Calculate the area of a sphere using an integral

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Homework Statement


I tried to calculate the area of a sphere using the function [tex]f(x) = \sqrt{1-x^2}[/tex]

Homework Equations



hmm?

The Attempt at a Solution


So I thought I could slice up the sphere in small cylinders, and then calculate the outer surface by circumference * height of all the cylinders and then add them together.

In this case, the radius would be equal to f(x) and the height of the cylinder would equal [tex]\Delta x[/tex]. So the surface of the whole sphere would be:

2pi \int f(x) dx with lower limit -1 and upper limit 1. However, this does not seem to equal 4pi*r^2, I checked this with r = 1. It gives two quite distinct values.

Where did I go wrong?
 

Answers and Replies

  • #2
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6,626

Homework Statement


I tried to calculate the area of a sphere using the function [tex]f(x) = \sqrt{1-x^2}[/tex]

Homework Equations



hmm?

The Attempt at a Solution


So I thought I could slice up the sphere in small cylinders, and then calculate the outer surface by circumference * height of all the cylinders and then add them together.

In this case, the radius would be equal to f(x) and the height of the cylinder would equal [tex]\Delta x[/tex]. So the surface of the whole sphere would be:

2pi \int f(x) dx with lower limit -1 and upper limit 1. However, this does not seem to equal 4pi*r^2, I checked this with r = 1. It gives two quite distinct values.

Where did I go wrong?

What sort of a knife could you use to slice a sphere into cylinders? When you slice something, you get flat surfaces along the cut.

The integral you arrived at is way off.
[tex]2\pi \int_{-1}^1 \sqrt{1 - x^2}dx[/tex]

The integral itself represents the area between the upper half circle of radius 1, centered at (0, 0). This area is [itex]\pi /2[/itex], so multiplying by [itex]2\pi [/itex] gives [itex]\pi ^2[/itex].

You started off with f(x) = sqrt(1 - x^2). If you revolve that curve around the y-axis, you get the upper half of a sphere of radius 1. You can slice this hemisphere into horizontal slices, and calculate the surface area of each slice, then integrate to get the surface area of the hemisphere. For the whole sphere, you'll need to double that result.
 
  • #3
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But won't you get really flat cylinders by approximation when you slice a hemisphere horizontally?

So using the formula S = 2pi \int y sqrt(1 + f'(x)^2) dx I can calculate the surface area of a sphere segment. Thus dS = 2pi y sqrt(1 + f'(x)^2). This is the surface area of a cylinder with radius = y and height = sqrt(1 + f'(x)^2), which is the change in the length of the arc for a small change in x. Does this formula not still use really flat cylinders to approximate the surface area of the sphere?

I tried the above formula on the graph of y = ax:

S = 2pi \int y sqrt(1 + a^2) dx
and
dS = 2pi y sqrt(1 + a^2)

What bothers me is that when x is increased by a very small amount, the increase in surface area is given as the surface area of a cylinder with radius y and height sqrt(1 + a^2), but the actual increase in surface area does not correspond with this at all because for a small increase in x a cone segment is added at the bottom of the cone and not a really flat cylinder. The difference in surface area of the cone segment and the cylinder becomes smaller as the increase in x becomes smaller (does this difference approach 0?), but this is counter-intuitive, as a staircase-like shape is used to approximate a straight line?

Thanks for the help = )
 
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  • #4
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6,626
But won't you get really flat cylinders by approximation when you slice a hemisphere horizontally?
None of the slices will be cylinders, since the top and bottom radii are different for each slice. Of course, the lower on the hemisphere the slice is, the closer the slice is to being a cylinder.
So using the formula S = 2pi \int y sqrt(1 + f'(x)^2) dx I can calculate the surface area of a sphere segment.
No, that's not quite right. You have y for the radius - it should be x.
Thus dS = 2pi y sqrt(1 + f'(x)^2).
See above. Also, you're missing the differential on the right side.
This is the surface area of a cylinder with radius = y and height = sqrt(1 + f'(x)^2), which is the change in the length of the arc for a small change in x. Does this formula not still use really flat cylinders to approximate the surface area of the sphere?
It effectively uses thin slices of a cone. That's where the sqrt(1 + (f'(x))^2)dx part comes in.
I tried the above formula on the graph of y = ax:

S = 2pi \int y sqrt(1 + a^2) dx
and
dS = 2pi y sqrt(1 + a^2)

What bothers me is that when x is increased by a very small amount, the increase in surface area is given as the surface area of a cylinder with radius y and height sqrt(1 + a^2), but the actual increase in surface area does not correspond with this at all because for a small increase in x a cone segment is added at the bottom of the cone and not a really flat cylinder. The difference in surface area of the cone segment and the cylinder becomes smaller as the increase in x becomes smaller (does this difference approach 0?), but this is counter-intuitive, as a staircase-like shape is used to approximate a straight line?
 
  • #5
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No, that's not quite right. You have y for the radius - it should be x.

But I revolved around the x-axis there, so the radius should be y?

The pdf file should be easier readable for you, if you don't mind.
 

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