How to Find the Area Enclosed by Curves Using Any Axes

To find the area between curves, choose the axis of integration carefully. Use vertical slices \(dx\) for functions defined as \( y = f(x) \) and \(y = g(x) \), integrating \([f(x) − g(x)]\) from \(x = a\) to \(x = b\). Use horizontal slices \(dy\) for \(x = f(y)\) and \(x = g(y)\), integrating \([f(y) − g(y)]\) from \(y = c\) to \(y = d\). When curves intersect multiple times, identify which curve is on top in each segment, set up separate integrals, and sum the areas.

Free tool: try the problems from this lesson in our Area & Perimeter Calculator.

For more advanced cases, when dealing with parametric or polar equations, the area between curves requires specialized integration techniques. For parametric curves \( (x(t), y(t)) \), the area is calculated using \( \int_{t_1}^{t_2} [x(t) y'(t) – y(t) x'(t)] \, dt \). In polar coordinates, the area between curves is found using \( \frac{1}{2} \int_{\theta_1}^{\theta_2} [r_2^2 – r_1^2] \, d\theta \).

Additionally, when dealing with implicit functions or regions bounded by complex curves, double integrals are used to compute the area. By setting up a double integral over the region \( R \), the area is \( \iint_R dA \). Green’s Theorem can also be applied to find areas by converting a line integral around a closed curve into a double integral over the region it encloses, which is especially useful in vector calculus.

Example: Finding the Area Between Curves Using Vertical and Horizontal Slices

Problem:

Find the area of the region enclosed by the curves \( y = x^2 \) and \( y = 2x \).

Solution:

Step 1: Sketch the Curves

First, sketch the graphs of the functions to understand the region we’re dealing with.

• \( y = x^2 \) is a parabola opening upwards.
• \( y = 2x \) is a straight line.

Step 2: Find Points of Intersection

Set the equations equal to find the points where the curves intersect.

\( x^2 – 2x =0 \)

So, \( x = 0 \) or \( x = 2 \).

Step 3: Determine Which Curve Is on Top

For \( x \) between \( 0 \) and \(2\):

• Choose \( x = 1 \):
• \( y = x^2 = 1^2 = 1 \)
• \( y = 2x = 2 \times 1 = 2 \)

Since \( y = 2x \) yields a higher value, \( y = 2x \) is the upper function, and \( y = x^2 \) is the lower function in this interval.

Step 4: Set Up the Integral Using Vertical Slices (dx)

The area \( A \) between the curves from \( x = 0 \) to \( x = 2 \) is:

\( [A = \int_{0}^{2} [\text{Upper function} – \text{Lower function}] \, dx = \int_{0}^{2} [2x – x^2] \, dx]\)

Step 5: Evaluate the Integral

Compute the integral:

\( [\begin{align} A &= \int_{0}^{2} (2x – x^2) \, dx \ &= \left[ x^2 – \frac{1}{3} x^3 \right]_0^2 \ &= \left( 2^2 – \frac{1}{3} \times 2^3 \right) – \left( 0^2 – \frac{1}{3} \times 0^3 \right) \ &= \left( 4 – \frac{8}{3} \right) – 0 \ &= \left( \frac{12}{3} – \frac{8}{3} \right) \ &= \frac{4}{3} \end{align}] \)

Answer:

The area of the region enclosed by \( y = x^2 \) and \( y = 2x \) is \( \frac{4}{3} \) square units.

Alternative Scenario with Horizontal Slices \(dy\):

Suppose we have functions defined in terms of \( y \):

Problem:

Find the area of the region enclosed by \( x = y^2 \) and \( x = y + 2 \).

Solution:

Step 1: Sketch the Curves

• \( x = y^2 \) is a parabola opening to the right.
• \( x = y + 2 \) is a straight line.

Step 2: Find Points of Intersection

Set the equations equal:

\( [\begin{align} y^2 &= y + 2 \ y^2 – y – 2 &= 0 \ (y – 2)(y + 1) &= 0 \end{align}]\)

So, \( y = -1 \) or \( y = 2 \).

Step 3: Determine Which Curve Is on the Right

For \( y \) between \( -1 \) and \( 2 \):

• Choose \( y = 0 \):
• \( x = y^2 = 0^2 = 0 \)
• \( x = y + 2 = 0 + 2 = 2 \)

Since \( x = y + 2 \) yields a higher value, it’s the rightmost function.

Step 4: Set Up the Integral Using Horizontal Slices \(dy\)

The area \( A \) is:

\( [A = \int_{-1}^{2} [\text{Rightmost function} – \text{Leftmost function}] \, dy = \int_{-1}^{2} [(y + 2) – y^2],dy] \)

Step 5: Evaluate the Integral

Compute the integral:

\( [\begin{align} A &= \int_{-1}^{2} (y + 2 – y^2) \, dy \ &= \int_{-1}^{2} (-y^2 + y + 2) \, dy \ &= \left[ -\frac{1}{3} y^3 + \frac{1}{2} y^2 + 2y \right]_{-1}^{2} \ \end{align}]\)

Evaluate at \( y = 2 \):

\( [\left( -\frac{1}{3} \times 8 + \frac{1}{2} \times 4 + 4 \right) = \left( -\frac{8}{3} + 2 + 4 \right) = \frac{10}{3}] \)

Evaluate at \( y = -1 \):

\( [\left( -\frac{1}{3} \times (-1) + \frac{1}{2} \times 1 – 2 \right) = \left( \frac{1}{3} + \frac{1}{2} – 2 \right) = -\frac{7}{6}] \)

Subtract:

\( [A = \frac{10}{3} – \left( -\frac{7}{6} \right) = \frac{10}{3} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}] \)

Answer:

The area of the region enclosed by \( x = y^2 \) and \( x = y + 2 \) is \( \frac{9}{2} \) square units.

Handling Multiple Intersections:

Problem:

Find the area between the curves \( y = x^2 \) and \( y = (x – 2)^2 \).

Solution:

Step 1: Find Points of Intersection

Set \( x^2 = (x – 2)^2 \):

\( [\begin{align} x^2 &= x^2 – 4x + 4 \ 0 &= -4x + 4 \ x &= 1 \end{align}] \)

They intersect at \( x = 1 \).

Step 2: Determine Regions of Integration

Since the curves intersect at \( x = 1 \) and are symmetric, we’ll consider the interval from \( x = 0 \) to \( x = 2 \), splitting at \( x = 1 \).

For \( x = 0 \) to \( x = 1 \):

• \( y = (x – 2)^2 \) is above \( y = x^2 \).

For \( x = 1 \) to \( x = 2 \):

• \( y = x^2 \) is above \( y = (x – 2)^2 \).

Step 3: Set Up the Integrals

First Integral \(( x = 0 ) to ( x = 1 )\):

\( [A_1 = \int_{0}^{1} [(x – 2)^2 – x^2] \, dx]\)

Simplify:

\([(x – 2)^2 – x^2 = x^2 – 4x + 4 – x^2 = -4x + 4]\)

So,

\([A_1 = \int_{0}^{1} (-4x + 4) \, dx]\)

Second Integral (( x = 1 ) to ( x = 2 )):

\([A_2 = \int_{1}^{2} [x^2 – (x – 2)^2] \, dx]\)

Simplify:

\([x^2 – (x – 2)^2 = x^2 – (x^2 – 4x + 4) = 4x – 4]\)

So,

\([A_2 = \int_{1}^{2} (4x – 4) \, dx]\)

Step 4: Evaluate the Integrals

First Integral:

\([\begin{align} A_1 &= \int_{0}^{1} (-4x + 4) \, dx \ &= \left[ -2x^2 + 4x \right]_0^1 \ &= \left( -2 \times 1^2 + 4 \times 1 \right) – \left( -2 \times 0^2 + 4 \times 0 \right) \ &= (-2 + 4) – 0 = 2 \end{align}]\)

Second Integral:

\([\begin{align} A_2 &= \int_{1}^{2} (4x – 4) \, dx \ &= \left[ 2x^2 – 4x \right]_1^2 \ &= \left( 2 \times 2^2 – 4 \times 2 \right) – \left( 2 \times 1^2 – 4 \times 1 \right) \ &= (8 – 8) – (2 – 4) = 0 – (-2) = 2 \end{align}] \)

Step 5: Sum the Areas

\([A = A_1 + A_2 = 2 + 2 = 4]\)

Answer:

The total area between \( y = x^2 \) and \( y = (x – 2)^2 \) from \( x = 0 \) to \( x = 2 \) is \( 4 \) square units.

Conclusion:

In each example, we carefully chose the axis of integration based on how the functions were defined and identified which curve was on top (or to the right) in each interval. When the curves intersected multiple times, we split the integrals accordingly and summed the areas to find the total area between the curves.

Frequently Asked Questions

What is the area enclosed by curves?

The area enclosed by curves is the size of the region trapped between two or more graphs. If y = f(x) lies above y = g(x) from x = a to x = b, the area is ∫ from a to b of [f(x) − g(x)] dx. The result is always positive, because you subtract the lower curve from the upper one, or the left curve from the right one when you integrate along the y-axis.

How do you find the area enclosed by curves?

First set the curves equal to find their intersection points, which become your limits. Then choose vertical slices (dx, top minus bottom) or horizontal slices (dy, right minus left), whichever is simpler. For x = y² and y = x − 2, horizontal slices work best: integrate (y + 2) − y² from y = −1 to y = 2 to get an area of 9/2.

What is the area enclosed by the curves y = x and y = x³?

The curves y = x and y = x³ meet at x = −1, 0 and 1, so the enclosed region has two pieces. On [0, 1], y = x is on top, giving ∫ from 0 to 1 of (x − x³) dx = 1/2 − 1/4 = 1/4. On [−1, 0], y = x³ is on top, and by symmetry that piece also has area 1/4. The total enclosed area is 1/2.

How do you find the area enclosed by 3 curves?

Sketch all three curves and find every intersection point, because the boundary changes wherever two curves cross. Split the region at those points so each piece has one top and one bottom curve, integrate each piece, and add. For the region bounded by y = x², y = 2 − x and y = 0, split at x = 1: ∫ from 0 to 1 of x² dx + ∫ from 1 to 2 of (2 − x) dx = 1/3 + 1/2 = 5/6.