Surface area of a shifted sphere in spherical coordinates

In summary, this problem can be solved using the spherical coordinate system and the sphere equation.
  • #1
MilkyWay2020
2
1

Homework Statement



find the surface area of a sphere shifted R in the z direction using spherical coordinate system.

Homework Equations



$$S= \int\int \rho^2 sin(\theta) d\theta d\phi$$

$$x^2+y^2+(z-R)^2=R^2$$

The Attempt at a Solution



I tried to use the sphere equation mentioned above and solve for ρ, using the definition of ρ and z in spherical coordinate system, this gives me:

$$x^2+y^2+z^2+R^2-2 R z= R^2$$
$$\rho^2-2 R \rho cos(\theta)= 0$$
$$\rho= 2 R cos(\theta)$$

then substitute in the integral:

$$S= \int\int 4 R^2 cos(\theta)^2 sin(\theta) d\theta d\phi$$

and integrate with the following boundaries:

θ: 0 → π/2
Φ: 0 → 2π

this gives a wrong answer of:

$$\frac{8 \pi R^2}{3}$$

What am I missing? This is a very basic problem but I cannot put my hand on the mistake.
 
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  • #2
MilkyWay2020 said:

Homework Statement



find the surface area of a sphere shifted R in the z direction using spherical coordinate system.

Homework Equations



$$S= \int\int \rho^2 sin(\theta) d\theta d\phi$$

$$x^2+y^2+(z-R)^2=R^2$$

The Attempt at a Solution



I tried to use the sphere equation mentioned above and solve for ρ, using the definition of ρ and z in spherical coordinate system, this gives me:

$$x^2+y^2+z^2+R^2-2 R z= R^2$$
$$\rho^2-2 R \rho cos(\theta)= 0$$
$$\rho= 2 R cos(\theta)$$

then substitute in the integral:

$$S= \int\int 4 R^2 cos(\theta)^2 sin(\theta) d\theta d\phi$$
With your choice of variables shouldn't that be a ##\sin\phi## in the integral?
 
  • #3
You need to recalculate the metric for the new parametrization. Your form of the integral implicitly assumes ##\rho=\text{constant}##.
 
  • #4
Haborix said:
You need to recalculate the metric for the new parametrization. Your form of the integral implicitly assumes ##\rho=\text{constant}##.

I was afraid someone would post this specific answer. I had my doubts, but I thought the problem is too easy to get complicated. I think I was wrong about that.

Thank you, I am able to find the solution I am looking for.
 
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Likes berkeman

1. What is the formula for finding the surface area of a shifted sphere in spherical coordinates?

The formula for finding the surface area of a shifted sphere in spherical coordinates is A = 4πR2, where R is the radius of the sphere.

2. How do you convert Cartesian coordinates to spherical coordinates?

To convert Cartesian coordinates (x, y, z) to spherical coordinates (r, θ, φ), you can use the following equations:

r = √(x2 + y2 + z2)

θ = arctan(y/x)

φ = arccos(z/r)

3. Can the surface area of a shifted sphere in spherical coordinates be negative?

No, the surface area of a shifted sphere in spherical coordinates cannot be negative. It is always a positive value.

4. How does the surface area of a shifted sphere change as the radius increases?

The surface area of a shifted sphere increases as the radius increases. This is because the surface area is directly proportional to the square of the radius.

5. Is there a difference between the surface area of a shifted sphere in spherical coordinates and a regular sphere?

Yes, there is a difference between the surface area of a shifted sphere in spherical coordinates and a regular sphere. A shifted sphere has its center shifted from the origin, while a regular sphere has its center at the origin. This results in a difference in the coordinates and therefore, a difference in the surface area calculation.

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