/sci/ - Science & Math

[Anonymous No.6370027]

I'm retarded and I forgot how to but,

How do you get a circle area with integrals?

How do you get the area of a sphere with integrals?

I want to remember before its too late

[Anonymous No.6370055]

<div class="math">A circle is x^2 + y^2 = 1</div>
<div class="math">y^2 = 1 - x^2</div>
<div class="math">y = \pm\sqrt{1 - x^2}</div>
<div class="math">2\cdot\int\limits_{-1}^{1}\sqrt{1 - x^2}\Delta x</div>

[Anonymous No.6370064]

Oh, you wanted sphere too.
Well, if you know the area of a circle is <span class="math">\pi r^2[/spoiler] then you can express a sphere as the sum of circles from z = -1 to z = 1 for <span class="math">r^2 + z^2 = 1[/spoiler]
<div class="math">Area = \int\limits_{-1}^{1}\pi(1 - z^2)\Delta z</div>
I think that should work.

[Anonymous No.6370067]

>circle area with integrals?

circle length = 2 * pi * r

integrate with respect to r => pi * r ^ 2

similar for sphere

[Anonymous No.6370077]

Consider the area of an infinitessimally small area, dA at radius r. The length is simply dr, and the width is r*dtheta.

dA = r*dr*dtheta

integrate r from 0 to R and theta from 0 to 2 pi.

For a sphere in coordinates, do the same thing, except you can either express r as a function of z (or vice versa) and set up the integral, or do it in terms of phi.

[Anonymous No.6370080]

>>6370077

Expounding upon this,

Consider a sphere.

dA = r*dr*dtheta
dV = dA * height

height is something like r*sin phi (draw this out)

dV = r^2 sin(phi) dtheta

integrate, V = R^3/3 * 2 * 2pi

The two is b/c phi goes from -pi/2 to pi/2