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MODULE- 4
Multivariable Calculus- I
(Multiple Integration)
Course Outcomes
Relate the multiple integral tools for Calculating area, volume, centre of mass and centre of
gravity.
TABLE OF CONTENTS
S.No. Topic Page No.
4.1 Course Objectives of Module 4 1
4.2 Introduction 2
4.3 Multiple Integration 2
4.3.1 Double Integral 3
4.3.2 Triple Integral 12
4.3.3 Change of order of integration 15
4.3.4 Change of variables 19
4.4 Applications 30
4.4.1 Area 30
4.4.2 Volume 36
4.4.3 Centre of mass & Centre of gravity
( Constant and variable densities)
4.5 Related Links 46
Syllabus: Multiple integral: Double integral, Triple integral, Change of order of integration, Change of
variables, Applications: Area, Volume, Centre of Mass & Centre of Gravity (constant and variable
densities).
4.2 INTRODUCTION
Multiple integral is a natural extension of a definite integral to a function of two variables (Double integral) or three variables (triple integral) or more variables.
4.2.1 APPLICATIONS
➢ Double and triple integrals are useful in finding Area. ➢ Double and triple integrals are useful in finding Volume. ➢ Double and triple integrals are useful in finding Mass. ➢ Double and triple integrals are useful in finding Centroid. ➢ In finding Average value of a function. ➢ In finding Distance, Velocity, Acceleration. ➢ Useful in calculating Kinetic energy and Improper Integrals. ➢ In finding Arc Length of a curve. ➢ The most important application of Multiple Integrals involves finding areas bounded by a curve and coordinate axes and area between two curves. ➢ It includes finding solutions to various complicated problems of work and energy. ➢ In mechanics, the moment of inertia is calculated as the volume integral (triple integral) of the density weighed with the square of the distance from the axis. ➢ Multiple integrals are used in many applications in physics. The gravitational potential associated with a mass distribution given by a mass measure on three-dimensional Euclidean space R3 is calculated by triple integration. ➢ In electromagnetism, Maxwell's equations can be written using multiple integrals to calculate the total magnetic and electric fields. ➢ We can determine the probability of an event if we know the probability density function using double integration.
4.3 MULTIPLE INTEGRAL
4.3.1 Double integral
A double integral is its counterpart in two dimensions. Let a single valued and bounded function f(x, y) of two independent variables x, y defined in a closed region R. Then double integral of f(x, y) over the region R is denoted by, Also we can express as or
(Here we have drawn the strip parallel to y axis, because variable limits [ are provided.) (iii) When are constants: If we have both the variables x and y with constant limits then we can first integrate with respect to any variable x or y in case of double integral, as follows: (Here we can draw the strip parallel to any of the axes, because both x and y are having constant limits.) From case no. (i) and (ii) discussed above, we observe that integration is to be performed w.r.t. the variable limits first and then w.r.t. the variable with constant limits.
4.3.1.3 Solved examples
Example1: Evaluate. Solution: = (Here we have constant limits for both x and y variables, so we may integrate w.r.t. any variable the first)
= = = (Answer) Example 2: Evaluate over the area bounded by the ellipse Solution: (Here we have area bounded by the curve , depending on variables x and y so we have to construct a strip parallel to any one axis to observe variable limits of one variable.) For the ellipse we may write or The region of integration R can be expressed as
- a , where we have chosen variable limits of y and constant limits of x. So first we will integrate w.r.t. y,
= dx = - ] dx = - ] dx = = (Answer) Example 4: Evaluate over the region R bounded by the parabolas. Solution: Solving , we have x = 0, 4 When x = 4, y = 4 Co-ordinates of A (intersection point of parabolas) are (4, 4) The region R can be expressed as 0 x 4,
y dx = = dx = = = (Answer) Example 5: Evaluate , where S is a triangle with vertices (0, 0), (10, 1) and (1, 1). Solution: Let OAB be the triangle formed by given vertices (0, 0), (10, 1) and (1, 1) as shown in the figure through shaded area. The equation of the line joining O (0, 0) and A (1, 1) can be find as follows, The equation of the line joining O (0, 0) and B (10, 1) can be calculated as follows, Here we have taken strip intentionally parallel to x-axis, so that the strip bounded by x = y and x = 10 y may cover the complete shaded area from y = 0 to y = 1. Hence the region of integration can be expressed as
Here we can see the equations x = y and xy = 16, on solving give intersection point at A (4, 4) Similarly on solving xy =16 and x = 8, we get the intersection point at B (8, 2) Drawing the curves we get the intersection area as shown in figure. Now we are to decide with respect to which variable we should first integrate, we construct strips in such a manner that the complete area may be covered. Here we cannot cover the whole shaded area using single strip (Neither parallel to x- axis nor parallel to y- axis). Because area is changing from dotted lines, if we plot strip parallel to y-axis and also area is changing from lines drawn, if we plot the strip parallel to x-axis. So in both cases we need to draw two strips. Here we are splitting the area OABNO in two parts by AM as shown in figure and plotted strips parallel to y- axis from x = 0 to x = 4 and from x= 4 to x = 8 Then, = =
= = = 64 +8(64 - 1 6) = 64 +384 = 448 (Answer)
4.3.1.4 Practice problems
1. Evaluate
2 (^1 0 ) x
x y
dydx
Ans :
2. Evaluate. Ans. 8( log8-2)+e
3. Evaluate:
(i) Ans. ½
(ii) Ans. 41/
(iii) Ans. log a log b
(iv) Ans.3/ 35
4. Evaluate over the first quadrant of the ellipse. Ans.
5. Evaluate over the area between. Ans. 3/
6. Evaluate , where A is the domain bounded by x-axis, ordinate x = 2a and the curve.
Ans.
4.3.1.5 Evaluation of Double Integrals in Polar Coordinates
In polar coordinates we know that, x = r cos and y = r sin Sometimes integration can be easier by converting Cartesian form to polar form. In such cases we may evaluate integral by polar coordinates using variable r and in the same manner as we done earlier. Here we draw radial strip to decide the limit in order to cover the whole area.
4.3.1.6 Solved Examples
Example 1: Evaluate , over the area bounded between the circles r = 2 cos and r = 4 cos.
= = = (Answer)
4.3.1.7 Practice problems
1. Evaluate
r drd
a
2 0 cos 0
sin Ans:
- Evaluate Ans:
3. Evaluate , over the area bounded between the circles r = 2 sin and r = 4 sin. Ans:
4.3.2 TRIPLE INTEGRALS
Consider a function f(x, y, z) which is continuous at every point of a finite region V of three dimensional space.
The triple integral of f(x, y, z) over the region and is denoted by
For purpose of evaluation, it can be expressed as the repeated integral
the order of integration depending upon the limits.
Working rule:
Let
Let.
Then the integral (i) is evaluated as follows
First f(x,y,z) is integrated w.t.r. x (keeping y and x constant)between the limits and.
The resulting expression, which is a function of y and z is then integrated w.r.t. y (keeping
z constant) between the limits and .The resulting expression , which of the function of
z only is then integrated w.r.t. z between the limits and. The order of integration is
from the innermost rectangle to the outermost rectangle.
Limits involving two variables are to be kept innermost, then the limits involving
one variable and finally the constant limits.
If are all constants, then the order of integration is immaterial, provided the limits are
changed accordingly. Thus
4.3.2.1 Solved examples
Example 1: Evaluate.
Solution: Let I =
= = = (Answer)
Example 2: Evaluate.
Solution: I=
4.3.3 CHANGE OF ORDER OF INTEGRATION
In double integral, if the limits of integration are constant, then the order of integration does not matter, provided the limits of integration are changed accordingly. Thus, But if the limits of integration are variable, then in order to change the order of limits of integration we have to construct the rough figure of given region of integration and re construct the strip parallel to that axis with respect to which we want to first integrate. Mostly we need this process to make our integration simpler if possible.
4.3.3.1 Solved examples
Example 1: Evaluate the following integral by changing the order of integration
Solution: The given limits show that the region of integration is bounded by the curves y = , y = e, x = 0, x = 1.
Plotting these curves we have the shaded region of integration as shown in figure. In given problem we had variable limits of y in terms of x, so we had to integrate w.r.t. y the first. But we are instructed to solve this problem changing the order of integration.
Now to integrate first w.r.t. x we have to find variable limits of x in terms of y. So we have to construct a strip parallel to x- axis in order to find variable limits of x. From the strip we can see lower limit lies on x = 0 and y = in between the constant limits of y from y = 1 to y = e. Hence = = = = = (e-1) (Answer). Example 2: Change the order of integration in I = and hence evaluate the same Solution: From the variable limits of integration, it is clear that we have to integrate first w.r. to y which varies from y = to y = 2-x and then with respect to x which varies from x = 0 to x = 1. The region of integration is divided into vertical strips. For changing the order of integration, we divide the region of integration into horizontal strips. Solving y = and y = 2-x, the co-ordinates of A are (1,1). Draw AM OY. The region of integration is divided into two parts, OAM and MAB. For the region OAM, x varies from 0 to 2-y and y varies from 1 to 2. = = = = (Answer) Example 3: Change the order of integration in the following integral and evaluate: y =2-x y Y = O
A(1,1)
X=
M
B
Y =
x
Solution: The given limits shows that the area of integration lies between , y = a, x = 0, x = a. We can consider it as lying between y =0, y = a, x = 0, x = by changing the order of integration. Hence the given integral, = = = = = = = (Answer) Example5: Evaluate the following integral by changing the order of integration: Solution: The given limits shows that the area of integration lies between y = x, y = ∞, x = 0 and x = ∞. We can consider it as lying between x = 0, x = y, y = 0 and y = ∞ by changing the order of integration. Hence the given integral, =
= 1-0 = 1 (Answer)
4.3.3.2 Practice problems
- Evaluate the integrals by changing the order of integration: (i) Ans: 1- (ii) Ans: (iii) Ans: 3/ (iv) Ans: ½(1- cos 1) (v) Ans: (vi) Ans: (vii) Ans: (viii) Ans: (ix) Ans: 27 (x) Ans:
4. 3 .4 CHANGE OF VARIABLES
In simple integration, we use substitution to make our integration simpler than before. Similarly, in double or triple integration we use suitable change of variables to make the evaluation of integration simple. In general, there are following four types of transformation:
i. To change Cartesian co-ordinates (^ x , y )to some given co-ordinates (^ u , v ).
ii. To change Cartesian co-ordinates ( x , y )to polar co-ordinates ( r , ).
iii. To change Cartesian co-ordinates ( x , y , z )to spherical polar co-ordinates ( r ,, ).
iv. To change Cartesian co-ordinates ( x , y , z )to cylindrical co-ordinates ( r , , z ).
4.3.4.1 To change Cartesian co-ordinates ( x , y ) to some given co-ordinates ( u , v ):
