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Homework Statement
Say the sphere of radius "a" is made out of various rings with height R(x) and thickness dx. Adding up all of the rings will form a sphere, and in order to do that, I have to integrate.
Homework Equations
Trigonometric Substitution:
[tex]\frac{x}{a}=sin \theta[/tex]
[tex]dx=a cos \theta d\theta[/tex]
Function of ring height related to position in cartesian plane:
[tex]R(x)=\sqrt(a^2-x^2)[/tex]
The Attempt at a Solution
Set up the integral, I just want to make my life simpler and integrate half a circle:
[tex]A=\int_0^a 2\pi R(x) dx[/tex]
Substitute R(x):
[tex]\int_0^a 2\pi \sqrt(a^2-x^2) dx[/tex]
Now using trigonometric substitution and factoring out a:
[tex]\int_0^a 2\pi a \sqrt(1-sin^2 \theta) dx[/tex]
Using pythagorean trig identity and trig substituting for dx:
[tex]\int_0^a 2\pi a cos\theta (a cos\theta d\theta)[/tex]
[tex]\int_0^a 2\pi a^2 cos^2\theta d\theta[/tex]
Putting out all the constants, and integrating, using tables of integral, I get:
[tex]2\pi a^2\int_0^a cos^2\theta d\theta[/tex]
[tex]2\pi a^2\int_0^a cos^2\theta d\theta[/tex]
[tex]2\pi a^2 (\frac{\theta}{2}+\frac{sin(2\theta)}{4})|_0^a[/tex]
The problem is, how do I solve the final part of the definite integral? What I know is that the parenthesis should be equal to 1 because the surface area is 4 pi r^2 and I integrated half a sphere.