Circle to sphere integration

  • Thread starter meemoe_uk
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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

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.
 
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Answers and Replies

  • #2
vela
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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

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.
 
  • #3
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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.
 
  • #4
Curious3141
Homework Helper
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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

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.
 

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