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x
y
z
P(,, )
is the distance of P from the origin (O)
4.5 The Triple Integral in Spherical Coordinates
O
is the measure of the angle which makes with the + side of the z-axis
OP
Q
is the measure of the angle which makes with the + side of the x-axis
OQπθ 0
πφ 20 0
Example Plot the following spherical points.
243 a.
,,P
422 b.
,,Q
x
y
z
3
2
P
x
y
z
4
2
Q
4
r
θsin
z
θcos
θsinr
θcosz
cosrx θsin cos
sinry θsin sin
22 zr
222 zyx
z
r
x
y
z
P(,, )
O
Q
P(x, y, z)
4
θcosz
cosθsinx
sinθsiny
2222 zyx
Example In spherical coordinates, an equation of the paraboloid given by
22 yxz
5
is 22 sinsincossinθcos
222222 sinsincossinθcos
2222 sincossinθcos
cosθsinx sinθsiny
θcosz
22 sinθcos 02 θcossin 0or 0 2 θcossin
2
sin
θcos
6
Example In spherical coordinates, an equation of the sphere given by
9222 zyx
6
is
9222 θcossinsincossin
cosθsinx sinθsiny
θcosz
922222222 θcossinsincossin
92222 θcossin
9222 θcossin
92
3 .3
7
x
y
z
Consider a solid S.
Subdivide S into n sub-solids by first drawing planes through the z-axis.
In doing so, we are actually forming a partition of the interval for .)
i
88
x
y
z
We then draw spheres centered at the origin,
In doing so, we are actually forming a partition of the interval for .)
i
99
x
y
z
We then draw circular cones with vertex at the origin and having the z-axis as axis of each cone.
In doing so, we are actually forming a partition of the interval for .)
i
x
y
z
Let d, d and d be increments of the coordinates , and .
These increments determine an element of volume, three of whose edges are of lengths d, d and sin d.
d
d
sin d
dsindddV dddsindV 2
solution:
Example Find the volume of a spherical solid of radius r.
An equation of a sphere of radius r is
.ror z
x
y
zThe volume of the spherical solid is 8 times the volume of the solid in the first octant.
2222 rzyx
20
2
0
rzyx 222
dddsindV 2
dddsinVr
8 22
0
0
2
0
dddsinr
8 2
0
0
22
0
ddsinr
3
8 2
0 0
32
0
ddsinr
3
82
0
2
0
3
dφθcos-r / 2
02
0
3
3
8
dφr
3
82
0
3
23
8 3
runits. cubic
3
4 3r
Example Let S be the solid bounded by the graph in the first octant of .zyx 4222
Evaluate
z
x
y
z
solution:
20
2
0
2222 zyx
.22 yx
dV
S
dddsindV 2
dddsinV 22
0
2
0
2
0
cosθsinx sinθsiny
θcosz
2222 sinθsincosθsinyx
222222 sinθsincosθsin
2222 sincosθsin θsin22
θsin
dddsinV 22
0
2
0
2
0 θsinyx 22
S yx
dV22
dddsin 22
0
2
0
2
0 θsin
1
ddd 2
0
2
0
2
0
dd 2
2
0
22
0
2
0
dd 2 2
0
2
0
d 2
2 2
0 2
0
d .22
2
Example Let S be the solid inside the graphs in the first octant of .yxzzyx 22222 and 4
z
x
y
zsolution:
Set-up the iterated integral which gives the value of
42222 yxyx
422 22 yx
222 yx
222 yxz
.32 dVzyxS
2
2
4
40
2
0 2222 zyx
dddsinV 24
0
2
0
2
0
cosθsinx sinθsiny
θcosz
dddsin24
0
2
0
2
0 θcossinθsincosθsin 32
.32 dVzyxS
Exercise Evaluate the following.
3 πa.Ans
15
122
.Ans
dddsincosa 24
0
2
0
2
0 a.
dzdydxzy yx
yx
21
0
1
0
2
2 22
22 b.
2
a.Ans
222
0
0
0
22 222
c.zyx
dzdydxa xa yxa
