please derive the area of a circle using calculus?

Answer 1

Here is a method that is often used for "solids of rotation," but also applies to this (simpler) situation:

Place the center of the circle at the origin, and consider the radius that lies along the x-axis, from 0 to r.

Think of the circle as being the sum of a series of rings of varying radius, each with negligible thickness (like a very thin letter O). The largest ring is r units in radius; the smallest ring is 0 units in radius (actually, it's a tiny, tiny circle). If you put them all together with their centers at the origin, their total area equals the area of the circle.

Now, we just write the integral that adds up the areas of the rings. The WIDTH of each ring is actually dx, and its area is its circumference (2 pi x) times its width, or 2 pi x dx.

To get the circle's area, you just integrate 2 pi x dx for all x values from the origin to the end of the radius:

Integral (from 0 to r) (2 pi x dx) = pi x^2 evaluated from 0 to 4
= pi r^2 - pi 0^2 = pi r^2

Answer 2

Implicit in the first 3 answers thus far is that your circle has a center at the origin (or pole) and the radius is fixed at R.

Further the equation of the circle is x^2 + y^2 = R^2
(which is symmetric abt the origin and the x-axis and y-axis).

Using polar coordinates and integrating {r dr }d(theta)between r=0 to R then the last integration from theta=0 and theta=pi/2 and multiplying by 4 (because you're only getting quarter of total circle area) is the easiest way.

In rectangular coordinates you can double integrate dxdy, which becomes (if first integration is with respect to y)

integral sq-rt [R^2 - x^2] dx where the integrand is managed by trig substitution: cos theta = x/R, etc.
(I can fill in further details if you really need them.)

Answer 3

Let y = sqrt(r^2 - x^2). As you know, this function forms a semicircle with radius r when bounded by the x-axis.

To find the area under the curve, we integrate it from -r to r (since -r and r are the points that intersect with the curve y = 0, or the x-axis). Therefore

A1 = Integral ([-r to r], sqrt (r^2 - x^2) dx)

This would get us the area of HALF of the circle; why not double the value to get the full circle?

A = 2 * Integral ([-r to r], sqrt (r^2 - x^2)dx)

Even better, let's use the fact that this graph is an even function. By symmetry, all we need to do is calculate the integral from 0 to r, and doubling the integral once more.

A = 2 * 2 * Integral ([0 to r], sqrt (r^2 - x^2)dx)
A = 4 * Integral ([0 to r], sqrt (r^2 - x^2)dx)

To solve this integral, we use trigonometric substitution.

Let x = rsin(theta) {It follows from here that sin(theta) = x/r}
dx = rcos(theta) (d-theta).

Since we're using substitution, we also substitute our upper and lower bounds.
If x = rsin(theta) and x = 0, then
0 = rsin(theta), 0 = sin(theta), and theta = arcsin(0) = 0.
If x = r, then r = rsin(theta), 1 = sin(theta), and theta = arcsin(1) = pi/2.

So now we have

A = 4 * Integral ([0, pi/2], sqrt {r^2 - r^2sin^2(theta)} rcos(theta)[dtheta]}

A = 4 * Integral ([0, pi/2], r * sqrt(1 - sin^2(theta)) rcos(theta)[dtheta])

A = 4 * Integral (0, pi/2] r^2 sqrt(cos^2(theta))cos(theta)[dtheta])

A = 4 * Integral (0, pi/2], r^2 cos^2(theta) [d-theta])

We can pull the constant r^2 out.

A = 4r^2 * Integral ([0, pi/2], cos^2(theta) [d-theta])

We solve this using the half-angle identity
cos^2(x) = [1 + cos(2x)]/2 = 1/2 + (1/2)cos(2x)

A = 4r^2 * Integral ([0, pi/2], 1/2 + (1/2)cos(2theta) [d-theta])

Which is actually easy to integrate.

A = 4r^2 * [theta/2 + sin(2theta)/4] {evaluated from 0 to pi/2}

A = 4r^2 * [ {(pi/2)/2 + sin(pi)/4} - {0 + 0} ]
A = 4r^2 * [ pi/4 + sin(pi)/4 ]
A = 4r^2 * [pi/4 + 0]
A = 4r^2 * [pi/4]
A = pi(r^2)

Answer 4

Are you allowed to use polar coordinates? This is easiest done by taking a double integral in polar coordinates:

The first integral goes from 0 to 2pi.
The second integral goes from 0 to r
And the thing you're integrating is r dr d(theta).

The inner integral gives you r^2/2
The outer integral multiplies that by 2pi
leaving you with pi*r^2, which is the area of a circle.

Answer 5

Here is another (albeit, quite odd) way. Notice that the area of a circle is the limit as n->infinity of the area of a regular n-gon. I will call the radius of the n-gon to be the distance from the center to the midpoint of a side (I do not know if that is technically called that). Let the radius be r. Now the area of the n-gon is n times the area of one of the n isoceles triangles (formed by two consecutive vertices of the n-gon and the center of the n-gon).

The area of one of these triangles is (r/2) times the length of a side of the n-gon, which can be found using the Law of Sines to be r*sin(2pi/n)/sin(pi/2 - pi/n). So the total area of the n-gon is (r^2)*(n/2)*sin(2pi/n)/sin(pi/2 - pi/n).

Now to take the limit of this as n->infinity, we first multiply and divide by pi to get:
(pi*r^2)*[sin(2pi/n)/(2pi/n)]*sin(pi/2 - pi/n)

Now since the limits of of both [sin(2pi/n)/(2pi/n)] (this is the same as limit as x->0 of sin(x)/x) and sin(pi/2 - pi/n) are both 1 as n->infinity. We can see that the limit of the area of a regular n-gon as n->infinity is pi*r^2. Since a circle is just the limit of a regular n-gon as n->infinity, the area of a circle must thus be pi*r^2.

Answer 6

I don't know about deriving the area, but I know that integrating the circumference, 2piR will give you the area, piR^2