Calculating Sphere Mass using Volume and Density - Homework Help Provided

[matt222]

Homework Statement

problem is attached

Homework Equations

The Attempt at a Solution

I confused with the question, to find the mass we have to find the volume of the sphere and then multiply it by the given density, what I did is we have the radius of the sphere is a and the sphere volume 4/3*pi*a^3, density as it specified c=1+rz,at this point I noticed that i am doing wrong thing, can you help me please to find it out

 

[Apphysicist]

Recall that the mass of a continuous object is given as int(dm) and dm=p*dV, in this case dm=c*dV . The nice thing about your problem is that it's a regular sphere with radius, a, centered at the origin. Start thinking about what your differential volume element for your sphere is...

 

[timthereaper]

If the center of the sphere is located on the origin, take the quarter that lies between the positive X and Y axes. Find the equation of a differential volume (f(x)*f(y)*dz), multiply it by your density function and integrate from -z to z. You should then get an equation for the mass of that quarter of the sphere. Multiply by 4 and you get your answer.

 

[matt222]

it should be dv=a^2sin(theata)dad(theata)d(phi), is it right

 

[lanedance]

or consider volume of top half (as z is positive, the total mass will be 2 times due to symmetry) in spherical coords
m = \int \int \int \rho(r, \theta, \phi) r sin(\theta) dr. d \theta. d \phi

i'll let you complete the limits and integrand, and the integration should straight forward

 

[matt222]

or consider volume of top half (as z is positive, the total mass will be 2 times due to symmetry) in spherical coords
m = \int \int \int \rho(r, \theta, \phi) r sin(\theta) dr. d \theta. d \phi

i'll let you complete the limits and integrand, and the integration should straight forward

i understad from you clearly, so when we take the half top on poistive z, the limit should goes from 0 to a for dr term, 0 to pi for theta and phi, is it true

one more thing in the density term in our question we have two terms one constant 1 and the other one is r*z, i confused now with rdr we have and z as well, constant 1 , so in this case we take two integration part or by letting the density c all the way and integrating without opening c

 

[HallsofIvy]

Yes.

(Note: I assume you are using "engineering" notation which swaps \theta and \phi from "mathematics" notation- \theta is the "co-latitude" and \phi is the "longitude".)

 

[matt222]

what about the density term i have two terms constant 1 and r*z, how we are going to relate this because we have now radius a , r any distance from the sphere z no idea about it,

 

[matt222]

is the final answer going to be 2*a^2*pi*c

 

[icystrike]

is the final answer going to be 2*a^2*pi*c

I'm looking at the "c" in ur answer

 

[matt222]

c is the density if you read the question attached

 

[icystrike]

but i suppose c isn't constant

 

[matt222]

i really confused with the term of density c=1+r*z

 

[vela]

it should be dv=a^2sin(theata)dad(theata)d(phi), is it right

No, the volume element depends on the radial coordinate r, not the radius of the sphere a. It should be

dv = r^2 \sin\theta \,dr\,d\theta\,d\phi

i understad from you clearly, so when we take the half top on poistive z, the limit should goes from 0 to a for dr term, 0 to pi for theta and phi, is it true

You have the correct limits for r, but not for the other variables.

what about the density term i have two terms constant 1 and r*z, how we are going to relate this because we have now radius a , r any distance from the sphere z no idea about it,

You just plug the expression for the density in. First, take the volume element dv and multiply by the density c to get dm:

dm = c\,dv = (1+r|z|)r^2\sin\theta \,dr\,d\theta\,d\phi

And then integrate:

M = \int dm = \int\int\int_0^a (1+r|z|)r^2\sin\theta \,dr\,d\theta\,d\phi

I'll leave it to you to determine the correct limits for the angular variables and to work out the integral itself. You'll have to express z in terms of the integration variables.

 

[matt222]

the limit for theta is from 0 to pi, for phi from 0 to 2*pi, z=rcos(theta), is it true the limits and the value for z

 

[vela]

Yes, that's correct. To actually evaluate the integral, you may want to use lanedance's suggestion to integrate over just the top half of the sphere and then double the result.

 

[matt222]

the final answer will be 4*pi*a^3/3, is it right

 

[matt222]

Yes, that's correct. To actually evaluate the integral, you may want to use lanedance's suggestion to integrate over just the top half of the sphere and then double the result.

i have problem with the limit of theta and phi, so from landance's said, theta if its top it will goes from 0 to pi/2, and for phi from 0 to pi, then integrating whole and multiply by 2, but can you check my intgeral limits are true or wrong

 

[vela]

the final answer will be 4*pi*a^3/3, is it right

No, the correct answer is

M = \frac{4}{3}\pi a^3 + \frac{2}{5}\pi a^5

i have problem with the limit of theta and phi, so from landance's said, theta if its top it will goes from 0 to pi/2, and for phi from 0 to pi, then integrating whole and multiply by 2, but can you check my intgeral limits are true or wrong

Your limits for θ are correct, but your limits for ϕ are not. ϕ should still go from 0 to 2π.

 

[matt222]

i got it ..,many thanks