So I had a problem asking to find the surface area of a sphere with radius=a

schattenjaeger

So the first thought that occured to me was to use a surface integral

I got 2pi*a^2, which is half the SA, if I used the surface integral is it possible or likely that I just found half the SA and can then multiply by 2? Or something. Actually, I think I see what I did, for my sec(gamma) I used the sphere and am pretty sure I did that stuff right, then(I converted to polar coordinates)I did r from 0-a, and theta from 0-2pi, which would be the circular region ON the xy plane, and would give me half the SA, right? Maybe? Please?

heh, but the big problem is the next part, find the centroid of the curved surface area of a hemisphere

hur?

Cyrus

can you show your work please.

schattenjaeger

I don't even know where to start

I never was good with spherical coordinates...

so for one of the CM points, it'd be 1/M * the triple integral of the density over half the sphere? The problem doesn't mention density, do I assume it's constant?

I also need to find M, so the triple integral of p(which again I guess I'll assume is constant, p is density, btw)times r^2sin(phi) drd(phi)d(theta), and the limits of integration being 0-a for r, and 0-2pi for phi and theta? so 2pi*p?

Cyrus

Here is how you calculate the surface area, I will just copy the solution from my textbook by Stewart. If it does not help, I can try to explain it further.

Example 1: Find the surface area of a sphere of radius a:

[tex] x=asin( \phi)cos ( \theta) [/tex]

[tex] y=asin( \phi)sin ( \theta) [/tex]

[tex] z =acos( \phi) [/tex]

where the parameter domain is:

[tex] D = { ( \phi, \theta) | 0 \leq \phi \leq \pi , 0 \leq \theta \leq 2 \pi } [/tex]

we first compute the cross product of the tangent vectors:

[tex] r_\phi \times r_\theta = \left(\begin{array}{ccc} \hat{i}& \hat{j} & \hat{k} \\ \frac{ \delta x}{ \delta \phi} &\frac{ \delta y}{ \delta \phi} & \frac{ \delta z}{ \delta \phi} \\ \frac{ \delta x}{ \delta \theta} & \frac{ \delta y}{ \delta \theta} & \frac{ \delta z}{ \delta \theta} \end{array}\right) = \left(\begin{array}{ccc} \hat{i}& \hat{j} & \hat{k} \\ acos( \phi) cos( \theta) & acos( \phi)sin( \theta) & -a sin ( \phi) \\ -a sin( \phi) sin ( \theta) & a sin ( \phi) cos ( \theta) & 0 \end{array}\right)
[/tex]


[tex] = a^2sin^2( \phi) cos ( \theta) \hat{i} + a^2 sin^2 ( \phi) sin ( \theta) \hat{j} + a^2 sin ( \phi) cos ( \phi) \hat{k} [/tex]

Thus:

[tex] | r_\phi \times r_\theta | = \sqrt{a^4 sin^4 ( \phi)cos^2( \theta) + a^4 sin^4( \phi) sin^2 ( \theta) + a^4 sin^2 ( \phi) cos^2 (\phi)} [/tex]

[tex] = \sqrt{ a^4sin^4 ( \phi) + a^4 sin^2 ( \phi)cos^2 ( \phi)} =a^2 \sqrt{sin^2 ( \phi) } = a^2 sin ( \phi) [/tex]

since [tex] sin ( \phi) >= 0 [/tex] for [tex] 0 \leq \phi \leq \pi [/tex] Therefore, by definition 4, the area of the sphere is:

[tex] A = \int \int_D |r_\phi \times r_\theta| dA = \int^{2 \pi}_0 \int^{\pi}_0 a^2 sin( \phi) d \phi d \theta [/tex]

[tex] = a^2 \int^{2 \pi} _ 0 d \theta \int^ { \pi}_ 0 sin( \phi) d \phi = a^2(2 \pi)2 = 4 \pi a^2 [/tex]

LoL 100 edits later, and finally an end product!

uart

If you were asked to prove the forumla for the SA of a sphere (S = 4 pi r^2) then by far the easiest method I've found is to first prove that the volume of the sphere is V = 4/3 pi r^3. BTW, this is quite an easy volume of revolution problem with a resultant integration that is very elementary indeed.

Once you've estabished the formula for V it's not hard to justify that [tex]dV = S dr[/tex] and hence [tex]S = dV/dr[/tex] which establishes [tex]S = 4 \pi r^2[/tex] very easily.