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

In summary, the first thought that occurred 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?
  • #1
schattenjaeger
178
0
So the first thought that occurred 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?
 
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  • #2
can you show your work please.
 
  • #3
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?
 
  • #4
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!
 
Last edited:
  • #5
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.
 

1. How do you find the surface area of a sphere with a given radius?

To find the surface area of a sphere with a given radius, you can use the formula A=4πr², where A represents the surface area and r represents the radius of the sphere. Simply plug in the value of the radius into the formula and solve for A.

2. Why is the formula for the surface area of a sphere A=4πr²?

The formula A=4πr² comes from the fact that the surface area of a sphere is equal to the sum of all the infinitesimal areas of the curved surface. This can be represented mathematically as A=∫∫ds, where A represents the surface area, s represents the infinitesimal area, and the integral is taken over the entire surface. Using calculus, this integral can be simplified to A=4πr².

3. Can you use the same formula to find the surface area of a hemisphere?

Yes, you can use the same formula A=4πr² to find the surface area of a hemisphere. This is because a hemisphere is essentially half of a sphere, so the same formula still applies. However, if you only want to find the curved surface area of the hemisphere, you can use the formula A=2πr² instead.

4. What units should the radius be in for the surface area formula to work?

The radius should be in the same units as the surface area for the formula A=4πr² to work. For example, if the surface area is measured in square centimeters, the radius should be in centimeters as well. It is important to keep the units consistent in order to get an accurate result.

5. Is there an alternative formula for finding the surface area of a sphere?

Yes, there is another formula for finding the surface area of a sphere, which is A=4πr². This formula is derived from the volume formula for a sphere, which is V=(4/3)πr³. By taking the derivative of this volume formula with respect to the radius, we get the surface area formula A=4πr². Both formulas will give the same result for the surface area of a sphere with a given radius.

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