4.5 The Triple Integral in Spherical Coordinatesz

P(V,U, J) y U V Ox y

V is the distance of P from the origin (O) U is the measure of the angle which OP makes with the + side of the z-axis J is the measure of the angle which O makes with the + side of the x-axis1

J

V

y Q

0e e0e e2 0eV

Example Plot the following spherical points. T T a. P 3, , 4 2z

T T b. Q 2 , , 2 4z

U!

T 4

yP3 y x

T U! 2y

yy yQ

x

T J! 2

T J! 4

z

P(x, y, z) P(V,U, J) y U V

r sin ! Vr

Ox

y

J

V

y Q

r ! V sin z cos ! V z ! V cos

z U V

V ! r z2

2

x ! r cosJ ! V sin cosJ y ! r sinJ ! V sin sinJ

V ! x y z

2

2

2

x ! V si y ! V siz ! V cos

cosJ si J

x y z !V2 2 2

2

4

Example In spherical coordinates, an equation of the paraboloid given by 2 2 z!x y is 2 V cos ! V si U cos J V si U si2 2 2 2 2

x ! V sin cos J y ! V sin sinJ z ! V cosJ2 2

V cos ! V si U cos J V si U si J V cos ! V 2 sin2 U cos2 J sin2 J V cos ! V 2 sin2 U V V si 2 U cos ! 0

2

V ! 0 or V si U cos ! 0cos V ! sin2 U5

Example In spherical coordinates, an equation of the sphere given by 2 2 2 x y z !9 is

x ! V si cos J y ! V si si J z ! V cos2

V si

U cos J V si U si J V cos2

2

!9

V 2 si 2 U cos2 J V 2 si 2 U si 2 J V 2 cos2 ! 9 V 2 sin2 U V 2 cos2 ! 9 V sin U cos V2 ! 9 V ! s3 V ! 3.6

2

2

2

! 9

Consider a solid S.z

Subdivide S into n sub-solids by first drawing planes through the z-axis.y x

( iJ

In doing so, we are actually forming a partition of the interval for J.)7

z

(i Vy x

We then draw spheres centered at the origin,

In doing so, we are actually forming a partition of the interval for V.)8

z

( iU

y x

We then draw circular cones with vertex at the origin and having the zaxis as axis of each cone.

In doing so, we are actually forming a partition of the interval for U.)9

z

These increments determine an element of volume, three of whose edges are of lengths dV, VdU and V sinU dJ. dVy

x

VdU V sinU dJ

Let dV, dU and dJ be increments of the coordinates V, U and J .

dV ! dV VdU V sin UdJ 2 dV ! V sin U dV dU dJ

Example Find the volume of a spherical solid of radius r. solution: An equation of a sphere of radius r is

x y z !r z2 2 2

2

or V ! r .

z

The volume of the spherical solid is 8 times the volume of the solid in the first octant.y

x

T 0 eJ e 22 2

T 0 eU e 22

V ! x y z !r

dV ! V sin U dV dU dJV ! 8T 2 0 T 2 0 T 2 0 3

2

T 2 0 T 2 0

T 2 0

r 0

V 2 sin U dV dU dJ

! 8

r V 2 sin U dV dU dJ 0 V sin U dU dJ 3 0T 2 0 3 r

! 8

8r ! 3 T 3 8r 2 ! 0 d 3

T 2 0

8r T/2 sin UdU dJ ! - cos 0 d 3 3 8r 3 T 4Tr ! ! cubic units. 3 2 3

3

T 2 0

Example Let S be the solid bounded by the graph in the first octant of x 2 y 2 z 2 ! 4. Evaluate

S

dV x2 y 2

.

solution:z z

T 0eJ e 2y x

T 0 eU e 2

V ! x2 y2 z2 ! 2

dV ! V sin U dV dU dJ

2

x ! V sin cosJ y ! V sin sinJ z ! V cos

V !2

T 2 02

T 2 0

2 0

V sin U dV dU dJcos J V sin sinJ 2 2 2

2

x y !

V sin2

! V sin ! V sin ! V sin2

cos J V sin2 2

2

2

2

sin J

2

2

cos J sin J !

V sin

2

2

V !

T 2 0

T 2 0

2 0

V 2 sin U dV dU dJT 2 0 T 2 0 T 2 0

x 2 y 2 ! V sin

S

dV x2 y2

! ! !

T 2 0

T 2 0

2 0

1 V sin

V 2 sin U dV dU dJ

T 2 0 T 2 0

2 0 2

V dV dU dJ2

V 2

dU dJ0

! 2 ! 2

T 2 0 T 2 0

dU dJ

T T T T2 d J ! T 2 dJ ! T ! . 0 2 2 2

Example Let S be the solid inside the graphs in the first octant of x 2 y 2 z 2 ! 4 and z ! x 2 y 2 . Set-up the iterated integral which gives the value of

2 x 3 y z dV .S z solution:z

x2 y2 x2 y2 ! 4 2x 2 y ! 4 x y !22 2 2 2

y x

z!

x y ! 2

2

2

2 2 T J! 4T 0 eU e 22 0

x ! V sin cosJ y ! V sin sinJ z ! V cosV ! x y z !22 2 2

T 0 eJ e 4 V !T 4 0

T 2 0

V 2 sin U dV dU dJ

2 x 3 y z dV .S

!

T 4 0

2V sin0

T 2 0

2

cos J 3V sin sinJ V cos V sin UdVdUdJ2

Exercise Evaluate the following.

a.

0 1

T 4 0

2 a cos J 0

2T 0

V sin JdUdVdJAns . a3

2

b.

1 y 2 0

2 x 2 y 2 x y2 2

z dzdydx dzdydx 2 2 2 x y zT 2 2 1 ns . 15

2

c.

0

a

a2 x2 0

a2 x2 y2 0

Ta Ans . 2