4: Integrate. $$2\int_0^2 (4-x^2)\,dx = 2\left[4x-\frac{x^3}{3}\right]_0^2$$

Step 5: Apply the limits. $$=2\left(8-\frac{8}{3}\right)=2\left(\frac{24-8}{3}\right)=2\cdot\frac{16}{3}=\frac{32}{3}$$

Answer: $\frac{32}{3}$ square units.

Example 2: Parabola and its Latus Rectum

Find the area bounded by the parabola $y^2 = 4ax$ and its latus rectum.

Solution: Step 1: Write the equation of the latus rectum. For the parabola $y^2=4ax$, the latus rectum is the vertical line $$x=a$$

Step 2: Find the points where this line meets the parabola. Substitute $x=a$ into $y^2=4ax$: $$y^2 = 4a(a)=4a^2$$ So, $$y=\pm 2a$$ The bounded region is symmetric about the x-axis.

Step 3: Express $y$ in terms of $x$. From $$y^2=4ax$$ we take the upper branch in the first quadrant: $$y=2\sqrt{ax}=2\sqrt{a}\sqrt{x}$$

Step 4: Find the upper half-area and double it. $$\text{Area} = 2\int_0^a y\,dx = 2\int_0^a 2\sqrt{a}\,x^{1/2}\,dx$$ $$=4\sqrt{a}\int_0^a x^{1/2}\,dx$$

Step 5: Integrate. $$=4\sqrt{a}\left[\frac{x^{3/2}}{3/2}\right]_0^a = 4\sqrt{a}\cdot \frac{2}{3}\left[a^{3/2}-0\right]$$ $$=\frac{8}{3}\sqrt{a}\,a^{3/2}=\frac{8a^2}{3}$$

Answer: $\frac{8a^2}{3}$ square units.

Example 3: Parabola and an Oblique Line

Find the area of the region bounded by the curve $y = x^2$ and the line $y = x$.

Solution: Step 1: Find the points of intersection. Set $$x^2=x$$ $$x(x-1)=0$$ So, $$x=0,\ 1$$ The intersection points are $(0,0)$ and $(1,1)$.

Step 2: Decide which graph lies above the other on $(0,1)$. Take a test point, say $x=\frac12$:

• line: $y=x=\frac12$
• parabola: $y=x^2=\frac14$ Thus the line lies above the parabola.

Step 3: Set up the area integral. $$\text{Area} = \int_0^1 (x-x^2)\,dx$$

Step 4: Integrate. $$\int_0^1 (x-x^2)\,dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1$$

Step 5: Evaluate. $$=\frac12-\frac13=\frac{3-2}{6}=\frac16$$

Answer: $\frac{1}{6}$ square units.

Example 4: Circle and a Vertical Line

Find the area of the smaller region bounded by the circle $x^2 + y^2 = 16$ and the line $x = 2$.

Solution: Step 1: Understand the region. The circle has radius $4$ and center at the origin. The vertical line $x=2$ cuts the circle and forms a smaller segment on the right side. This segment is symmetric about the x-axis.

Step 2: Use the upper semicircle. From $$x^2+y^2=16$$ we get $$y=\sqrt{16-x^2}$$ for the upper half.

Step 3: Set up the integral. The smaller segment lies between $x=2$ and $x=4$. So the required area is $$\text{Area} = 2\int_2^4 \sqrt{16-x^2}\,dx$$

Step 4: Use the standard integral: $$\int \sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$$ with $a=4$. Hence, $$\text{Area} = 2\left[\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\left(\frac{x}{4}\right)\right]_2^4$$

Step 5: Evaluate at the upper limit $x=4$. $$\frac{4}{2}\sqrt{16-16}+8\sin^{-1}(1)=0+8\cdot\frac{\pi}{2}=4\pi$$

Step 6: Evaluate at the lower limit $x=2$. $$\frac{2}{2}\sqrt{16-4}+8\sin^{-1}\left(\frac12\right)=\sqrt{12}+8\cdot\frac{\pi}{6}=2\sqrt3+\frac{4\pi}{3}$$

Step 7: Subtract and multiply by 2. $$\text{Area} = 2\left(4\pi-2\sqrt3-\frac{4\pi}{3}\right)$$ $$=2\left(\frac{12\pi-4\pi}{3}-2\sqrt3\right)=2\left(\frac{8\pi}{3}-2\sqrt3\right)$$ $$=\frac{16\pi}{3}-4\sqrt3$$

Answer: $\frac{16\pi}{3}-4\sqrt3$ square units.

Example 5: Integration along the y-axis

Find the area bounded by the curve $x = y^2$ and the line $x = y + 2$.

Solution: Step 1: Choose the variable of integration. Since both equations are already in the form $x=$ function of $y$, integrating with respect to $y$ is easiest.

Step 2: Find the points of intersection. Set $$y^2 = y+2$$ $$y^2-y-2=0$$ $$(y-2)(y+1)=0$$ So, $$y=-1,\ 2$$

Step 3: Determine which curve is to the right. Take a test value, say $y=0$:

• parabola: $x=y^2=0$
• line: $x=y+2=2$ So the line is to the right, and the parabola is to the left.

Step 4: Set up the integral. $$\text{Area} = \int_{-1}^{2} \Big[(y+2)-y^2\Big] \, dy$$

Step 5: Integrate. $$\int (y+2-y^2)\,dy = \frac{y^2}{2}+2y-\frac{y^3}{3}$$ So, $$\text{Area} = \left[\frac{y^2}{2}+2y-\frac{y^3}{3}\right]_{-1}^{2}$$

Step 6: Evaluate. At $y=2$: $$\frac{4}{2}+4-\frac{8}{3}=2+4-\frac{8}{3}=\frac{10}{3}$$ At $y=-1$: $$\frac{1}{2}-2+\frac{1}{3}=\frac{1}{2}+\frac{1}{3}-2=\frac{5}{6}-2=-\frac{7}{6}$$ Thus, $$\text{Area} = \frac{10}{3}-\left(-\frac{7}{6}\right)=\frac{20}{6}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}$$

Answer: $\frac{9}{2}$ square units.

Example 6: Area in the First Quadrant

Find the area in the first quadrant enclosed by the x-axis, the line $y = x$, and the circle $x^2 + y^2 = 32$.

Solution: Step 1: Understand the region. The circle has radius $$\sqrt{32}=4\sqrt2$$ and center at the origin. The line $y=x$ meets the circle in the first quadrant, and the enclosed region consists of:

• a triangular-type part under $y=x$ from $x=0$ to the intersection point,
• a curved part under the circle from the intersection point to the positive x-intercept of the circle.

Step 2: Find where the line meets the circle. Substitute $y=x$ into the circle: $$x^2+x^2=32$$ $$2x^2=32$$ $$x^2=16$$ $$x=4$$ So the point of intersection is $(4,4)$.

Step 3: Split the area. From $x=0$ to $x=4$, the upper boundary is the line $y=x$. From $x=4$ to $x=4\sqrt2$, the upper boundary is the upper semicircle $$y=\sqrt{32-x^2}$$ Hence, $$\text{Area} = \int_0^4 x\,dx + \int_4^{4\sqrt2} \sqrt{32-x^2}\,dx$$

Step 4: Evaluate the first integral. $$\int_0^4 x\,dx = \left[\frac{x^2}{2}\right]_0^4 = 8$$

Step 5: Evaluate the second integral using the standard formula. $$\int \sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$$ with $a=4\sqrt2$. So, $$\int_4^{4\sqrt2}\sqrt{32-x^2}\,dx = \left[\frac{x}{2}\sqrt{32-x^2}+16\sin^{-1}\left(\frac{x}{4\sqrt2}\right)\right]_4^{4\sqrt2}$$

At $x=4\sqrt2$: $$0+16\cdot\frac{\pi}{2}=8\pi$$ At $x=4$: $$\frac{4}{2}\sqrt{16}+16\sin^{-1}\left(\frac{1}{\sqrt2}\right)=2\cdot4+16\cdot\frac{\pi}{4}=8+4\pi$$ Thus the second integral is $$8\pi-(8+4\pi)=4\pi-8$$

Step 6: Add both parts. $$\text{Area} = 8+(4\pi-8)=4\pi$$

Answer: $4\pi$ square units.

Example 7: Line passing through the vertex of a parabola

Find the area bounded by the parabola $y^2 = 4x$ and the line $y = 2x$.

Solution: Step 1: Find the points of intersection. Substitute $y=2x$ into $y^2=4x$: $$(2x)^2=4x$$ $$4x^2=4x$$ $$4x(x-1)=0$$ So, $$x=0,\ 1$$ Hence the intersection points are $(0,0)$ and $(1,2)$.

Step 2: Identify upper and lower boundaries between $x=0$ and $x=1$. The parabola gives $$y=2\sqrt{x}$$ for the upper branch. Take a test point such as $x=\frac14$:

• parabola: $y=2\sqrt{1/4}=1$
• line: $y=2\cdot\frac14=\frac12$ So the parabola lies above the line.

Step 3: Set up the integral. $$\text{Area} = \int_0^1 (2\sqrt{x}-2x)\,dx = 2\int_0^1 (x^{1/2}-x)\,dx$$

Step 4: Integrate. $$=2\left[\frac{x^{3/2}}{3/2}-\frac{x^2}{2}\right]_0^1 = 2\left[\frac{2}{3}-\frac12\right]$$

Step 5: Simplify. $$=2\left(\frac{4-3}{6}\right)=2\cdot\frac16=\frac13$$

Answer: $\frac{1}{3}$ square units.

Example 8: Area between Ellipse and a Line

Find the area of the smaller region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the straight line $\frac{x}{a} + \frac{y}{b} = 1$.

Solution: Step 1: Identify the intersection points. The line meets the axes at $(a,0)$ and $(0,b)$. These points also satisfy the ellipse equation, so they are the intersection points. The smaller enclosed region lies in the first quadrant.

Step 2: Write both curves in the form $y=$ function of $x$. From the ellipse: $$\frac{y^2}{b^2}=1-\frac{x^2}{a^2}$$ $$y=\frac{b}{a}\sqrt{a^2-x^2}$$ for the upper half.

From the line: $$\frac{y}{b}=1-\frac{x}{a}$$ $$y=\frac{b}{a}(a-x)$$

Step 3: Determine upper and lower curves. In the first quadrant between $x=0$ and $x=a$, the ellipse lies above the line. So, $$\text{Area} = \int_0^a \left[\frac{b}{a}\sqrt{a^2-x^2}-\frac{b}{a}(a-x)\right]dx$$ $$=\frac{b}{a}\int_0^a \left(\sqrt{a^2-x^2}-a+x\right)dx$$

Step 4: Integrate each part. Using $$\int_0^a \sqrt{a^2-x^2}\,dx = \frac{\pi a^2}{4}, \qquad \int_0^a a\,dx = a^2, \qquad \int_0^a x\,dx = \frac{a^2}{2}$$ we get $$\text{Area} = \frac{b}{a}\left(\frac{\pi a^2}{4}-a^2+\frac{a^2}{2}\right)$$ $$=\frac{b}{a}\left(\frac{\pi a^2}{4}-\frac{a^2}{2}\right)$$ $$=ab\left(\frac{\pi}{4}-\frac12\right)$$ $$=\frac{ab}{4}(\pi-2)$$

Answer: $\frac{ab}{4}(\pi-2)$ square units.

Example 9: Modulus Function and a Line

Find the area bounded by the curve $y = |x|$ and the line $y = 3$.

Solution: Step 1: Find the points of intersection. Set $$|x|=3$$ So, $$x=\pm 3$$ Thus the bounded region extends from $x=-3$ to $x=3$.

Step 2: Use symmetry. The graph $y=|x|$ is symmetric about the y-axis, so we can find the area from $0$ to $3$ and double it.

Step 3: In the first quadrant, $|x|=x$. Hence, $$\text{Area} = 2\int_0^3 (3-x)\,dx$$

Step 4: Integrate. $$=2\left[3x-\frac{x^2}{2}\right]_0^3$$ $$=2\left(9-\frac92\right)=2\cdot\frac92=9$$

This matches the geometry too: the region is a triangle of base $6$ and height $3$, so $$\text{Area}=\frac12\cdot 6\cdot 3=9$$

Answer: $9$ square units.

Example 10: Offset Parabola and a Line

Find the area bounded by the parabola $y = x^2 + 1$ and the line $y = x + 3$.

Solution: Step 1: Find the points of intersection. Set $$x^2+1 = x+3$$ $$x^2-x-2=0$$ $$(x-2)(x+1)=0$$ So, $$x=-1,\ 2$$

Step 2: Determine which graph lies above the other. Take $x=0$:

• parabola: $y=1$
• line: $y=3$ So the line lies above the parabola between $x=-1$ and $x=2$.

Step 3: Set up the integral. $$\text{Area} = \int_{-1}^{2} \Big[(x+3)-(x^2+1)\Big]dx = \int_{-1}^{2} (x+2-x^2)\,dx$$

Step 4: Integrate. $$\int (x+2-x^2)\,dx = \frac{x^2}{2}+2x-\frac{x^3}{3}$$ Hence, $$\text{Area} = \left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}$$

Step 5: Evaluate. At $x=2$: $$\frac{4}{2}+4-\frac{8}{3}=6-\frac{8}{3}=\frac{10}{3}$$ At $x=-1$: $$\frac12-2+\frac13=\frac{1}{2}+\frac{1}{3}-2=-\frac{7}{6}$$ Thus, $$\text{Area}=\frac{10}{3}-\left(-\frac{7}{6}\right)=\frac{20}{6}+\frac{7}{6}=\frac{27}{6}=\frac92$$

Answer: $\frac{9}{2}$ square units.