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Unit: 3 MULTIPLE INTEGRAL
Unit-III: MULTIPLE INTEGRAL
Sr. No. Name of the Topic Page No.
1 Double integrals 2
2 Evaluation of Double Integral 2
3 To Calculate the integral over a given region 6
4 Change of order of integration 9
5 Change of variable 11
6 Area in Cartesian co-ordinates 13
7 Volume of solids by double integral 15
8 Volume of solids by rotation of an area
(Double Integral)
16
9 Triple Integration (Volume) 18
10 Reference Book 21
MULTIPLE INTEGRALS
RAI UNIVERSITY, AHMEDABAD 1
Unit: 3 MULTIPLE INTEGRAL
1.1 DOUBLE INTEGRALS:
We Know that
∫a
b
f ( x )dx= limn→ ∞δx →0
[f ( x1 ) δ x1+ f ( x2 ) δ x2+ f ( x3 ) δ x3+…+f ( xn ) δ xn ]
Let us consider a function f (x , y ) of two variables x and y defines in the finite region A of xy- plane. Divide the region A into elementary areas.
δ A1, δ A2, δ A3, …δ An
Then ∬A
f ( x , y )dA=¿ limn→ ∞δA → 0
[ f ( x1 , y1 ) δ A1+f ( x2 , y2 ) δ A2+…+ f ( xn , yn ) δ An ] ¿
2.1 Evaluation of Double Integral:
Double integral over region A may be evaluated by two successive integrations.If A is described as f 1(x)≤ y ≤ f 2(x ) [ y1 ≤ y ≤ y2 ]And a ≤ x≤ b ,
Then ∬A
f ( x , y )dA=∫a
b
∫y1
y2
f (x , y )dx dy
2.1.1 FIRST METHOD:
RAI UNIVERSITY, AHMEDABAD 2
Unit: 3 MULTIPLE INTEGRAL
∬A
f ( x , y )dxdy=∫x1
x2 [∫y1
y2
f ( x , y )dy ]dx
f (x , y ) is first integrated with respect to y treating x as constant between the limits y1 and y2 and then the result is integrated with respect to x between the limits x1 and x2.
In the region we take an elementary area δxδy . Then integration w.r.t to y (x
keeping constant) converts small rectangle δxδy into a stripPQ ( y δx). While the integration of the result w.r.t x corresponds to the sliding to the strip, from AD to BC covering the whole regionABCD.
2.1.2 SECOND METHOD:
∬A
f ( x , y )dxdy=∫y1
y2 [∫x1
x2
f ( x , y )dx ]dy
Here f (x , y ) is first integrated w.r.t x keeping y constant between the limits x1 and x2 and then the resulting expression is integrated with respect to y between the limits y1 and y2 and vice versa.
NOTE: For constant limits, it does not matter whether we first integrate w.r.t x and then w.r.t y or vice versa.
2.2 Examples:
RAI UNIVERSITY, AHMEDABAD 3
Unit: 3 MULTIPLE INTEGRAL
Example 1: Find ∫0
1
∫0
x2
ey / x dy dx
Solution: Here, we have
∫0
1 [∫0
x2
e y/ x dy ]dx=∫0
1
[ e y / x
1/ x ]0
x2
dx
¿∫0
1 (ex−1 )1/ x
dx
¿∫0
1
x ex dx−∫0
1
x dx
¿ [ x ex−ex ]01¿−[ x2
2 ]0
1
¿
¿e1−e1+1−12
¿12
∴∫0
1
∫0
x2
e y / x dy dx=12
________ Answer
Example 2: Evaluate ∫0
∞
∫0
∞
e−x2(1+ y2) x dxdy
Solution: Here, we have
∫0
∞
∫0
∞
e−x2(1+ y2) x dxdy=∫0
∞
dy∫0
∞
e−x2(1+ y2) x dx
¿∫0
∞
dy [ e−x2(1+ y2 )
−2(1+ y2) ]0∞
¿∫0
∞
[0+ 1
2 (1+ y2 ) ]dy
¿12
[ tan−1 y ]0∞¿¿
¿12
[ tan−1 ∞−tan−10 ]
¿12 ( π
2 )
RAI UNIVERSITY, AHMEDABAD 4
Unit: 3 MULTIPLE INTEGRAL
¿π4
∴∫0
∞
∫0
∞
e− x2(1+ y2) x dxdy=π4
________ Answer
Example 3: Sketch the area of integration and evaluate∫1
2
∫−√2− y
√2− y¿
2 x2 y2 dxdy
¿¿.
Solution: Here we have
∫1
2
∫−√2− y
√2− y¿
2x2 y2 dxdy=2
∫1
2
y2 dy ∫−√2− y
√2− y¿
x2 dx ¿
¿¿
¿4∫1
2
y2dy ∫0
√2− y¿
x2 dx¿
¿
[ ∵∫−a
a
f ( x )dx=2∫0
a
f (x ) dx
w here x2is an even function ] ¿4∫
1
2
y2dy [ x3
3 ]0
√2− y
¿ 43∫
1
2
y2 dy (2− y )32
¿ 43∫
1
0
(2− y )2 t3 /2(−dt) [ put 2− y=t∴dy=−dt ]
¿ 43 [(2−t)2( 2 t
52
5 )−(−2 ) (2−t ) 25
.27
t72+(2) 2
5.
27
.29
t92 ]
0
1
¿43 [ 2
5+2( 2
5.27 )+ (2 )( 2
5.27
.29 )]
¿43 [ 2
5+ 8
35+ 16
315 ] ¿
415 [2+ 8
7+16
63 ]RAI UNIVERSITY, AHMEDABAD 5
Unit: 3 MULTIPLE INTEGRAL
¿4
15 [ 126+72+1663 ]
¿4
15 ( 21463 )
¿856945
∴∫1
2
∫−√2− y
√2− y¿
2 x2 y2 dxdy=856945
¿¿ ________ Answer
3.1 To Calculate the integral over a given region:
Sometimes the limits of integration are not given but the area of the integration is given.
If the area of integration is given then we proceed as follows:
Take a small areadx dy. The integration w.r.t x between the limits x1 , x2 keeping y fixed indicates that integration is done, alongPQ. Then the integration of result w.r.t to y corresponds to sliding the strips PQ from BC to AD covering the whole regionABCD.
We can also integrate first w.r.t ‘ y’ then w.r.t x , which ever is convenient.
Example 4: Evaluate ∬ xy dxdy over the region in the positive quadrant for
whichx+ y≤ 1.
Solution: x+ y=1 represents a line AB in the figure.
x+ y<1 represents a plane OAB.
The region for integration is OAB as shaded in the figure.
RAI UNIVERSITY, AHMEDABAD 6
Unit: 3 MULTIPLE INTEGRAL
By drawing PQ parallel to y-axis, p lies on the line AB
i .e . ,(x+ y=1) & Q lies on x-axis. The limit for y is 1−x and 0.
Required integral =
∫0
1
dx ∫0
1−x
y dy=∫0
1
x dx [ y2
2 ]0
1− x
¿ 12∫
0
1
( xdx )(1−x)2
¿ 12∫
0
1
( x−2 x2+x3 ) dx
¿12 [ x2
2−
2 x3
3+
x4
4 ]0
1
¿ 12 [ 1
2−2
3+ 1
4 ]¿ 1
24 ________ Answer
Example 5: Evaluate ∬R
xy dxdy , where Rthe quadrant of the circle is
x2+ y2=a2 where x≥ 0∧ y ≥0.
Solution: Let the region of integration be the first quadrant of the circleOAB.
Let I=∬R
xy dx dy (x2+ y2=a2 , y=√a2−x2)
First we integrate w.r.t y and then w.r.t x.
The limits for y are 0 and √a2−x2 and for x, 0 to a.
RAI UNIVERSITY, AHMEDABAD 7
Unit: 3 MULTIPLE INTEGRAL
I=∫0
a
x dx ∫0
√a2−x2
y dy
¿∫0
a
x dx [ y2
2 ]0
√a2−x2
¿ 12∫
0
a
x ( a2−x2 ) dx
¿12 [ a2 x2
2−
x4
4 ]0
a
¿ a4
8________ Answer
Example 6: Evaluate ∬A
x2dx dy , where A is the region in the first quadrant
bounded by the hyperbola xy=16 and the linesy=x , y=0∧x=8.
Solution: The line OP , y=x and the curve PS , xy=16 intersect atp(4,4).
The line SN ,x=8 intersects the hyperbola at S(8,2). Andy=0 is x-axis.
The area A is shown shaded.
Divide the area into two parts by PM perpendicular to OX.
For the area OMP , y varies from 0 to x, and then x varies from 0 to 4.
For the area PMNS , y varies from 0 to 16x and then x varies from 4 to 8.
∴∬A
x2dx dy=∫0
4
∫0
x
x2 dxdy+∫4
8
∫0
16x
x2 dxdy
¿∫0
4
x2dx∫0
x
dy+∫4
8
x2 dx∫0
16x
dy
¿∫0
4
x2 [ y ]¿0x
dx+∫4
8
x2 [ y ]¿0
16x dx
RAI UNIVERSITY, AHMEDABAD 8
Unit: 3 MULTIPLE INTEGRAL
¿∫0
4
x3 dx+∫4
8
16 x dx
¿ [ x4
4 ]0
4
+16 [ x2
2 ]4
8
¿64+8(82−42)
¿64+384
¿448 ________ Answer
3.2 EXERCISE:
1) Find ∫0
1
∫0
y
x y e−x2
dx dy .
2) Evaluate the integral∫1
log 8
∫0
log y
ex+ y dx dy.
3) Evaluate∫0
a
∫xa
x¿
x dxdyx2+ y2
¿¿.
4) Evaluate∫0
1
∫0
1dx dy
√ (1−x2 ) (1− y2 ) .
5) Evaluate∬S
√xy− y2 dy dx, where S is a triangle with vertices (0, 0), (10,
1), and (1, 1).
6) Evaluate ∬ ( x2+ y2) dxdy over the area of the triangle whose vertices
are (0, 1), (1, 0), (1, 2).
7) Evaluate ∬ y dxdy over the area bounded by x=0 , y=x2 ,
x+ y=2 in the first quadrant.
8) Evaluate ∬ xy dxdy over the region R given by
x2+ y2−2 x=0 , y2=2 x , y=x.
4.1 CHANGE OF ORDER OF INTEGRATION:
On changing the order of integration, the limits of the integration change. To find the new limits, draw the rough sketch of the region of integration.
RAI UNIVERSITY, AHMEDABAD 9
Unit: 3 MULTIPLE INTEGRAL
Some of the problems connected with double integrals, which seem to be complicated can be made easy to handle by a change in the order of integration.
4.2 Examples:
Example 1: Evaluate ∫0
∞
∫x
∞e− y
ydx dy .
Solution: We have, ∫0
∞
∫x
∞e− y
ydx dy
Here the elementary strip PQ extends from y=x to y=∞ and this vertical strip slides fromx=0¿ x=∞. The shaded portion of the figure is, therefore, the region of integration.
On changing the order of integration, we first integrate w.r.t x along a horizontal strip RS which extends from x=0 tox= y. To cover the given region, we then integrate w.r.t ' y ' fromy=0¿ y=∞.
Thus
∫0
∞
dx∫x
∞e− y
ydy=∫
0
∞e− y
ydy∫
0
y
dx
¿∫0
∞e− y
ydy [ x ]0
y¿ ¿
¿∫0
∞
ye− y
ydy
¿∫0
∞
e− y dy
¿ [ e− y
−1 ]0
∞
¿−[ 1e y ]
0
∞
RAI UNIVERSITY, AHMEDABAD 10
Unit: 3 MULTIPLE INTEGRAL
¿−[ 1∞
−1]¿1 ________ Answer
Example 2: Change the order of integration in I=∫0
1
∫x2
2−x
xy dx dy and hence evaluate
the same.
Solution: We have I=∫0
1
∫x2
2−x
xy dx dy
The region of integration is shown by shaded portion in the figure bounded by parabola y=x2 , y=2−x , x=0( y−axis).The point of intersection of the parabola y=x2 and the line y=2−x is B (1,1 ).In the figure below (left) we draw a strip parallel to y-axis and the strip y, varies from x2 to 2−x and x varies from 0 to 1.
On changing the order of integration we have taken a strip parallel to x-axis in the area OBC and second strip in the areaCBA. The limits of x in the area OBC are 0 and √ y and the limits of x in the area CBA are 0 and2− y.
So, the given integral is
¿∫0
1
y dy∫0
√y
x dx+∫1
2
y dy ∫0
2− y
x dx
¿∫0
1
y dy [ x2
2 ]0
√ y
+∫1
2
ydy [ x2
2 ]0
2− y
¿ 12∫
0
1
y2 dy+ 12∫1
2
y (2− y )2dy
¿12 [ y3
3 ]0
1
+12∫1
2
(4 y−4 y2+ y3)
¿ 16+ 1
2 [ 96−128+48−24+16−312 ]
¿ 16+ 5
24
RAI UNIVERSITY, AHMEDABAD 11
Unit: 3 MULTIPLE INTEGRAL
¿ 924
¿ 38 ________ Answer
4.3 EXERCISE:
1) Change the order of the integration∫0
∞
∫0
x
e−xy y dydx.
2) Evaluate ∫0
2
∫1
ex
dxdy by changing the order of integration.
3) Change the order of integration and evaluate∫0
2
∫√2 y
2x2 dx dy
√x 4−4 y2 .
5.1 CHANGE OF VARIABLE:
Sometimes the problems of double integration can be solved easily by
change of independent variables. Let the double integral be as∬R
f ( x , y )dxdy.
It is to be changed by the new variablesu , v.The relation of x , y with u , v are given asx=∅ (u , v ) , y=ψ (u , v ). Then the double integration is converted into.
1. ∬R
f ( x , y )dx dy=∬R'
f {ϕ (u , v ) , ψ (u , v )}|J|du dv ,
[dx dy=∂(x , y )∂(u , v )
du dv] ∬
R
f ( x , y )dx dy=∬D
f {x (u , v ) , y (u , v)}|∂(x , y)∂(u , v)|du dv
5.2 Example 1: Using x+ y=u , x− y=v , evaluate the double integral over the square R
∬R
( x2+ y2) dxdy
Integration being taken over the area bounded by the lines x+ y=2 , x+ y=0 , x− y=2 , x− y=0.
RAI UNIVERSITY, AHMEDABAD 12
Unit: 3 MULTIPLE INTEGRAL
Solution: x+ y=u ________(1)x− y=v ________(2)
On solving (1) and (2), we get
x=12
(u+v ) , y=12
(u−v )
J=∂( x , y )∂ (u , v)
=|∂ x∂ u
∂ x∂ v
∂ y∂ u
∂ y∂ v
|=|12
12
12
−12
| ¿−14−1
4 ¿−12
∬R
( x2+ y2) dxdy=∫0
2
∫0
2
[ 14(u+v)2+ 1
4(u−v)2]|∂(x , y )
∂(u , v )|du dv
¿∫0
2
∫0
212
(u2+v2 )|−12 |du dv
¿−14∫
0
2
dv∫0
2
(u2+v2 ) du
¿−14∫0
2
dv [ u3
3+u v2]
0
2
¿−14∫
0
2
dv ( 83+2v2)
¿−14∫
0
2
( 83+2 v2)dv
¿−14 [ 8
3v+ 2
3v3]
0
2
¿−14 [ 8
3(2 )+ 2
3(2 )3]
¿−14 ( 16
3+16
3 )¿−1
4 ( 323 )
RAI UNIVERSITY, AHMEDABAD 13
Unit: 3 MULTIPLE INTEGRAL
¿−83 ________ Answer
5.3 EXERCISE:
1) Using the transformation x+ y=u , y=uv show that ∫0
1
∫0
1− x
e y/(x + y)dy dx=12(e−1)
2) Evaluate ∬R
(x+ y )2dx dy , where R is the parallelogram in the xy-plane with
vertices (1,0), (3,1), (2,2), (0,1), using the transformation u=x+ y and v=x−2 y.
6.1 AREA IN CARTESIAN CO-ORDINATES:
Area = ∫a
b
∫y1
y2
dxdy
6.2 Example 1: Find the area bounded by the lines
y=x+2
y=− x+2x=5
Solution: The region of integration is bounded by the lines
y=x+2 _________(1)
y=− x+2 _________(2)
x=5 _________(3)
On solving (1) and (2), we get the point A(0,2)
On solving (2) and (3), we get the point C (5 ,−3)
On solving (1) and (3), we get the point E(5,7)
We draw a strip parallel to y-axis.
On this strip the limits of y are y=− x+2 and y=x+2, and the limit of x are x=0 and x=5.
RAI UNIVERSITY, AHMEDABAD 14
Unit: 3 MULTIPLE INTEGRAL
Required area = Shaded portion of the figure
¿∬dxdy
¿∫0
5
dx ∫– x+2
x+2
dy
¿∫0
5
dx [ y ]−x+2
x+2¿ ¿
¿∫0
5
dx [ x+2−(−x+2)]
¿∫0
5
[ 2 x ] dx
¿ [ 2 x2
2 ]0
5
¿ [ x2 ]05¿¿
¿ [ 25−0 ]
¿25 Sq. units ________ Answer
Example 2: Find the area between the parabolas y2=4 ax andx2=4 ay.
Solution: We have, y2=4 ax ________ (1)
x2=4 ay ________ (2)
On solving the equations (1) and (2) we get the point of intersection (4a, 4a).
Divide the area into horizontal strips of width δy , x varies from
P ,y2
4a¿Q ,√4 ay and then y varies from O ( y=0 )¿ A( y=4 a).
∴The required area ¿∫0
4 a
dy ∫y2 /4 a
√4 ay
dx
RAI UNIVERSITY, AHMEDABAD 15
Unit: 3 MULTIPLE INTEGRAL
¿∫0
4 a
dy [ x ]¿ y2
4 a
√4 ay
¿∫0
4 a
dy [√4 ay− y2
4 a ]
¿ [√4 ay3 /2
32
− y3
12 a ]0
4 a
¿ 4√a3
(4 a)3 /2−(4 a)3
12 a
¿ 163
a2 ________ Answer
7.1 VOLUME OF SOLIDS BY DOUBLE INTEGRAL:
Let a surface S' be z=f (x , y )
The projection of s ' on x− y plane be S .
Take infinite number of elementary rectanglesδx δy. Erect vertical rod on the δx δy of height z .
Volume of each vertical rod ¿ Areaof the base× height
¿δx δy . z
Volume of the solid cylinder on S ¿ lim
δx → 0δy → 0
∑∑ z dx dy
¿∬ z dx dy
¿∬ f ( x , y ) dx dy
Here the integration is carried out over the area S.
RAI UNIVERSITY, AHMEDABAD 16
Unit: 3 MULTIPLE INTEGRAL
Example 1: Find the volume bounded by the xy-plane, the paraboloid 2 z=x2+ y2 and the cylinderx2+ y2=4.
Solution: Here, we have
2 z=x2+ y2⇒2 z=r2⇒ z= r2
2 (Paraboloid) ______ (1)
x2+ y2=4⇒r=2 , z=0 , (circle) ______ (2)
Volume of one vertical rod ¿ z .r dr dθ
Volume of the solid ¿∬ z r dr dθ
¿2∫0
π
dθ∫0
2r 2
2r dr
¿ 22∫
0
π
dθ∫0
2
r3 dr
¿∫0
π
dθ (r 4
4 )0
2
¿∫0
π
dθ (164 )
¿4∫0
π
dθ
¿4 [θ ]0π¿¿
¿4 π ________ Answer
8.1 VOLUME OF SOLID BY ROTATION OF AN AREA (DOUBLE INTEGRAL):
When the area enclosed by a curve y=f (x ) is revolved about an axis, a solid is generated; we have to find out the volume of solid generated.
Volume of the solid generated about x-axis ¿∫a
b
∫y1( x)
y2( x)
2 π PQ dxdy
RAI UNIVERSITY, AHMEDABAD 17
Unit: 3 MULTIPLE INTEGRAL
Example 1: Find the volume of the torus generated by revolving the circle x2+ y2=4 about the linex=3.
Solution: x2+ y2=4
V=∬ (2 π PQ ) dxdy
¿2 π∬ (3−x ) dx dy
¿2 π∫−2
2
dx (3 y−xy )¿−√4− x2
+√4−x2
(3− x ) dy
¿2 π∫−2
+2
dx [3√4−x2−x √4−x2+3√4−x2−x √4−x2 ]
¿4 π∫−2
+2
[ 3√4−x2−x √4−x2 ] dx
¿4 π [3 x2
√4−x2+3×42
sin−1 x2+ 1
3(4−x2 )3 /2]
−2
+2
¿4 π [6×π2
+6 ×π2 ]
¿24 π2 ________ Ans.
8.2 EXERCISE:
1) Find the area of the ellipse x2
a2 + y2
b2 =1
2) Find by double integration the area of the smaller region bounded by x2+ y2=a2 andx+ y=a.
3) Find the volume bounded byxy−plane, the cylinder x2+ y2=1 and the planex+ y+z=3.
4) Evaluate the volume of the solid generated by revolving the area of the parabola y2=4 ax bounded by the latus rectum about the tangent at the vertex.
RAI UNIVERSITY, AHMEDABAD 18
Unit: 3 MULTIPLE INTEGRAL
9.1 TRIPLE INTEGRATION (VOLUME) :
Let a function f (x , y , z) be a continuous at every point of a finite region S of three dimensional spaces. Consider n sub-spaces δ s1 , δ s2 , δ s3 , …. δ sn of the space S.If (xr , yr , zr) be a point in the rth subspace.
The limit of the sum ∑r=1
n¿
f (x r , y r , zr)δ sr ,as n →∞ ,δ sr →0¿
¿ is known as the triple integral of
f (x , y , z) over the space S.
Symbolically, it is denoted by
∭S
f (x , y , z ) dS
It can be calculated as∫x1
x2
∫y1
y2
∫z 1
z 2
f (x , y , z ) dx dy dz. First we integrate with respect to
z treating x , y as constant between the limitsz1∧z2. The resulting expression (function ofx , y) is integrated with respect to y keeping x as constant between the limits y1∧ y2. At the end we integrate the resulting expression (function of x only) within the limitsx1∧x2.
First we integrate from inner most integral w.r.t z, and then we integrate w.r.ty, and finally the outer most w.r.tx.
RAI UNIVERSITY, AHMEDABAD 19
Unit: 3 MULTIPLE INTEGRAL
But the above order of integration is immaterial provided the limits change accordingly.
Example 1: Evaluate ∭R
( x−2 y+z ) dx dydz , w h ere R :0 ≤ x ≤10≤ y≤ x2
0 ≤ z≤ x+ y
Solution: ∭R
( x−2 y+z ) dx dydz
¿∫0
1
dx∫0
x2
dy ∫0
x+ y
(x−2 y+z)dz
¿∫0
1
dx∫0
x2
dy (xz−2 yz+ z2
2 )0
x+ y
¿∫0
1
dx∫0
x2
dy [ x ( x+ y )−2 y ( x+ y )+ (x+ y)2
2 ] ¿∫
0
1
dx∫0
x2
dy [ x2+xy−2xy−2 y2+(x+ y )2
2 ] ¿∫
0
1
dx∫0
x2
dy [ x2−xy−2 y2+ x2
2+xy+ y2
2 ] ¿∫
0
1
dx∫0
x2
dy [ 3 x2
2−3 y2
2 ] ¿ 3
2∫0
1
dx∫0
x2
( x2− y2 ) dy
¿32∫0
1
dx (x2 y−y3
3 )0
x2
¿ 32∫0
1
dx (x4− x6
3 ) ¿ 3
2 ( x5
5− x7
21 )0
1
RAI UNIVERSITY, AHMEDABAD 20
Unit: 3 MULTIPLE INTEGRAL
¿ 32 ( 1
5− 1
21 ) ¿ 8
35 ________ Answer
Example 2: Evaluate ∫0
log 2
∫0
x
∫0
x+ y
ex+ y+z dx dy dz
Solution: I=∫0
log 2
∫0
x
ex+ y [ez ]0x+ y¿
dx dy¿
¿ ∫0
log 2
∫0
x
ex+ y(ex+ y−1)dx dy
¿ ∫0
log 2
∫0
x
[e2(x+ y)−ex + y ] dxdy
¿ ∫0
log 2
[e2 x .e2 x
2−ex . e y ]
0
x
dx
¿ ∫0
log 2
[ e4 x
2−e2x−
e2 x
2+ex]
0
x
dx
¿ [ e4 x
8−
e2x
2−
e2x
4+ex ]
0
log 2
¿ [ e4 log 2
8− e2 log 2
2− e2 log2
4+e log2]−( 1
8−1
2−1
4+1)
¿ [ e log16
8− e log 4
2− elog 4
4+e log2]−( 1
8−1
2−1
4+1)
¿( 168
− 42−4
4+2)−( 1
8−1
2− 1
4+1)
¿ 58 ________ Answer
9.2 EXERCISE:
RAI UNIVERSITY, AHMEDABAD 21
Unit: 3 MULTIPLE INTEGRAL
1) Evaluate ∭R
¿ ¿( x+ y+z ) dxdy dz ,where R : 0≤ x ≤1 , 1≤ y ≤2 , 2≤ z≤ 3.
¿¿
2) Evaluate∫0
log 2
∫0
x
∫0
x+ log y
ex+ y +z dz dy dx.3) Evaluate ∭
R
( x2+ y2+z2 ) dxdy dz where R :x=0 , y=0 , z=0
x+ y+z=a ,(a>0)
10.1 REFERENCE BOOK:1) Introduction to Engineering MathematicsBy H. K. DASS. & Dr. RAMA VERMA2) www.bookspar.com/wp-content/uploads/vtu/notes/1st- 2nd-sem/m2-21/Unit-5-Multiple-Integrals.pdf3) http://www.mathstat.concordia.ca/faculty/cdavid/ EMAT212/solintegrals.pdf4) http://studentsblog100.blogspot.in/2013/02/anna- university-engineering-mathematics.html
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