Help w/ Seaborn's Mathematics for the Physical Sciences

[bcjochim07]

Homework Statement

In spherical polar coordinates, a vector F is given by F=(r*costheta*cosphi)rhat + (r*costheta*sinphi)thetahat + (r*sintheta*cosphi)phihat

Check the validity of the divergence theorem for this vector and the volume that is one octant of a sphere radius b.

Homework Equations

\ointF \cdot dA = \int_Vgrad\cdotFd³r

The Attempt at a Solution

The solution for this problem is in my book, and I am having difficulty following it, I think because it is in polar coordinates, and I am still trying to get used to them.

The author proceeds to divide the surface into four area elements:

spherical surface : dA = rhat(b^2)*sintheta*dtheta*dphi

before I go on to any of the other elements, could someone please help me understand how the author derived this expression? Thanks.

 

[tiny-tim]

spherical surface : dA = rhat(b^2)*sintheta*dtheta*dphi

before I go on to any of the other elements, could someone please help me understand how the author derived this expression? Thanks.

Hi bcjochim07!

Mark a "rectangle" on the surface, with change in latitude dθ and change in longitude dφ.

Then its width is bsinθ dθ, and its height is bdφ, so its area is b²sinθdθdφ.

 

[bcjochim07]

ok,

So I think I understand where the bdφ comes from; isn't that an arc length? Then I would think that the other dimension would be bdθ, so where does that sinθ come from?

Thanks.

 

[tiny-tim]

Just look at a globe of the Earth …

a circle of latitude has radius bsinθ

(while a circle of longitude has radius b).

 

[bcjochim07]

Thanks. Great explanation. Now I am trying to figure out how the book derives the other surfaces. For example, the area element in the xz plane is in the negative phi hat direction with dA = r dr dθ. Considering the area formula for a circular sector, I thought it should be (1/2)rdrdθ.

 

[bcjochim07]

Ok... I think I see it now. I just need to divide the area up into little "rectangles" just like I did for the spherical surface. Each of these rectangles has dimensions dr by rdθ.

 

[tiny-tim]

Ok... I think I see it now. I just need to divide the area up into little "rectangles" just like I did for the spherical surface. Each of these rectangles has dimensions dr by rdθ.

Yup! … it's all simple geometry …

just draw the right diagram, and it becomes obvious! ​