Surface area of a sphere, I know the value is 4*pi*R^2, can some one show how to find this by Calculus ?

Question

Please explain the idea for the integral, I will try to work out the integral myself.

Thanks

Answer 1

Think of the boundary of the sphere as a collection of rings, with radius varying from 0 to R and then back to 0. So focus on half of the sphere, and double the answer.

Assume the sphere is centered at the origin, and you're viewing it so that you're looking at the xy-plane. To compute the surface area of the right half-sphere, for each value of x between 0 and R, you have a ring of some radius p(x). You know how to compute its circumference, given the radius, so that is what you're integrating: 2*pi*p(x), from 0 to R. The question is, what is p(x)? When x=0, p(x) should be R, and when x=R, p(x) should be 0. Use the Pythagorean theorem to figure out what p(x) should be.

Keep in mind that you need to take the curvature of the sphere, so the thickness of each ring is not dx, it's actually somewhat greater. Using right-triangle trig it can be figured out.

Answer 2

Consider the sphere, it can be subdivided up into an infinite number of 2 dimensional circles. We also know that a cirlce (in a cartesian coordinate system) is represented by x^2 + y^2 = r^2,
therefore we can say that x = sqrt(r^2 - y^2)

Now, in this case the surface area of the solid of revolution is defined as:
2pi times the integral from a to b of (x)sqrt(1-dx/dy)^2 dx. Since we are rotating around the y-axis. If you are unfamiliar with this expression lookup Arc Lenght in a Calculus textbook, there you'll be able to see how this is derived.

So in this case, set the interval of you intergral to be from [-a, a] (notice that a is equal to the radius).

from here we get that the surface area is equal too:

2pi times the integral from -a to a of (x)sqrt(1-(d sqrt(r^2 - y^2)/dy)^2 dx.

dx/dy is equal too -2y/2sqrt(r^2 - y^2) which is equal too:
-y/sqrt(r^2 - y^2).

Now we have to square this expression, giving y^2/(r^2 - y^2).

Remember the expression we are using for the surface area says 1-(dx/dy)^2 so...

1 - y^2/(r^2 - y^2)

which is equal too:

r^2/(r^2 - y^2)

Now we have to take the square root, giving us:

r/sqrt(r^2 - y^2). Now we plug that into our expression for surface area, getting:

(x) (r/sqrt(r^2 - y^2) Now remember that we stated at the beginning that x = sqrt(r^2 - y^2), so we replace x in the expression above with this and get r.

So we can plug that back into our integral and get:

2pi times the integral from -a to a of r dx.

Which is equal to:

4pi times the integral from 0 to a of r dx.

Now we take the integral and get:

4pi(r)(a). Now remember that we noted that in our interval -a to a, a represented the radius...so we get:

4pi(r)(r).

Therefore, the surface area of a sphere is equal to 4pi(r^2).

Alternatively, we could us parametric equations,

A sphere can be created by rotating a semicircle, defined by:

x = r cos t
y = r sin t
where 0 less than or equal to t which is less than or equal to pi.
around the x-axis. (notice that the formula we have to use in this case is not the same as the one above, since we have to modify it to use parametric equations). If you aren't familiar with this, you can look it up in a calculus textbook probably under parametric curves. Anyway, we get:

S = 2pi times the integral from o to pi of (r sin(t)) (sqrt((-r sin(t))^2 + (r cos(t))^2) dt

which is equal too:

2pi times the integral from 0 to pi of (r sin(t)) (sqrt(r^2 (sin^2 (t) + cos^2 (t)) dt.

which becomes,

2pi times the integral from o to pi of r sin(t) (r) dt

which equals:

2pi r^2 times the integral from o to pi of sin (t) dt

which becomes,

2pi r^2 [(-cos(t)] from 0 to pi

which equals too

4pi(r^2)

This is might not be all that easy to understand without seeing it written out properly (i.e. with proper symbols instead of words and what not). It is also a lot easier to visualize if you make a drawing of the sphere. If you would like to see it written out properly, you could search around the net a bit, I'm sure you could find it all here http://mathworld.wolfram.com/. Or you can email me and I'll type it up properly.