# Area Between Curves

Find the area between curves using definite integrals. Tutorials, on the applications of integrals to calculate areas between curves, with examples and detailed solutions are presented . This tutorial is a continuation to the tutorial on area under a curve. It starts from some obvious examples to more challenging one ones which should be attempted if good preparation to applications of integrals is to be attained.

## Formulas of Areas Between Curves

We can find the area between two curves by subtracting the area corresponding the lower curve from the area of the upper curve as follows:

1) If f and h are functions of x such that f(x) ≥ h(x) for all x in the interval [ x_{1} , x_{2} ], the area shown below (in blue) is given by

2 - If Z and X are functions of y such that Z(y) ≥ X(y) for all y in the interval [ y1 , y2 ], the area shown below, in blue, is given by

## Examples with Solutions

### Example 1

Find the area of the region enclosed between the curves defined by the equations y = x^{ 2} - 2x + 2 and y = - x^{ 2} + 6 .

### Solution to Example 1

We first graph the two equations and examine the region enclosed between the curves.

The region whose area is in question is limited above by the curve y = - x

^{ 2}+ 6 and below by the curve y = x

^{ 2}- 2x + 2. The left endpoint and the right endpoint of the region are the point of intersection of the curves and can be found by solving the system of equations

y = x

^{ 2}- 2x + 2 and y = - x

^{ 2}+ 6.

which gives

x

^{2}- 2x + 2 = - x

^{ 2}+ 6

rewrite in standard form

2x

^{2}- 2x - 4 = 0

factor left side

which gives solutions

x = -1 and x = 2

Between the points of intersection, -x

^{ 2}+6 is greater than or equal to x

^{ 2}- 2x + 2.

Let f(x) = -x

^{ 2}+ 6 and h(x) = x

^{ 2}- 2x + 2 and apply formula 1 above to find the area A of the region between the two curves.

The limits of integration are the x coordinates of the points of intersection found above: -1 and 2 \( \)\( \) \( \) \( \)

\( \displaystyle \text{A} = \int_{-1}^{2} (f(x) - h(x)) dx = \int_{-1}^{2} ((-x^2+6) - (x^2 - 2x + 2)) dx \\ = \int_{-1}^{2} (-2x^2 + 2x + 4) dx = [-(2 / 3) x^3 + x^2 + 4x]_{-1}^2 = 9 \)

The area of the region enclosed between the curves y = x

^{ 2}- 2x + 2 and -x

^{ 2}+ 6 is equal to 9.

### Example 2

Find the area of the region enclosed between the curves defined by the equations \( y = \sqrt{x + 2} \) , y = x and y = 0.### Solution to Example 2

We first graph all three curves and examine the region enclosed.

__Two methods to solve this problem__

__Method 1__Use the equations of the curves as y as a function of x and integrate on x using the first formula above.

The region from x = -2 to x = 0 is under the curve y = √(x + 2) and therefore its area A1 may be calculated as follow

\( \displaystyle \text{A1} = \int_{-2}^{0} \sqrt{x+2} dx = \left [ \dfrac{2}{3} (x+2)^{3/2} \right ]_{-2}^0 = \dfrac{2}{3} (2^{3/2}) \)

The region from x = 0 to the right is limited by the point of intersection of the two curves which is found by solving the system of equation

\( y = \sqrt{x + 2} \) and y = x.

which gives

\( \sqrt{x + 2} = x \)

Square both side of the above equation to obtain

x + 2 = x

^{2}

Solve the above to find one valid solutions

x = 2

The area A2 of the region from x = 0 to x = 2 (point of intersection) is found using the formula for the area between two curves as follows

\( \displaystyle \text{A2} = \int_{0}^{2} (\sqrt{x+2} - x ) dx = \left [ \dfrac{2}{3} (x+2)^{3/2} - x^2 / 2 \right ]_{0}^2 = \dfrac{2}{3} (4^{3/2}) - 2 - \dfrac{2}{3} (2^{3/2}) \)

The total area is given by

\( \displaystyle \text{A1 + A2} = \dfrac{2}{3} (2^{3/2}) + \dfrac{2}{3}(4^{3/2}) - 2 - \dfrac{2}{3} (2^{3/2}) = 10/3 \; \text{unit}^2 \)

__Method 2__Use the equations of the curves as x as a function of y and integrate on y using the second formula above.

The equation x in terms of y of the curve on the right is x = y; the equation of the curve on the right is given as

\( y = \sqrt {x + 2} \)

Square both sides and solve for x o obtain

x = y

^{2}- 2

The region whose area we need to find is limited by y = 0 to y = 2 (y coordinate of the point of intersection). The total area of this region is given by

\( \displaystyle \text{Area} = \int_{0}^{2} ( y - (y^2 - 2) ) dy = \left [ y^2 / 2 - y^3 / 3 + 2y \right ]_{0}^2 = 10/3 \; \text{unit}^2 \)

Note that the second method is much faster.

### Example 3

Find the area of the region enclosed by the curves \( y = \sin(x) \) , \( y = \cos(x) \), x = 0 and x = 2π.

### Solution to Example 3

We first graph \( sin(x) \) and \( cos(x) \) from 0 to 2π.

We first need to find the x coordinates of the points of intersection by solving the system of equations

\( y = \sin(x) \) and \( y = \cos(x) \)

which gives

\( sin(x) = cos(x) \)

Which may be written as

\( \tan(x) = 1 \)

The solutions between x = 0 and x = 2π to the above equation are

x = π/4 and 5 π/4 as shown in the graph.

We now identify 3 regions limited by x = 0, x = 2π and the points of intersections as follows

__Region 1__

from x = 0 to x = π/4 and its area A1 is given by (taking into account that cos x ≥ sin x in this region)

\( \displaystyle \text{A1} = \int_{0}^{\pi/4} ( \cos x - \sin x ) dx = \left[\sin x + \cos x \right]_{0}^{\pi/4} = \sqrt{2} - 1 \; \text{unit}^2 \)

__Region 2__

from x = π/4, to x = 5π/4 and its area A2 is given by (taking into account that sin x ≥ cos x in this region)

\( \displaystyle \text{A2} = \int_{\pi/4}^{5\pi/4} ( \sin x - \cos x ) dx = \left [ - \cos x - \sin x \right ]_{\pi/4}^{5\pi/4} = 2\sqrt{2}\; \text{unit}^2 \)

__Region 3__

from x = 5π/4, to x = 2π and its area A3 is given by (taking into account that cos x ≥ sin x in this region)

\( \displaystyle \text{A3} = \int_{5\pi/4}^{2\pi} ( \cos x - \sin x ) dx = \left [ \sin x + \cos x \right ]_{5\pi/4}^{2\pi} = 1+ \sqrt{2}\; \text{unit}^2 \)

Total area A is obtained by adding A1, A2 and A3

\( \displaystyle \text{A3= A1+A2+A3} = 4 \sqrt 2 \; \text{unit}^2 \)

### Example 4

Below are graphed y = 3x - x^{2} and y = 0.5 x. Find the ratio of the area of region A to the area of region B.

### Solution to Example 4

We first calculate the area A of region A as being the area of a region between two curves y = 3 x - x

^{2}and y = 0.5 x, x = 0 and the point of intersection of the two curves. First find the point of intersection by solving the system of equations

y = 3x - x

^{2}and y = 0.5 x

which gives

3x - x

^{2}= 0.5 x

2.5x - x

^{2}= 0

x(2.5 - x) = 0

Two solutions: x = 0 and x = 2.5

The area of region A is calculated as follows

\( \displaystyle \text{A} = \int_{0}^{2.5} ( 3x - x^2 - 0. 5x ) dx = \left [ 2.5 x^2 / 2 - x^3/3 \right ]_{0}^{2.5} = 125/48 \; \text{unit}^2 \)

We now need to find the area B of region B which may be easily calculated by subtracting area of A from the total area At under the curve of y = 3x - x

^{2}which is given by

\( \displaystyle \text{At} = \int_{0}^{3} ( 3x - x^2 ) dx = \left [ 3 x^2 / 2 - x^3/3 \right ]_{0}^{3} = 9/2 \; \text{unit}^2 \)

The ratio r of the area of A to the area of B

\( \displaystyle \text{r} = \dfrac{A}{At - A} = \dfrac{125/48}{9/2 - 125/48} = 125 / 91 \)

Example 5

Find the area of the overlapping region of the circles with equations: x^{ 2} + y^{ 2} = 4 and
x^{ 2} + (y - 2)^{ 2} = 1.

### Solution to Example 5

We first graph the equations of the given circles in order to identify the overlapping region which colored in light blue. The region whose area is to be found is limited by the point of intersection of the two circles, hence we need to find the x coordintes of the two points of intersection by solving the system of equations given by the equations of the circles.

x

^{ 2}+ y

^{ 2}= 4 and x

^{ 2}+ (y - 2)

^{ 2}= 1

expand the left side of equation x

^{ 2}+ (y - 2)

^{ 2}= 1

x

^{2}+ y

^{2}- 4y + 4 = 1

Now subtract side by side the equations x

^{ 2}+ y

^{ 2}= 4 and x

^{2}+ y

^{2}- 4y + 4 = 1 to obtain an equation in one variable

- 4 y + 4 = -3

Solve for y

y = 7/4

We substitute y by 7/4 in the equation x

^{ 2}+ y

^{ 2}= 4 and solve for x to obtain two solutions which are the x-coordinates of the points of intersection of the two circles.

\( x = \sqrt{15} / 4 \) and \( x = - \sqrt{15} / 4 \)

We now need to find the equations of the curves that make the region whose area is to be found.

Solve x

^{ 2}+ y

^{ 2}= 4 for y to obtain 2 solutions.

\( y = \sqrt{4 - x^2} \) and \( y = - \sqrt{4 - x^2} \)

We select the upper half \( y = \sqrt{4 - x^2} \) of this (large) circle as shown in the graphs above.

Solve x

^{ 2}+ (y - 2)

^{ 2}= 1 for y to obtain 2 solutions.

\( y = 2 - \sqrt{1 - x^2} \) and \( y = 2 + \sqrt{1 - x^2} \)

We select the lower half \( y = 2 - \sqrt{1 - x^2} \) of this (small) circle as shown in the graphs above.

Because of the symmetry of the two circle with respect to the y axis, we can calculate half the area A': from x = 0 to \(x = \sqrt{15} / 4 \) of the overlapping region and then mutliply by 2. A' is calculated as the area between two curves as follows .

\( \displaystyle \text{A'} = \int_{0}^{\sqrt{15} / 4} ( (\sqrt{4 - x^2}) - (2 - \sqrt{1 - x^2}) ) dx \)

Use the sum and difference rule of integration to write the above integral as.

\( \displaystyle \text{A'} = \int_{0}^{\sqrt{15} / 4} \sqrt{4 - x^2} \; dx - \int_{0}^{\sqrt{15} / 4} 2 \; dx + \int_{0}^{\sqrt{15} / 4} \sqrt{1 - x^2} \; dx \)

Use the formula for integration .

\( \displaystyle \int \sqrt{a^2 - x^2}dx = (x/2)\sqrt{a^2-x^2} + \dfrac{a^2}{2} \arcsin(x/a) + C \)

to obtain

\( \displaystyle \text{A'} = [ \dfrac{x}{2} \sqrt{4-x^2} + 2 \arcsin(x/2) - 2x + \dfrac{x}{2} \sqrt{1-x^2} + \dfrac{1}{2} \arcsin(x)]_{0}^{\sqrt{15} / 4} \approx 0.70153 \)

The total area A of the overlapping region is twice

\( \displaystyle A \approx 1.4 \; \text{unit}^2 \)

## Exercises

1) Find the area of the region enclosed by y = (x-1)^{ 2}+ 3 and y = 7

2) Find the area of the region bounded by x = 0 on the left, x = 2 on the right, y = x

^{ 3}above and y = -1 below.

3) Find the area enclosed by the curves y = 1/x

^{2}, y = x and y = 3.

## Solutions to Above Exercises

1) 32 / 32) 6

3) 6 - 2√3

### More References and links

integrals and their applications in calculus.Area under a curve .

Volume of a Solid of Revolution .

Volume by Cylindrical Shells Method .