Surface area of a sphere with calculus and integrals

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To find the surface area of a sphere using integrals, the initial approach involves rotating a semicircle defined by the equation y = √(15² - x²) around the x-axis. The surface area is calculated by integrating the expression dA = 2πy dx, leading to the integral ∫[0,15](2π√(15² - x²))dx, which accounts for only half the sphere. The mistake identified is treating the surface area of the cylinder as parallel to the x-axis, neglecting the slanted nature of the sphere's surface. A more accurate method involves approximating the sphere's surface with truncated cones or using polar coordinates for better accuracy. The discussion emphasizes the importance of correctly accounting for the geometry of the sphere in the calculations.
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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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.
 
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?
 
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?
 
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.
 
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
 
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.
 
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?
 
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.
 

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