double integrals - wordpress.com · multiple integrals to calculate quantities that vary over two...
TRANSCRIPT
Multiple Integrals In this chapter we consider the integral of a function of two variables f(x, y) over a region in the plane and the integral of a function of three variables f(x, y, z) over a region in space. These integrals are called Multiple integrals. We can use multiple integrals to calculate quantities that vary over two or three dimensions, such as the total mass or the angular momentum of an object of varying density and the volume of solids with general curved boundaries.
Double Integration
We know that [ ]nn
b
a
xxfxxfxxf
x
n
Limit
dxxf δδδ
δ
)(...........)()(
0
)( 2211 +++
→
∞→=∫
Letus consider a function f(x,y) of two variables x and y defined in the finite
region A of xy-plane. Divide the region A into elementary areas 1Aδ , 2Aδ ,……….
Anδ . Then
[ ]nn
A
AxfAxfAxf
A
n
it
dAyxf δδδ
δ
)(...........)()(
0
lim
),( 2211 +++
→
∞→=∫∫
Evaluation: Double integration over region A may be evaluated by two successive integrations.
If A is described as )()( 21 xfyxf ≤≤ and bxa ≤≤ ,then
∫∫ ∫ ∫=A
b
a
xf
xf
dxdyyxfdAyxf
)(
)(
2
1
),(),(
First Method:
dAyxfA
b
a
f
f
∫∫ ∫ ∫
=
(
(
2
1
),(
f(x,y) is
first integrated with respect to y treating x as constant between the limits
)()( 21 xfandxf and then resulting expression is integrated with respect to x
between the limits a and b. Second Method:
dAyxfA
d
c
f
f
∫∫ ∫
=
2
1
),(
Here f(x,y) is first integrated with respect to x treating y as constant between the
limits ( )yfandyf 21 )(
respect to y between the limits c and d. Note: For constant limits, it does not or with respect to x. Problems:
1) Evaluate dydxe
x
x
y
∫ ∫1
0 0
Solution:
dxdyyxf
x
x
∫
)(
)
),(
first integrated with respect to y treating x as constant between the limits
and then resulting expression is integrated with respect to x
between the limits a and b.
dydxyxf
y
y
∫
)(
)(
),(
Here f(x,y) is first integrated with respect to x treating y as constant between the
and then resulting expression is integrated with
respect to y between the limits c and d.
For constant limits, it does not matter whether we first integrate with respect to y
first integrated with respect to y treating x as constant between the limits
and then resulting expression is integrated with respect to x
Here f(x,y) is first integrated with respect to x treating y as constant between the
and then resulting expression is integrated with
matter whether we first integrate with respect to y
Let dydxeI
x
x
y
∫ ∫=1
0 0
since the limits are functions of x first
we have to integrate w.r.to y, then w.r.to x.
dxdyeI
x
x
y
∫ ∫
=
1
0 0
dxx
eI x
y
∫∞
=
1
0 0
/1
/
[ ] ∫∫ −=−=1
0
1
0
)1(1 xdxedxeI
( )1
0
2
21
−=
xeI
( )12
1−= eI
2. Evaluate ∫ ∫+
++
1
0
1
0
22
2
1
xx
yx
dydx
Solution: Let dxyx
dyI
x
∫ ∫
++=
+1
0
1
0
22
2
1
dxx
y
xI x 21
0
1
0
2
1
2 1tan
1
1 +−
∫
++=
( )dxx
I ∫−− −
+=
1
0
11
20tan1tan
1
1
dxx
I ∫
+=
1
0
2 41
1 π
[ ]10
2 1log(4
++= xxIπ
( )[ ]021log4
−+=π
I
( )21log4
+=π
I
3. Evaluate ∫ ∫a ay
xydxdy0 0
Let ∫ ∫=a ay
xydxdyI0 0
dyxydxyI
a ay
∫ ∫
=
0 0
dyx
yI
aya
00
2
2∫
=
[ ]dyayyI
a
∫ −=0
02
1
dyya
I
a
∫=0
2
2
a
yaI
0
3
32
=
=
32
3aa
I
6
4a
I =
4. Evaluate ∫ ∫−a ya
dxdy0 0
22
Let ∫ ∫−
=a ya
dxdyI0 0
22
= dydx
a ya
∫ ∫
−
0 0
22
[ ] dyxI
aya
∫−
=0
0
22
= dyya
a
∫ −0
22
Let θsinay = therefore θθdady cos=
Limits when y=0 , 0=θ
y=a , 2
πθ =
∫ −=2
0
22 sin
π
θθ daI = ∫=2
0
22 cos
π
θθ da
( )∫
−=
2
0
2
2
2cos1
π
θθ
daI =2
0
2
2
2sin
2
π
θθ
−=
a
4
022
22 ππ aaI =
−=
5. Evaluate ( )dxdyyx∫ ∫ +1
0
3
0
22 3
Solution: Let ( )dxdyyxI ∫ ∫ +=1
0
3
0
22 3 = dyxyx
3
0
1
0
23
33∫
+=
[ ]dyyI ∫ +=1
0
299 = [ ]dyy∫ +=1
0
219
1
0
1
0
3
39∫
+=
yyI =
+=
3
119 =12
6. Evaluate ∫ ∫ xydxdy over the region in the positive quadrant for which
1≤+ yx .
Solution: x+y=1 represents a line AB in the figure. 1<+ yx represents the plane
OAB. Therefore the region for integration is OAB as shown in the figure By drawing pQ parallel to y-axis, P lies on the line AB (x+y=1) and Q lies on x-axis. The limits for y are 0 and (1-x). Also limits for x are 0 to 1 as the strip moves from left to right.
Let ∫ ∫−
=1
0
1
0
x
xydxdyI = dxy
x
x−
∫
=
1
0
1
0
2
2
( ) dxxxI2
1
0
12
1−= ∫ = ( )dxxxx 12
2
1 2
1
0
+−= ∫
( 23
1
0
22
1+−= ∫ xxxI
242
1
3
2
4
1
2
1=
+−=I
7. Evaluate ∫∫R
xydxdy where R is the quadrant of the circle
0,0 ≥≥ yx .
Solution: Given that 0≥x
quadrant of the circle 2x +
Therefore, the limits for y are y=0 to
∴ ∫∫R
xydxdy ∫ ∫−
=a xa
xydxdy0 0
22
∫
=
ay
x0
2
2
[∫=a
ax0
2
2
1
[∫ −=a
xa0
2
2
1
x
a2
2
22
1
=
422
1 4aa
−=
8. Evaluate ∫∫A
xydxdy , where A is the domain bounded by x
x=2a and the curve 42 ayx =
)1
0
234
232
42
1
+−=
xxxdxx
24
1
where R is the quadrant of the circle 22 yx =+
0,0 ≥y , therefore, region of integration be the first 22 ay = .
Therefore, the limits for y are y=0 to 22 xay −= and x=0 to x=a
xydxdy
−xa
dx
0
22
]− dxx22
]− dxx3
a
x
0
42
42
−
84
44aa
=
, where A is the domain bounded by x-axis, ordinate
.ay
X
2a , and
, therefore, region of integration be the first
axis, ordinate
Solution: The region of integration be OAB and limits for y are
a
xytoy
40
2
== and x=0 to x=2a.
∴ ∫∫A
xydxdy ∫ ∫=a a
x
xydxdy
2
0
4
0
2
∫
=
ay
x
2
0
2
2
∫
=
ax
x0
162
1
[∫=a
a0
232
1
x
a 2 632
1
=
=
32
1 6
2
a
a
3
4a
=
9. By double integration , find the area of the circle
Solution: The circle is symmetrical about the coboth x & y occur with even powers.The area of the circle is
Solution: The region of integration be OAB and limits for y are
and x=0 to x=2a.
xydxdy
a
x
dx4
0
2
dx
a
x2
4
16
[ ]dxx5
a
x2
0
6
6
6
66x
By double integration , find the area of the circle 222
ayx =+ .
Solution: The circle is symmetrical about the co-ordinate axes x and y, since both x & y occur with even powers.
ordinate axes x and y, since
∴ ∫∫ dxdy ∫= 4
[∫=a
0
4
∫=a
0
4
x
4
=
04
−=
2
2
4 a×=
10. Using double integration, find the area of the ellipse
1
2
2
2
2=+
b
y
a
x
Solution: The equation of the ellipse is
terms having even powers of x and y. Therefore the ellipse is sx and y.
∫ ∫−a xa
dxdy0 0
22
[ ] − xadxy 0
22
− dxxa 22
a
a
xa
a
xa
0
1222
sin2
+
− −
a
a
0
12
01sin4
−− −
2
2aπ
π=×
Using double integration, find the area of the ellipse
1.
Solution: The equation of the ellipse is 1
2
2
2
2=+
b
y
a
x, the equation contain the
terms having even powers of x and y. Therefore the ellipse is symmetrical about
, the equation contain the
ymmetrical about
The area bounded by the ellipse
=A
=A
=A
Aa
b4=
=4
aA
A =
11. Find the area of the triangle formed by the line y=x, x=3 and
double integration. Solution: Here area of integration is OAB and integrating first w.r.to y and then w.r.to x. The limits of integration areY=0 to y=x x=0 to x=3
The area bounded by the ellipse 1
2
2
2
2=+
b
y
a
x is given by
∫ ∫−
=a
xaa
b
dydx0 0
22
4
[ ]∫−
axa
a
b
dxy0
0
22
4
∫ −a
dxxaa
b
0
224
a
a
xa
a
xax
a
b
0
1222
sin2
+
− −
+ −
1sin2
04 1
2a
a
b
aba
a
bπ
π=××
22
4 2
Find the area of the triangle formed by the line y=x, x=3 and y=0 using
Solution: Here area of integration is OAB and integrating first w.r.to y and then
The limits of integration are
y=0 using
Solution: Here area of integration is OAB and integrating first w.r.to y and then
Therefore Area A = x
∫ ∫3
0 0
= =
12. Find the area between the parabola
Solution: Given axy 42 =
and x 42 =
on solving equations (1) and (2), we get the point of intersection
from (2) a
xy
4
2
= and
substituting it into (1), we get
( )
axa
x4
42
4
=
i.e. ( )33 4ax =
therefore ax 4= similarly
the point of intersection is
dividing the area into horizontal strips of width
x varies from a
y
4
2
to ay4
y varies from y=0 to y=4a The required area
dxdy = [ ] dxx
y0
3
0
∫=
dxx∫=3
0
= 0
3
2
2
=
x
2
9=
Find the area between the parabola axy 42 = and ayx 42 = .
ax (1)
ay4 (2)
on solving equations (1) and (2), we get the point of intersection
( )2
42
4a
xy =
substituting it into (1), we get
similarly ay 4=
the point of intersection is ( )aa 4,4
dividing the area into horizontal strips of width yδ ,
and
∫ ∫=a ay
a
y
dxdyA
4
0
4
4
2
∫ ∫=a ay
a
y
dxdyA
4
0
4
4
2
A
4
=
A ∫=4
A
=
=A
=A
16
A =
Y
O
A(4a,4a)
y 2
=4ax
P
Q
dxdy
[ ] dy
a
y
ay
x
a
4
42
4
0
∫
dya
yay
a
∫
−
4
0
2
44
a
y
a
ya
4
0
32
3
34
1
2
34
−
( ) ( )
−
32
3
412
144
3
2a
aaa
− 22
5
3
16
3
2aa
2
3
16a
X
A(4a,4a)
=4ax
x 2
=4ay
The Cartesian coordinates ( x,y) and the polar coordinates
transformation equation x
transformation is given by
( )( )
θ
θθ
∂
∂
∂
∂∂
∂
∂
∂
=∂
∂=
y
r
y
x
r
x
r
yxJ
,
,=
sin
cos=
1. Transform the integral into polar coordinates and hence evaluate
∫ ∫a a
0 0
2
Solution: Given
Here upper limit of y is a
22.,. xayei ⇒−=
(1) represents the circle whose center is (0,0) and radius a. Lower limit of y is zero i.e. x-axis. Lower limit of x is zero i.e. yfirst quadrant of the circle.
Letus convert (1) into polar co
Therefore 22 ayx =+
2 ar =
In the first quadrant
Limits for r are 0=r to
Change of variables The Cartesian coordinates ( x,y) and the polar coordinates ( )θ,r are related by
θcosr= , θsinry = . The Jacobian of the
transformation is given by
0cossin
sincos≠=
−r
r
r
θθ
θθ
Transform the integral into polar coordinates and hence evaluate
∫−
+x
dydxyx0
22
22
∫ ∫−
+a xa
dydxyx0 0
22
22
22xa −
222222ayxxay =+⇒+= (1)
(1) represents the circle whose center is (0,0) and radius a. Lower limit of y is axis. Lower limit of x is zero i.e. y-axis. Region of integration is the
Letus convert (1) into polar co-ordinates by putting θcosrx = , y2a
2a or ar =
to ar = and
are related by
. The Jacobian of the
Transform the integral into polar coordinates and hence evaluate
(1) represents the circle whose center is (0,0) and radius a. Lower limit of y is axis. Region of integration is the
θsinr= .
Limits for θ are 0=θ to 2
πθ =
Therefore ∫ ∫−
+a xa
dydxyx0 0
22
22
( )∫ ∫=2
0 0
π
θa
rdrdr
[ ] drr
a
2
0
20∫=π
θ
drr
a
∫=0
2
2
π
=632
3
0
3 ara
ππ=
=
2. Change into polar coordinates and evaluate
( )
∫ ∫∞ ∞
+−
0 0
22
dydxe yx
Solution: The given limits of integration are 0=y to ∞=y and
0=x to ∞=x i.e. the entire first quadrant,
let θcosrx = and θsinry =
therefore 222
ayx =+
Limits for r are 0=r to ∞=r and
Limits for θ are 0=θ to 2
πθ =
Therefore
( )
∫ ∫∞ ∞
+−
0 0
22
dydxeyx
∫ ∫∞
−=0
2
0
2
π
θrdrder
[0
∞
∫=
=2
π
Let tr =2 therefore
Limits are same
( )
∫ ∫∞ ∞
+−
0 0
22
dydxe yx
3. Evaluate ∫ ∫ +
a a
yx
xdxdy
0
2
coordinates. Solution: Given
∫ ∫a a
yx
0
[ ] drerr
2220
−π
θ
drerr
∫∞
−
0
2 2
dtrdr =2 and 2
dtrdr =
dydx22
0
dte
t
∫∞
−−=π
∞−
−=
014
teπ
[ ]104
+=π
4
π=
+ y
xdxdy2 by changing to polar
+ yx
xdxdy22
Limits for x are yx = to ax = and
Limits for y are 0=y to ay =
The region of integration is OAB
The lower limit for 0=r
Upper limit ar =θcos
Therefore θcos
ar =
The lower limit for 0=θ which is varying from OA to OB
∴
= −
x
y1tanθ at B ay = and ax =
( )4
1tan 1 πθ == −
Therefore θ varies from 0=θ to 4
πθ =
Then ∫ ∫ +
a a
yyx
xdxdy
0
22 = ∫ ∫4
0
cos
0
2
cos
π
θ θθa
r
rdrdr
= ∫ ∫4
0
cos
0
cos
π
θ
θθ
a
drd
Y
X O
A(a,0)
B( a,a)
x=y x = a
4. Using polar coordinates find the area of the circle
integration
Solution: The circle 2x +
the elementary area of the circle in the first quadrant.
Therefore
Area = ∫∫R
dA4
= ∫∫R
4
= 4
= [ ]∫4
0
cos0
π
θ θdr
a
= θθ
θ
π
da
∫
4
0cos
cos
= ∫4
0
π
θda
= [ ]40
π
θa
= 4
aπ
4. Using polar coordinates find the area of the circle 222
ayx =+
22 ay = about the coordinate axis. Let dA=dxdy be
the elementary area of the circle in the first quadrant.
∫∫R
dxdy
∫ ∫−a ya
dxdy0 0
22
by double
about the coordinate axis. Let dA=dxdy be
= ∫ ∫2
0 0
4
π
θa
rdrd
= θ
π
dr
a
∫
2
0 0
2
24
= θ
π
da
∫
2
0
2
24
= [ ] 20
22
π
θa
= 2
2 2 πa
= 2aπ
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. 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 on integration.
1. Evaluate dxdyy
e
x
y
∫ ∫∞ ∞ −
0
Solution: Given dxdyy
e
x
y
∫ ∫∞ ∞ −
0
Here the elementary strip PQ extends from y=x to ∞=y and this vertical strip
slides from x=0 to ∞=x . On changing the order of integration, we first integrate w.r.to x along a horizontal strip RS which extends from x=0 to x=y. To cover the given region, we then
integrate w.r.to y from y=0 to ∞=y .
dxdyy
e
x
y
∫ ∫∞ ∞ −
0
=
2. Change the order of integration in
Solution: Given
Limits are for y are xy = x are x=0 , x=1. The region of integration is OBA, if we change the order of integration, the region of integration will be OBC and ABC.
Y
R S
P
Q
O y = x
dxdy = dxdyy
ey y
∫ ∫∞ −
0 0
= [ ] dyxy
e yy
0
0
∫∞ −
= [ ]dyyy
ey
∫∞ −
0
= dye y
∫∞
−
0
=
∞−
−0
1
ye
= [ ]00 e+
= 1.
2. Change the order of integration in dxdyxy
x
x
∫ ∫−1
0
2
2
.
dxdyxy
x
x
∫ ∫−1
0
2
2
2x , y = 2-x
x are x=0 , x=1.
The region of integration is OBA, if we change the order of integration, the region of integration will be OBC and ABC.
X
The region of integration is OBA, if we change the order of integration, the region
i.e., dxdyxy
x
x
∫ ∫−1
0
2
2
= ∫1
02
1
3. Evaluate ∫ ∫ +
a a
yx
xdxdy
0
2
Solution: Limits are
The limits of integration asserts that the region of integration is bounded by the
lines yx = , ax = ,
= ∫∫OBC
xydxdy + ∫∫ABC
xydxdy
= ∫ ∫1
0 0
y
xydxdy + ∫ ∫−2
1
2
0
y
xydxdy
= ∫
1
0 0
2
2dy
xy
y
+ ∫
2
1
2
2
xy
= [ ]∫1
02
1dyyy + ( )∫ −
2
1
22
2
1dyyy
∫1
0
2dyy + ( )∫ +−
2
1
2442
1dyyyy
= ( )
+−+ ∫∫
2
1
32
1
0
2 442
1dyyyydyy
=
+−+
2
1
4321
0
3
43
4
2
4
32
1 yyyy
=
+−+
4
15
3
286
3
1
2
1
=8
3
+ y
xdxdy2 by changing the order of integration.
yx = , ax = and
0=y , ay =
The limits of integration asserts that the region of integration is bounded by the
0=y , ay = .
xydxdy
−
2
0
dy
y
dy
dy
dy
integration.
The limits of integration asserts that the region of integration is bounded by the
The region of integration is the triangle OPQ.
If the order of integration is changed the limits for y are
DC, the limits for x are =x
Therefore ∫ ∫ +
a a
yx
xdxdy
0
2
4. By changing the order of integration, evaluate
(x
y
∫ ∫−3
0
4
0
Solution: Given
The region of integration is OABC. If we change the order of integration the region will divide into two parts, ODBC and ABD, then the limits are
Y
Q O Y = 0
The region of integration is the triangle OPQ.
If the order of integration is changed the limits for y are 0=y , y
0= , ax = .
y2 ∫ ∫ +
=a x
yx
xdxdy
0 0
22
∫
= −
a y
dxx
y
0 0
1tan
[ ]∫ −= −a
dx0
1 0)1(tan
∫=a
dx0
4
π
[ ]ax 0
4
π=
4
aπ=
4. By changing the order of integration, evaluate
)dxdyy+
( )dxdyyx
y
∫ ∫−
+3
0
4
0
The region of integration is OABC. If we change the order of integration the region will divide into two parts, ODBC and ABD, then the limits are
X
xy = along
The region of integration is OABC. If we change the order of integration the
In the region ODBC x
y
In the region ABD x
y
Thus ( )dxdyyx
y
∫ ∫−
+3
0
4
0
( )dxdyyx
x
∫ ∫−
++2
1
4
0
2
dyxyx
∫
+=
3
0
1
0
2
2
dyy∫
+=
3
0
1
02
1+
3
0
2
22
+=
yy2
+
60
1587=
5. Evaluate dxdyy
x
∫ ∫−1
0
1
0
2
2
Solution: The region of integration is OAB,If we change the order of integration, vertical strip AB will convert into horizontal strip CD. Then the new limits are
10
30
==
==
xto
ytoy
21
40 2
==
−==
xto
xytoy
)dxdy ( )dxdyyx∫ ∫ +=3
0
1
0
dxdy
dy dyy
xy
x
∫−
++
2
1
4
0
22
2
( ) ( ) dyxxx∫
−+−
2
1
222 42
14
5342
53
816
2
1
4
+−+−
xxx
xx
dxdy by changing the order of integration.
Solution: The region of integration is OAB, If we change the order of integration, vertical strip AB will convert into horizontal strip CD. Then the new limits are
2
1
by changing the order of integration.
If we change the order of integration, vertical strip AB will convert into horizontal
( 210
10
yxtox
ytoy
−==
==
Therefore dxdyy
x
∫ ∫−1
0
1
0
2
2
Let θsin=y , therefore
Limits 1
0
=
=
theny
thenywhen
Therefore y∫ −1
0
2 1
(θ
π
sin1sin2
0
2
∫ −=
)
dxdy dxdyy
y
∫ ∫−
=1
0
1
0
2
2
[ ] dyxyy
∫−
=1
0
1
0
22
dyyy∫ −=1
0
22 1
, therefore θθ ddy cos=
2
0
πθ
θ
=
=
then
then
dyy2
=
) θθθ dcossin2
θ
π
22
0
2 cossin∫=
32
π= .
θθ d
Triple Integration Let a function f(x,y,z) be a continuous at every point of a finite region S of three dimensional space. Consider n subspaces
nssss δδδδ ...........,, 321 of the space
S.
If ( )rrr zyx ,, be a point in the rth space.
The limit of the sum ,),,(1
rrr
n
r
r Szyxf δ∑=
as ∞→n , 0→rSδ is known as
the triple integral of ( )rrr zyxf ,, over the space S.
Symbolically, it is denoted by
∫∫∫S
dSzyxf ),,(
It can be calculated as ∫ ∫ ∫2
1
2
1
2
1
),,(
x
x
y
y
z
z
dzdydxzyxf , first we integrate with
respect to z treating x,y as constant between the limits 21 zandz . The
resulting expression is integrated with respect to y keeping x as constant
between the limits 21 yandy . At the end we integrate the resulting
expression within the limits 21 xandx .
i.e.
∫∫∫
=
=
=
=
=
=
),(
),(
)(
)(
22
11
22
11
2
1
),,(),()(
yxfz
yxfz
xy
xy
bx
ax
dzzyxfdyyxdxx
φ
φ
φϕ
First we integrate from inner most integral w.r.t z, then we integrate with respect to y and finally the outer most with respect to x. Examples:
1. Evaluate ( )∫∫∫ ++R
dxdydzzyx where 10 ≤≤ x , 21 ≤≤ y ,
32 ≤≤ z .
Solution: Given ( )∫∫∫ ++R
dxdydzzyx
( )dzzyxdydx∫ ∫ ∫ ++=1
0
2
1
3
2
( )
3
2
21
0
2
12
++= ∫ ∫
zyxdydx
( )∫ ∫ ++=1
0
2
1
522 dyyxdx
(
∫
+=
1
0
2 x
(∫=1
0
48
1x
(∫ +=1
0
4x
2
2
=
x
2
9=
2. Evaluate (∫∫∫ +R
x2
bounded by x=0,y=0,z=0 and
Solution: Given
(∫∫∫R
x
azyx =++ or
On xy plane, zyx =++Upper limit of xay −=
Upper limit of ax = .
dydx
a xa xa
∫ ∫ ∫− −
=0 0 0
)
++2
1
2
2
52dx
y
)+ 216 dx
)4 dx
1
0
2
4
+ x
)+ dxdydzzy22
where R denote the region
bounded by x=0,y=0,z=0 and azyx =++ , 0>a .
)++ dxdydzzyx222
yxaz −−= which the upper limit for z
a= becomes ayx =+ as shown in the figure.
x
( ) dzzyx
y−
++ 222
where R denote the region
as shown in the figure.
yxaa xaz
zyzxdydx
−−−
++= ∫ ∫
0
222
0 02
( ) ( ) ( )dy
yxayyxayxxaxdx
a xa
∫ ∫−
−−+−−+−−=
0 0
3
3222
3
( ) ( ) ( )xa
ayxayy
xay
xyxaxdx
−
−−+−−+−−= ∫
0
443222
04432
( ) ( )∫
−+−=
a
dxxa
xax
0
42
2
42
( ) ( )∫
−++−=
a
dxxa
xaxxa0
4
4322
42
2
1
( )
a
xaxxa
xa
0
45432
4542
32
1
−+
+−=
301046
5555aaaa
++−=
20
5a
=
Integration by changing of Cartesian coordinates into spherical coordinates Sometimes it is easy to integrate by changing the Cartesian coordinates into spherical coordinates. The relation between the Cartesian and spherical polar are given by the relations φθ cossinrx = ,
φθ sinsinry = ,
θcosrz =
φθ ddrdJdxdydz =
φθθ ddrdr sin2=
Note.
1. Spherical coordinates are very useful if the expression 222
zyx ++ is
involved in the problem. 2. In a sphere 2222
azyx =++ the limits of r are 0 to a and limits of θ
are π,0 and that of φ are π2,0 .
Examples
1. Evaluate the integral ( )∫∫∫ ++v
dxdydzzyx222
taken over the
volume enclosed by the sphere 1222 =++ zyx .
Solution: Letus convert the given integral into spherical coordinates.
( )∫∫∫ ++v
dxdydzzyx222
( )∫ ∫ ∫=π π
φθθ2
0 0
1
0
22 sin drddrr
∫ ∫ ∫=π π
θθφ2
0 0
1
0
4sin drrdd
∫ ∫
=
π π
θθφ2
0 0
1
0
5
5sin
rdd
[ ]∫ −=π
πθφ
2
0
0cos5
1d
∫=π
φ2
05
2d
[ ] πφ
2
05
2=
5
4π= .
2. Evaluate the integral ∫∫∫ ++v
zyx
dxdydzz222
2
over the volume of the
sphere 2222 =++ zyx .
Solution: Convert the given integral into spherical coordinates, we have
∫∫∫ ++v
zyx
dxdydzz222
2
( )φθ
θθπ π
ddrdr
r∫ ∫ ∫=2
0
2
0
2
0
2
24 sincos8
Because sphere 2222 =++ zyx lies in 8 quadrants, which are the limits in first
octant.
∫ ∫ ∫=2
0
2
0
2
0
22 sincos8
π π
θθθφ drrdd
∫ ∫
=
2
0
2
0
2
0
2
3
3sincos8
π π
θθθφr
dd
[ ]2
0
2
0
3
20
3
3
3
cos8
−=
rπ
π θφ
3
22
3
1
28
π=
9
28π= .
Application of Triple integration:
Volume: The elementary volume vδ is zyx δδδ ,, and therefore the volume of
the whole solid is obtained by evaluating the triple integral.
∫∫∫== dxdydzVvolume
Example: 1. Find the volume of the tetrahedron bounded by the planes x=0,y=0,z=0 and
azyx =++ .
Solution: Here we have a solid which is bounded by x=0,y=0,z=0 and
azyx =++ planes.
The limits of z are 0 to yxaz −−= , the limits of y are 0 to xay −= , the
limits of x are 0 and ax= .
Therefore ∫ ∫ ∫− −−
==a xa yxa
dxdydzVVolume0 0 0
[ ]∫ ∫−
−−=
a xayxa
dxdyz0 0
0
( )∫ ∫−
−−=a xa
dxdyyxa0 0
∫−
−−=
a xa
dxy
xyay0 0
2
2
∫
−−−−−=
a
dxxa
xaxxaa0
2
2
)()()(
∫
+−=
a
dxx
axa
0
22
22
a
xaxxa
0
322
322
+−=
+−=
322
333 aaa
6
3a
=
2. Find the volume of the cylindrical column standing on the area common to
the parabolas xy =2, yx =2
and cut off by the surface 212 xyz −+= .
Solution: we have xy =2, yx =2
, 212 xyz −+= .
∫ ∫ ∫−+
=1
0
12
02
x
x
xy
dzdydxV
[ ]dyxydx
x
x
∫ ∫ −+=1
0 2
12
dxyxy
y
x
x 2
221
02
12
−+= ∫
dxxx
xxx
x
+−−−+= ∫
44
22
521
02
122
12
1
0
5532
72
2
3
5104
7
2
412
3
2
+−−−+=
xxxx
xx
+−−−+=
5
1
10
14
7
2
4
18
140
569=