Surface area of a sphere with calculus and integrals

In summary, the conversation discusses using integrals to find the surface area of a sphere (r=15), with a focus on the mistake of using cylinders with a parallel surface rather than a slanted surface. The conversation also suggests using polar coordinates as an alternative approach. The conversation ends with a question about the software used to draw/upload the accompanying diagram.
  • #1
the_dane
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Homework Statement


How do I find the surface area of a sphere (r=15) with integrals.

Homework Equations


Surface area for cylinder and sphere A=4*pi*r2.

The Attempt at a Solution


I draw the graph for y=f(x)=√(152-x2). A circle for for positive y values which I rotate. I will create infinite many cylinders with the height dx and radius y. The surface area of those dA=2*pi*y*dx then. I know that √(152-x2) so ∫dA=∫[0,15](2*pi*√(152-x2))dx. It's only a half sphere for I multiplie by 2.

If I calculate the value by 4*pi*r2. What is my mistake?
 
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  • #2
the_dane said:

Homework Statement


How do I find the surface area of a sphere (r=15) with integrals.

Homework Equations


Surface area for cylinder and sphere A=4*pi*r2.

The Attempt at a Solution


I draw the graph for y=f(x)=√(152-x2). A circle for for positive y values which I rotate. I will create infinite many cylinders with the height dx and radius y. The surface area of those dA=2*pi*y*dx then. I know that √(152-x2) so ∫dA=∫[0,15](2*pi*√(152-x2))dx. It's only a half sphere for I multiplie by 2.

If I calculate the value by 4*pi*r2. What is my mistake?

Your mistake is that the when you take the surface of a cylinder of radius y and height dx (actually, width dx, since the x-axis is horizontal), you are taking the surface to be parallel to the x-axis. However, on an actual sphere between x and x + dx the true surface is slanted at an angle to the x-axis. For given values of y and dx the slanted surface area will be larger than that of the unslanted surface area.
 
  • #3
Ray Vickson said:
Your mistake is that the when you take the surface of a cylinder of radius y and height dx (actually, width dx, since the x-axis is horizontal), you are taking the surface to be parallel to the x-axis. However, on an actual sphere between x and x + dx the true surface is slanted at an angle to the x-axis. For given values of y and dx the slanted surface area will be larger than that of the unslanted surface area.
Hi. Can you illustrate with some equations?
 
  • #4
Forget about an infinite number of cylinders. Consider a series of ##n## very small cylinders, centred on the x axis, with radius f(x) and X dimension equal to ##\Delta_n \equiv\frac{30}{n}##.
Your approach is summing the areas of all those cylinders, and then taking the limit as ##n## increases to infinity.
The ##k##th cylinder lies between the planes ##x=x_{k-1}^n## and ##x=x_k^n## where ##x_k^n\equiv 30\frac{k}{n}##, and that cylinder has surface area ##2\pi f(x_{k-1}^n)\Delta_n ##.

What is the true shape of the sphere in between those two planes?
Can you think of an approximation to the shape of the sphere between those two planes for which you can still give a reasonably simple formula (although slightly more complex than the one above), but which is much closer to the true shape than the cylinder is?
 
  • #5
the_dane said:
Hi. Can you illustrate with some equations?

No, but you can easily draw a picture yourself. Alternatively, do a Google search on "surface area of sphere" or something similar, to find that it has already been done hundreds of times by numerous other people. For example, the article
http://math.oregonstate.edu/home/pr...usQuestStudyGuides/vcalc/surface/surface.html
has a derivation of the surface-area formula.
 
  • #6
the_dane said:
Hi. Can you illustrate with some equations?
To get the surface of the sphere, we slice it as shown in the figure, and approximate that shape with a truncated cone, the piece of the surface of the sphere with the area of the side of that truncated cone.
spheresurface.png
 
  • #7
Perhaps it would be beneficial if you attempted to derive the expression using polar coordinates instead, it should be pretty straight-forward figure out what ##ds## is by looking at ehild's diagram.
 
  • #8
ehild said:
To get the surface of the sphere, we slice it as shown in the figure, and approximate that shape with a truncated cone, the piece of the surface of the sphere with the area of the side of that truncated cone.
View attachment 93487

Nice picture. What package did you use do draw/upload it?
 
  • #9
Ray Vickson said:
Nice picture. What package did you use do draw/upload it?
I use Paint, included in Windows. And I just do Upload. Or I copy the picture and paste into the post.
 

FAQ: Surface area of a sphere with calculus and integrals

What is the formula for calculating the surface area of a sphere using calculus and integrals?

The formula for calculating the surface area of a sphere using calculus and integrals is:
S = ∫2πr√(1+(dy/dx)^2)dx
where r is the radius of the sphere and dy/dx is the derivative of the equation of the sphere.

How is this formula derived?

This formula is derived by using the concept of surface area as the limit of the sum of infinitely small areas. By dividing the surface of the sphere into infinitely small strips, we can use the integral to sum up all of these strips and calculate the total surface area.

Can this formula be used to find the surface area of any sphere?

Yes, this formula can be used to find the surface area of any sphere, regardless of its size or position in space.

Can this formula be used to find the surface area of a hemisphere?

Yes, this formula can also be used to find the surface area of a hemisphere. However, the limits of the integral will need to be adjusted to only include half of the sphere's surface.

Are there any other methods for calculating the surface area of a sphere?

Yes, there are other methods for calculating the surface area of a sphere, such as using the formula S = 4πr^2 or using surface area approximation techniques. However, the calculus and integral method is the most accurate and precise method for calculating the surface area of a sphere.

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