TOPICS DISCUSSED INTRODUCTION REDION OF INTEGRATION CHANGING THE ORDER OF INTEGRATION PLANE AREA USING DOUBLE INTEGRATION CARTISEAN FORM POLAR FORM

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MULTIPLE INTEGRALS

TOPICS DISCUSSED INTRODUCTION REDION OF INTEGRATION CHANGING THE ORDER OF INTEGRATION PLANE AREA USING DOUBLE INTEGRATION CARTISEAN FORM POLAR FORM

INTRODUTION When function f() is integrted with respect to between the limits nd b, we get the double integrl b f d. If the integrnd is function f, y nd if it is integrted with respect to nd y repetedly between the limits nd (for ) nd between the limits y nd y (for y ) we get double integrl tht is denoted by the symbol y f, y ddy. y Etending the concept of double integrl one step further, we get the triple integrl, denoted by z y f, y, z ddydz. z y

EVALUATION OF DOUBLE AND TRIPLE INTEGRALS To evlute y y f, y ddy first integrte f, y with respect to prtilly, treting y s constnt temporrily, between the limits nd. Then integrte the resulting function of y with respect to y between the limits y nd y s usul. In nottion y y f, y d dy ( for double integrl) z z integrl). y y f, y, z d dy dz ( for triple Note: Integrl with vrible limits should be the innermost integrl nd it should be integrted first nd then the constnt limits.

REGION OF INTEGRATION Consider the double integrl d c φ (y) φ (y) f, y ddy, vries from φ y to φ (y) nd y vries from c to d. (i.e) φ y φ y nd c y d. These inequlities determine region in the y plne, which is shown in the following figure.this region ABCD is known s the region of integrtion

EXAMPLE : Evlute y dy d y dy d = y/ = 8 d d = 8 = 4

EXAMPLE : Evlute y dy d y dy d = log dy y =(log log) dy y =log [logy] =log(log log) =log. log (/)

EXAMPLE : Evlute y zdzdyd y zdzdyd = z y dyd = = y y dyd d = 6 = 6

EXAMPLE :4 Evlute d dy y zdz d dy y zdz = d dy z y = y d = d = =

EXAMPLE :5 Evlute r sin θ drdθd r sin θ drdθd = sin θ r dθd = = = sin θdθd cosθ d d =

EXAMPLE :6 Evlute d dy dy d = y d = d = =

EXAMPLE :7 Evlute y yzddydz I = = y [ zdz]ydyd z y ydyd = y ydyd = y dy d = y 4 8 d = 6 = 6 48 48

EXAMPLE :8 Evlute y ddydz y z I = sin z y y ddy = ddy = [y] d = d = + sin = 8

EXAMPLE :9 Evlute sin θ rdrdθ I= = r sin θ dθ sin θ dθ = cos θ dθ = X θ sin θ = 4

PROBLEMS FOR PRACTICE Evlute the following. 4y ddy Ans: 4. b ddy y Ans: log.logb. ddy sin θ 4. r drdθ 5. yzddydz z y+z 6. dzdyd Ans: / Ans: /4 Ans: 9/ Ans: ½

EXAMPLE : Sketch the region of integrtion for f(, y) dyd. Given = nd = ; y = nd y = y = nd + y =

EXAMPLE : Sketch the region of integrtion for f(, y) dyd. Given = ; = nd y = ; y =. Y = y = X y =

EXAMPLE : Evlute D yz ddydz where D is the region bounded by the positive octnt of the sphere + y + z = I = y yz dzdyd = y z y dyd

= y ( y )dyd = ( y y y )dyd = y y y 4 4 d = 8 ( ) d = 8 ( 4 + 5 ) d = 8 4 4 6 = 6 4 6 48.

PROBLEMS FOR PRACTICE.Sketch the region of integrtion for the following (i) 4 y y 4 yddy +y (ii) (iii) yddy +y y dyd.evlute V y + yz + z ddydz, where V is the region of spce bounded by =,=,y=,y=,z= nd z=. Ans: /. Evlute V ddydz (++y+z) bounded by =,y=,z= nd +y+z= Ans: (8log 5) 6, where V is the region of spce 4. Evlute ddydz, where V is the region of spce bounded V by =,,y=,,z= nd +y+4z=. Ans:

CHANGE of ORDER OF INTEGRATION If the limits of integrtion in double integrl re constnts, then the order of integrtion cn be chnged, provided the relevnt limits re tken for the concerned vribles. When the limits for inner integrtion re functions of vrible, the chnge in the order of integrtion will result in chnges in the limits of integrtion. i.e. d c g y g y f, y ddy will tke the form b h () h () f, y dyd This process of converting given double integrl into its equivlent double integrl by chnging the order of integrtion is clled the chnge of order of integrtion.

EXAMPLE : Evlute integrtion. y y yd dy by chnging the order of X (,) =y =-y D D Y Given y : to nd : y to -y By chnging the order of integrtion, In Region D : to nd y : to. In Region D : to nd y : to -.

y y yd dy = ydy d + ydy d = y d + y d = = + 4 4 d + 4 + 4 4 4 + d 4 = 8 + 5 4 =

EXAMPLE :4 y Evlute ye y ddy Y by chnging the order of integrtion. = =y X Given =, = y, y =, y =. By chnging the order of integrtion y: to, : to

y ye y ddy = ye y = ye y = e y / dyd d y d = d e d Tke u =, dv = e d implies du = d, v = e, by integrtion by prts, = e e =

EXAMPLE :5 Evlute 4 y + y ddy by chnging the order of integrtion. Y y= = y=4- D X Given y=,y= nd =, = 4 y By chnging the order of integrtion, In region D, : to nd y : to 4-

4 y 4 + y ddy = + y dyd = y + y 4 d = 4 + (4 ) d = 4 = 5 4 = 4 8 4 4 + 4 + 8 d 4 4 + + 8

EXAMPLE :6 Evlute / y dy d by chnging the order of integrtion. Given y : / to nd : to By chnging the order of integrtion, In Region D : to y nd y : to. In Region D : to y nd y : to.

y dy d / y = ydy d y + ydy d = y y dy + y y dy = = y y dy + + y 4y 4 y 4y + y dy + y 4 4 = 4 6 + 5 4 4 = 4 8.

EXAMPLE :7 Evlute +y dy d by chnging the order of integrtion. Given =, = nd y =, y = - By chnging the order of integrtion In Region D, y : to, : to y In Region D, y : to, : to y

I = y +y d dy + y +y d dy = + y dy + + y y dy = y y dy + y dy = ( ) y + y y = -

PROBLEMS FOR PRACTICE Evlute the following by chnging the order of integrtion. ( + y ) dyd Ans: 4. y dyd Ans: 4 8 y y. y ddy Ans: 6 y 4. y ddy Ans: y

PLANE AREA USING DOUBLE INTEGRAL CARTESIAN FORM

EXAMPLE :8 Find by double integrtion, the re enclosed by the ellipse + y b = b A = 4 dyd = 4 dyd b = 4 y d = 4b = 4b d + sin = 4b = b sq.units.

EXAMPLE :9 Find the re between the prboly = 4 nd the line y =. Given y = 4 nd y =, solving for, = 4 => = => = => =, 4 4 A = dyd = y d = ( )d = = 9

EXAMPLE : Find the re between the prbol y = nd the line y = +. Given y = nd y = +. solving for, = + => =, + y + d A = dyd = = ( + )d = + =

PLANE AREA USING DOUBLE INTEGRAL POLAR FORM

EXAMPLE : Find the re bounded by the circle r = sin θ nd r = 4 sin θ. θ = / θ = θ = 4 sin θ sin θ A = r dr dθ = = 6 sin θ dθ = ( cos θ) dθ = θ sin θ =. r 4 sin θ dθ sin θ

EXAMPLE : Find the re enclosed by the leminiscte r = cos θ by double integrtion. If r = then cos θ = implies θ = 4. A = 4 4 cos θ r dr dθ = 4 r 4 cos θ dθ = 4 4 cos θ dθ = 4 sin θ 4 4 =.

EXAMPLE : Find the re tht lies inside the crdioids r = ( + cos θ) nd outside the circle r =, by double integrtion. Solving r = ( + cos θ) nd r = => ( + cos θ) = => cos θ = => θ =.

(+cos θ) A = r dr dθ = r = [ ( + cos θ) ]dθ = [ cos θ + cos θ ]dθ = = [4 cos θ + + cos θ] dθ θ + sin θ + 4 sin θ = (+cos θ) dθ + 8.

EXAMPLE :4 Find the common re to the circles r =, r = cos θ. Given r =, r = cos θ, solving = cos θ cos θ = θ=/ when r = => cos θ = => θ = /

A = rdrdθ cos θ = r dr dθ + r dr dθ = r dθ + r cos θ dθ = dθ + cos θ dθ = θ + sin θ θ + = + =

PROBLEMS FOR PRACTICE.Find by double integrtion, the re bounded by the prbols = 4y nd y = 4. Ans: 6 sq. units..find by double integrtion, the smllest re bounded by the circle + y = 9 nd the line + y =. Ans: 9 4 sq. units..find by double integrtion, the re common to the prbol y = nd the circle + y =. Ans: + sq units. 4.Find by double integrtion, the re lying inside the circle r = sin θ nd outside the coordinte r = ( cosθ). Ans: 4 sq. units.

THE END

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