Surface area of a shifted sphere in spherical coordinates

[MilkyWay2020]

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.

 

[LCKurtz]

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?

 

[Haborix]

You need to recalculate the metric for the new parametrization. Your form of the integral implicitly assumes ##\rho=\text{constant}##.

 

[MilkyWay2020]

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.