# Express the volume of V in Cartesian, cylindrical and spherical coordinates.?

Let V be the region in R^3 inside the sphere x^2+y^2+z^2=1 and above te plane z=0.

Express the volume of V in Cartesian, cylindrical and spherical coordinates.

I did some work on this but not sure whether they are right. Can anybody help me with this one?

Thanks!

### 2 Answers

- JLv 77 years agoBest Answer
Cartesian

V= INT dV = INT dz dydx

limits

0<z<sqrt (1- (x^2+y^2))

Since the Region is at z=0 , then it is x^2+y^2 =1 , so

-sqrt (1-x^2)<y<+sqrt (1-x^2)

-1<x<1

You can use dxdy also -sqrt (1-y^2) <x< +sqrt (1-y^2) and -1<y<+1

Cylindricals

x=rcosT

y=rsinT

z=z

The sphere is ( plug x and y ) r^2+z^2 =1 , z= sqrt ( 1-r^2)

Note that we are using r , not R=1 , because we are moving inside the Region x^2+y^2=1

So , V= INT_V dz dA

Over the Region , dA = rdrdT ( Polar coordinates )

V= INT _R z rdrdT

0<z<sqrt ( 1-r^2)

0<r<R=1

0<T<2pi

Solving , just in case , V=INT INT rsqrt (1-r^2) drdT

u=1-r^2

du= -2r dr

V= INT INT u^(1/2) (-1/2) du dT

V= INT (2/3) u^(3/2) (-1/2) dT

V= INT (-1/3) (1-r^2) ^(3/2) dT

V= INT (-1/3) ( 0- 1) dT herical

V= INT (1/3) dT

V= 2pi/3

Spherical

x=rsinZcosT

y=rsinZsinT

z=rcosZ

The sphere ( Plug x, y,z ) is r=1

dV= r^2sinZ dZdrdT

V= INT_V dV = INT INT INT r^2sinZ drdZdT

Limits

0<r<1

0<Z<pi/2

0<T<2pi

V= INT INT r^3/3 sinZ dZdT

V= (1/3) INT INT sinZ dZdT

V= (1/3) (-cosZ) INT dT

V= (1/3) (-) (0-1) INT dT

V= (1/3) *2pi

V= 2pi/3

God bless you

- Let'squestionLv 77 years ago
Spherical: very simple r<1 and θ = 0 to π/2 and φ = 0 to 2π

Cylindrical:-1<r<1; 0<z<1 and θ = 0 to 2π