SMT5 ENGINEERING MATHS- II UNIT-I MULTIPLE INTEGRALS
INTRODUTION When a function () is integrated with respect to x between the limits a and b, we get the double integral. If the integrand is a function, and if it is integrated with respect to x and y repeatedly between the limits and (for x ) and between the limits and (for y ) we get a double integral that is denoted by the symbol,. Extending the concept of double integral one step further, we get the triple integral, denoted by,,.
EVALUATION OF DOUBLE AND TRIPLE INTEGRALS To evaluate, first integrate, with respect to x partially, treating y as constant temporarily, between the limits and. Then integrate the resulting function of y with respect to y between the limits and as usual. In notation, ( for double integral) integral).,, ( for triple Note: Integral with variable limits should be the innermost integral and it should be integrated first and then the constant limits.
REGION OF INTEGRATION Consider the double integral () (),, varies from () and varies from. (i.e) and. These inequalities determine a region in the, which is shown in the following figure.this region ABCD is known as the region of integration
EXAMPLE : Evaluate = / = 8 = 8 = 4
EXAMPLE : Evaluate = log =( ) = [] =( ) =. log (/)
EXAMPLE : Evaluate = = = = 6 = 6
EXAMPLE :4 Evaluate = = = = =
EXAMPLE :5 Evaluate sin sin = sin = = = sin =
EXAMPLE :6 Evaluate = = = =
EXAMPLE :7 Evaluate I = = [ ] = = = 4 8 = 6 = 6 48 48
EXAMPLE :8 Evaluate I = sin = = [] = = + sin = 8
EXAMPLE :9 Evaluate sin I= = sin = = = 4
PROBLEMS FOR PRACTICE Evaluate the following. 4 Ans: 4. Ans: loga.logb. sin 4. 5. + 6. Ans: / Ans: π/4 Ans: 9/ Ans: ½
EXAMPLE : Sketch the region of integration for (, ). Given = = ; = = = and + =
EXAMPLE : Sketch the region of integration for (, ). Given = ; = = ; =. Y x = y x= X y =
EXAMPLE : Evaluate where D is the region bounded by the positive octant of the sphere + + = = =
= ( ) = ( ) = 4 4 = 8 ( ) = 8 ( 4 + 5 ) = 8 4 4 6 = 6 4 6 48.
PROBLEMS FOR PRACTICE.Sketch the region of integration for the following (i) 4 4 + (ii) (iii) +.Evaluate + +, where V is the region of space bounded by x=,x=,y=,y=,z= and z=. Ans: /. Evaluate (+++) bounded by x=,y=,z= and x+y+z= Ans: (8 5) 6, where V is the region of space 4. Evaluate, where V is the region of space bounded by x=,,y=,,z= and x+y+4z=. Ans:
CHANGE of ORDER OF INTEGRATION If the limits of integration in a double integral are constants, then the order of integration can be changed, provided the relevant limits are taken for the concerned variables. When the limits for inner integration are functions of a variable, the change in the order of integration will result in changes in the limits of integration. i.e., will take the form h () h (), This process of converting a given double integral into its equivalent double integral by changing the order of integration is called the change of order of integration.
EXAMPLE : Evaluate integration. by changing the order of X (,) x=y x=-y D D Y Given y : to and x : y to -y By changing the order of integration, In Region D x : to and y : to x. In Region D x : to and y : to -x.
= + = + = = + 4 4 + 4 + 4 4 4 + 4 = 8 + 5 4 =
EXAMPLE :4 Evaluate Y by changing the order of integration. x = x=y X Given x=, x = y, y =, y =. By changing the order of integration y: x to, x : to
= = = / = =, = =, =, by integration by parts, = =
EXAMPLE :5 Evaluate 4 + by changing the order of integration. Y y= x= y=4-x D X Given y=,y= and x=, x= 4 By changing the order of integration, In region D, x : to and y : to 4-x
4 4 x + = + = + 4 x = 4 x + (4 x ) = 4 = 5 4 = 4 8 4 4 + 4 + 8 4 4 + + 8
EXAMPLE :6 Evaluate / by changing the order of integration. Given y : / to and x : to a By changing the order of integration, In Region D x : to and y : to a. In Region D x : to and y : a to a.
/ = + = + = = + + 4 4 4 + + 4 4 = 4 6 + 5 4 4 = 4 8.
EXAMPLE :7 Evaluate + by changing the order of integration. Given x =, x = and y = x, y = -x By changing the order of integration In Region D, y : to,x : to y In Region D, y : to, x : to
I = + + + = + + + = + = ( ) + = -
PROBLEMS FOR PRACTICE Evaluate the following by changing the order of integration. ( + ) Ans: 4. Ans: 4 8. Ans: 6 4. Ans:
PLANE AREA USING DOUBLE INTEGRAL CARTESIAN FORM
EXAMPLE :8 Find by double integration, the area enclosed by the ellipse + = A = 4 = 4 = 4 = 4 = 4 + sin = 4 x x = sq.units.
EXAMPLE :9 Find the area between the parabola = 4 and the line =. Given = 4 =, solving for x, = 4 => = => = => =, 4 4 A = = = ( ) = = 9
EXAMPLE : Find the area between the parabola = and the line = +. Given = and = +. solving for, = + => =, + + A = = = ( + ) = + =
PLANE AREA USING DOUBLE INTEGRAL POLAR FORM
EXAMPLE : Find the area bounded by the circle = sin = 4 sin. = / = = 4 sin sin A = = = 6 sin = ( cos ) = sin =. 4 sin sin
EXAMPLE : Find the area enclosed by the leminiscate = cos by double integration. If r = then cos = implies = 4. A = 4 4 cos = 4 4 cos = 4 4 cos = 4 sin 4 4 =.
EXAMPLE : Find the area that lies inside the cardioids = ( + cos ) and outside the circle =, by double integration. Solving = ( + cos ) = => ( + cos ) = => cos = => =.
(+cos ) A = = = [ ( + cos ) ] = [ cos + cos ] = = [4 cos + + cos ] + sin + 4 sin = (+cos ) + 8.
EXAMPLE :4 Find the common area to the circles =, = cos. Given =, = cos, solving = cos cos = θ=π/ when = => cos = => = /
A = cos = + = + cos = + cos = + + sin = + =
PROBLEMS FOR PRACTICE.Find by double integration, the area bounded by the parabolas = 4 and = 4. Ans: 6...Find by double integration, the smallest area bounded by the circle + = 9 and the line + =. Ans: 9 4...Find by double integration, the area common to the parabola = and the circle + =. Ans: +. 4.Find by double integration, the area lying inside the circle = sin and outside the coordinate = ( ). Ans: 4..