I really appreciate the advice! The last time I computed double integrals was in Calc 3 a couple years ago. I now see that using integrals in practice is slightly different than when computing them in a class. I understand the basic concepts, but am rusty on the details.
Thanks alot!
One question I have for you though...How did you come up with the 11.8%? I'm also curious where f(theta) = sin(theta) comes from in this formula?
Thanks for your input.
Ok...I think I see how to handle this.
What I did was computed the dbl integral above symbolically to get,
\int _ {\phi = 0}^{2\pi}d\phi \int _ {\theta = 0}^{\pi}sin \theta d\theta = -(phi-2*pi)*(cos(theta)+1) = 4\pi
Now I can describe phi and theta in terms of the offset distance...
Thank you for your response.
Percentage = 100_cm^2/(11.9_cm^2) = 8.40 = 840%
On the other hand, if I flip the ratio,
Percentage = 11.9_cm^2/100_cm^2 = 11.9%
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