Definite Integral - Area Of Circle And Ellipse Using Integration

Definite Integral: Area of Circle and Ellipse using Integration

In this lecture, we will learn how to find the area of a circle and an ellipse using definite integration.
This method is based on the concept of integrating a constant function over a given interval.
Let’s start with finding the area of a circle using integration.

Area of a Circle

Consider a circle with radius ‘r’.
The equation of the circle is given by x^2 + y^2 = r^2.
To find the area enclosed by the circle, we will integrate the function “1” with respect to ‘x’ over the interval [0, r]. Example:
Let’s find the area of a circle with radius 5.
Integrate the constant function “1” with respect to ‘x’ over the interval [0, 5].
The area of the circle can be found using the definite integral: A = ∫[0, 5] 1 dx.

Result of Integration

The indefinite integral of a constant function “1” with respect to ‘x’ is simply ‘x’.
Evaluating this integral over the interval [0, 5] yields the area of the circle.
Let’s evaluate the definite integral using the fundamental theorem of calculus. Example:
∫[0, 5] 1 dx = [x] [0, 5] = 5 - 0 = 5.
Therefore, the area of the circle with radius 5 is 5 square units.

Area of an Ellipse

An ellipse is another conic section that can be defined by an equation of the form (x/a)^2 + (y/b)^2 = 1.
Here, ‘a’ represents the semi-major axis and ‘b’ represents the semi-minor axis.
To find the area enclosed by an ellipse, we will integrate the function “1” with respect to ‘x’ over the interval [-a, a]. Example:
Let’s find the area of an ellipse with semi-major axis 4 and semi-minor axis 3.
Integrate the constant function “1” with respect to ‘x’ over the interval [-4, 4].
The area of the ellipse can be found using the definite integral: A = ∫[-4, 4] 1 dx.

Result of Integration

Similar to the circle case, the indefinite integral of the constant function “1” with respect to ‘x’ is ‘x’.
Evaluating this definite integral over the interval [-4, 4] will give us the area of the ellipse.
Let’s calculate the definite integral using the fundamental theorem of calculus. Example:
∫[-4, 4] 1 dx = [x] [-4, 4] = 4 - (-4) = 8.
Therefore, the area of the ellipse with semi-major axis 4 and semi-minor axis 3 is 8 square units.

General Formula for Circle and Ellipse

From the previous examples, we can observe a general formula to find the area of a circle or an ellipse.
The area can be calculated as twice the definite integral of “√(r^2 - x^2)” with respect to ‘x’ over the interval [-r, r].
This formula holds for both circles (where a=b=r) and ellipses (where a>r, b<r).
Let’s apply this formula to find the area of a circle and an ellipse. Example:
Find the area of a circle with radius 2 and an ellipse with semi-major axis 3 and semi-minor axis 2.

Area of Circle Example

For the circle with radius 2, we have r = 2.
Applying the general formula, the area can be calculated as twice the definite integral of “√(2^2 - x^2)” with respect to ‘x’ over the interval [-2, 2].
Let’s evaluate this definite integral. Example:
A = 2 * ∫[-2, 2] √(2^2 - x^2) dx.

Area of Ellipse Example

For the ellipse with semi-major axis 3 and semi-minor axis 2, we have a = 3 and b = 2.
Applying the general formula, the area can be calculated as twice the definite integral of “√(3^2 - x^2)” with respect to ‘x’ over the interval [-3, 3].
Let’s evaluate this definite integral. Example:
A = 2 * ∫[-3, 3] √(3^2 - x^2) dx.

Integration of Polynomial Functions

Polynomial functions can be integrated using the power rule of integration.
The power rule states that the integral of x^n with respect to x is equal to (x^(n+1))/(n+1) + C, where C is the constant of integration.
This rule is applicable for all real values of n, except -1.
Let’s look at some examples of integrating polynomial functions. Example:
∫x dx = (x^2)/2 + C
∫x^2 dx = (x^3)/3 + C
∫x^3 dx = (x^4)/4 + C

Integration of Trigonometric Functions

Trigonometric functions can also be integrated using specific integration formulas.
Here are some of the commonly used integration formulas for trigonometric functions:
- ∫sin(x) dx = -cos(x) + C
- ∫cos(x) dx = sin(x) + C
- ∫sec^2(x) dx = tan(x) + C
- ∫cosec^2(x) dx = -cot(x) + C
These formulas can be used to integrate various trigonometric functions. Example:
∫sin(x) dx = -cos(x) + C
∫cos(x) dx = sin(x) + C
∫sec^2(x) dx = tan(x) + C
∫cosec^2(x) dx = -cot(x) + C

Techniques of Integration: Integration by Substitution

Integration by substitution is a powerful technique used to simplify integration problems.
It involves substituting a new variable or expression to transform the integral into an easier form.
The substitution is typically chosen in such a way that it simplifies the integrand or cancels out certain terms.
Let’s look at an example of integration by substitution. Example:
Consider the integral ∫2x(x^2 + 1)^5 dx.
Let’s substitute u = x^2 + 1.
Differentiating both sides gives du/dx = 2x.
Rearranging, we have x dx = (1/2) du.
Substituting these values, the integral becomes ∫(u^5) (1/2) du.
This simplifies to (1/2) ∫u^5 du.
Which can be further simplified using the power rule of integration.

Techniques of Integration: Integration by Parts

Integration by parts is another technique used to simplify integration problems.
It is based on the product rule of differentiation, which states that (u v)’ = u’ v + v’ u.
The formula for integration by parts is given by: ∫u v dx = u ∫v dx - ∫u’ (∫v dx) dx
The choice of u and v’ is crucial for the success of this method.
Let’s look at an example of integration by parts. Example:
Consider the integral ∫x sin(x) dx.
Choosing u = x and v’ = sin(x), we can differentiate to obtain u’ = 1 and ∫v dx = -cos(x).
Applying the formula, we have ∫x sin(x) dx = (-x)(cos(x)) - ∫(1)(-cos(x)) dx.
Simplifying further, we get -x cos(x) + ∫cos(x) dx.
The integral of cos(x) is sin(x), giving us the final result.

Techniques of Integration: Integration by Partial Fractions

Integration by partial fractions is a method used to simplify integration problems involving rational functions.
In this method, a rational function is decomposed into simpler fractions by breaking down the denominator.
The decomposed fractions are then integrated individually.
This technique is useful when the denominator of the rational function has distinct linear or irreducible quadratic factors.
Let’s look at an example of integration by partial fractions. Example:
Consider the integral ∫(3x + 5)/(x^2 + 4x + 3) dx.
The denominator can be factored as (x + 1)(x + 3).
We can now express the fraction as A/(x + 1) + B/(x + 3), where A and B are constants to be determined.
The next step is to find the values of A and B using algebraic manipulation or a system of equations.
Once we have the decomposed fractions, we can integrate them individually.

Techniques of Integration: Integration by Trigonometric Substitution

Integration by trigonometric substitution is used to simplify integration problems involving radical expressions.
It involves substituting a trigonometric function into the integral to simplify the expression.
The choice of substitution depends on the form of the integral and the presence of radicals.
There are three commonly used trigonometric substitutions:
- x = a sin(theta) for square roots of the form √(a^2 - x^2)
- x = a sec(theta) for square roots of the form √(x^2 - a^2)
- x = a tan(theta) for square roots of the form a^2 + x^2
Let’s look at an example of integration by trigonometric substitution. Example:
Consider the integral ∫(1 - x^2)^(3/2) dx.
We can simplify this integral using the substitution x = sin(theta).
Differentiating both sides gives dx = cos(theta) d(theta).
Substituting these values, the integral becomes ∫(1 - sin^2(theta))^(3/2) cos(theta) d(theta).
This can be further simplified using trigonometric identities.

Techniques of Integration: Integration of Improper Integrals

Improper integrals are defined as integrals with infinite limits or discontinuities in the interval of integration.
They often arise in problems involving infinite areas, infinite volumes, or diverging functions.
To evaluate improper integrals, we either manipulate the integral to obtain a convergent form or apply special techniques.
There are two types of improper integrals:
1. Type 1: Infinite limits of integration
2. Type 2: Discontinuities in the interval of integration
Let’s look at an example of each type. Example (Type 1):
∫[1, ∞] e^(-x) dx
To evaluate this integral, we take the limit as the upper bound approaches infinity. Example (Type 2):
∫[0, 1] 1/x dx
This integral has a discontinuity at x = 0. We split the integral into two parts and evaluate each separately.

Techniques of Integration: Special Techniques

Apart from the previously mentioned techniques, there are some special integration techniques for specific types of functions.
These techniques include but are not limited to the following:
1. Integration of rational functions using long division or partial fractions
2. Integration of inverse trigonometric functions using trigonometric identities
3. Integration of logarithmic functions using the logarithmic rules
These special techniques can significantly simplify integration problems and make them more manageable.
Let’s look at an example of each special technique. Example (Rational Function):
∫(x^2 + 2x + 1)/(x + 1) dx
We can either apply long division or use partial fractions to simplify the integrand. Example (Inverse Trigonometric Function):
∫(1 + sin(x))/cos(x) dx
By using trigonometric identities, we can transform the expression into a more manageable form. Example (Logarithmic Function):
∫(1/x) ln(x) dx
We can apply the logarithmic rules to simplify the integrand.

Applications of Integration: Area between Curves

Integration can be used to find the area between two curves in a given interval.
To calculate the area between curves, we subtract the lower curve from the upper curve and integrate over the interval.
The result is the net area enclosed by the curves.
This technique is useful when finding the area of irregular shapes or regions bounded by curves.
Let’s look at an example of finding the area between curves. Example:
Find the area enclosed by the curves y = x^2 and y = 2x - 1 over the interval [0, 2].
We subtract the lower curve (y = x^2) from the upper curve (y = 2x - 1) and integrate over the interval [0, 2].

Applications of Integration: Volume of Revolution

Integration can also be used to find the volume of a solid obtained by rotating a curve or region about an axis.
The volume of such a solid can be calculated using either the disc method or the shell method.
The disc method involves integrating the cross-sectional area of the solid perpendicular to the axis of revolution.
The shell method involves integrating the circumference multiplied by the height of each cylindrical shell.
By applying these methods, we can find the volume of various shapes and objects.
Let’s look at an example of finding the volume of revolution using the disc method. Example:
Find the volume of the solid obtained by rotating the region bounded by y = x^2, y = 0, and x = 2 about the x-axis.
We calculate the volume by integrating the cross-sectional area of the solid perpendicular to the x-axis.

Definite Integral - Area of Circle and Ellipse using Integration

In the previous slides, we discussed how to find the area of a circle and an ellipse using definite integration.
By integrating a constant function over a given interval, we were able to determine the area enclosed by these shapes.
Let’s review the steps involved in finding the area and work through some more examples.

Steps for Finding the Area of a Circle

Identify the equation of the circle in terms of ‘x’ and ‘y’.

Determine the radius of the circle.

Set up the definite integral by integrating the function “1” with respect to ‘x’ over the interval [a, b].

Evaluate the definite integral using the fundamental theorem of calculus.

The result will be the area of the circle.

Steps for Finding the Area of an Ellipse

Identify the equation of the ellipse in terms of ‘x’ and ‘y’.

Determine the semi-major axis (‘a’) and semi-minor axis (‘b’) of the ellipse.

Set up the definite integral by integrating the function “1” with respect to ‘x’ over the interval [a, b].

Evaluate the definite integral using the fundamental theorem of calculus.

The result will be the area of the ellipse.

Example: Finding the Area of a Circle

Let’s find the area of a circle with radius 3.

The equation of the circle is x^2 + y^2 = 9.

The radius of the circle is 3.

Set up the definite integral: A = ∫[a, b] 1 dx, where a = -3 and b = 3.

Evaluate the definite integral: A = ∫[-3, 3] 1 dx.

The result is the area of the circle.

Example: Finding the Area of an Ellipse

Let’s find the area of an ellipse with semi-major axis 4 and semi-minor axis 2.

The equation of the ellipse is (x/4)^2 + (y/2)^2 = 1.

The semi-major axis is 4 and the semi-minor axis is 2.

Set up the definite integral: A = ∫[a, b] 1 dx, where a = -4 and b = 4.

Evaluate the definite integral: A = ∫[-4, 4] 1 dx.

The result is the area of the ellipse.

General Formula for Circle and Ellipse

We can observe a general formula for finding the area of a circle or an ellipse using definite integration.
For a circle with radius ‘r’, the area can be calculated as:
- A = 2 * ∫[-r, r] √(r^2 - x^2) dx.
For an ellipse with semi-major axis ‘a’ and semi-minor axis ‘b’, the area can be calculated as:
- A = 2 * ∫[-a, a] √(a^2 - x^2) dx.
These formulas can be applied to find the area of any circle or ellipse.

Example: Finding the Area of a Circle and an Ellipse

Find the area of a circle with radius 2.

Find the area of an ellipse with semi-major axis 3 and semi-minor axis 2.

Apply the general formula for area using definite integration.

Evaluate the definite integrals using the fundamental theorem of calculus.

The results will be the areas of the circle and the ellipse.

Summary

Definite integration can be used to find the area of a circle and an ellipse.
By integrating a constant function over a given interval, we can determine the area enclosed by these shapes.
We discussed the steps involved in finding the area and worked through examples.
The general formulas for circles and ellipses can be applied to find the area of any specific instance.