How can the surface area of a sphere be derived using integration of circles?

[meemoe_uk]

Homework Statement

Derive the formula for surface area of a sphere using integration of circles

Homework Equations

Need to get : S = 4πr²

The Attempt at a Solution

Consider a sphere of radius r centred on the origin of a 3D space. Let y be an axis thru the origin. The sphere can be sliced into a row of circles with the y-axis at their centres. Consider one of these circles. The circumference c of the circle depends on it centre's distance from the origin such that
c = 2πr cos θ , where θ is the angle from the origin between the (x,z) plane to any point on the circle's circumference. r cosθ is the radius of the circle.

For -r to r, y= r sinθ
The Integration the circles circumferences along y from -r to r is
2πr cos θ ∫dy
Since y = r sinθ, then dy / dθ = r cosθ , so dy = r cos θ dθ
so the integration formula is
2πr² ∫ (cos θ)² dθ

I don't get the right formula from that integral.
There's something wrong with my setting up the integral. I can't see what though. Please give me a hint or two.

 

[vela]

Homework Statement

Derive the formula for surface area of a sphere using integration of circles

Homework Equations

Need to get : S = 4πr²

The Attempt at a Solution

Consider a sphere of radius r centred on the origin of a 3D space. Let y be an axis thru the origin. The sphere can be sliced into a row of circles with the y-axis at their centres. Consider one of these circles. The circumference c of the circle depends on it centre's distance from the origin such that
c = 2πr cos θ , where θ is the angle from the origin between the (x,z) plane to any point on the circle's circumference. r cosθ is the radius of the circle.

For -r to r, y= r sinθ
The Integration the circles circumferences along y from -r to r is
2πr cos θ ∫dy

The problem is in this last step. You're using $dA = 2\pi r \cos \theta\,dy$, but that's only good when $\theta \approx 0$, where a strip of the sphere looks like a piece of a cylinder. As you near the top and bottom of the sphere, the strip becomes flatter and flatter. Its area is no longer well approximated by that of cylinder of radius r cos θ and height dy.

Since y = r sinθ, then dy / dθ = r cosθ , so dy = r cos θ dθ
so the integration formula is
2πr² ∫ (cos θ)² dθ

I don't get the right formula from that integral.
There's something wrong with my setting up the integral. I can't see what though. Please give me a hint or two.

 

[meemoe_uk]

thks. Looks like I was hoping the circles would sort themselves out. Now I think I need to start by approximating a slice of the sphere as a frustum. The lower limit ( no min element ) of frustum of height h is a circle, and an array of frustums can approximate a sphere.

 

[Curious3141]

Homework Statement

Derive the formula for surface area of a sphere using integration of circles

Homework Equations

Need to get : S = 4πr²

The Attempt at a Solution

Consider a sphere of radius r centred on the origin of a 3D space. Let y be an axis thru the origin. The sphere can be sliced into a row of circles with the y-axis at their centres. Consider one of these circles. The circumference c of the circle depends on it centre's distance from the origin such that
c = 2πr cos θ , where θ is the angle from the origin between the (x,z) plane to any point on the circle's circumference. r cosθ is the radius of the circle.

For -r to r, y= r sinθ
The Integration the circles circumferences along y from -r to r is
2πr cos θ ∫dy
Since y = r sinθ, then dy / dθ = r cosθ , so dy = r cos θ dθ
so the integration formula is
2πr² ∫ (cos θ)² dθ

I don't get the right formula from that integral.
There's something wrong with my setting up the integral. I can't see what though. Please give me a hint or two.

If I understand your attempt correctly, you're basically summing the surface area of open cylindrical elements with radius $r\cos\theta$ and an infinitesimal height.

Unfortunately, the element you used for the height is not correct. I suggest you use the arc length along the sphere given by $\mathrm{d s} = r\mathrm{d \theta}$ as a better approximation for the height of the cylindrical element.