Question

Volume of sphere triple integration

Answer 1

I'm asked to find the volume of a sphere using a triple integral in spherical and cylindrical coordinates. I get different answers for each coordinate system, and I can't figure out where my mistake is

Answer 2

@phi

Answer 3

The formula is 4/3 pi r^3
hopefully you get that result for one of the coord systems?

Answer 4

sorry about that this should help since they're vertical

Answer 5

I'm not

Answer 6

and it would be 4/3pir^3 for a sphere but this would be density since it's asking to evaluate the triiple integral \[\int\limits_{}^{}\int\limits_{}^{}\int\limits_{}^{}xyzdV\]

Answer 7

the big picture is you want to do
\[ \int \int \int dV \]
in cylindrical coords that is
\[ \int \int \int r\ dz \ dr \ d \theta \]

Answer 8

I've got that. I'm just not sure where my error in my work is. I can't find it, so I was hoping someone else could because I'm getting different answers for my integrals

Answer 9

oh, you don't want the volume, you want something else

Answer 10

yes I realize now I have the wrong title

Answer 11

I have x=rcostheta and y=rsintheta and z=z

Answer 12

since it is 1/8 of the sphere I have the radius is 0 to 2 and theta is 0 to pi/2 and then z is zero to sqrt(r^2-4)

Answer 13

when I evaluate that I get 4/3 but when I convert to spherical and evaluate it I get 4/5

Answer 14

I just did cyl and got 4/3
now to do the sph

Answer 15

btw, both of your pics look like cylindrical i.e. no work for the spherical

Answer 16

alright here we go sorry this should be my sphereical

Answer 17

what are you using for dV in spherical ?

Answer 18

\[\rho \sin \phi d \rho d \phi d \theta \]

Answer 19

it should be
\[ dV= \rho^2 \sin \phi\ d \rho \ d \phi \ d \theta\]

Answer 20

ok that's it thank you

Answer 21

Yes, it works out

Answer 22

lot of corrections to make now. thanks