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juantweaver Group TitleBest ResponseYou've already chosen the best response.0
do you have an equation and what coordinates are you in?
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
say x^2 + y^2 = 1
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
\[Area = 2 \pi \int\limits_{0}^{R}r dr\] where R is radius of circle
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
In doing so I am trying to find pi
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
haha prob not what you're looking for
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
\[x^{2} + y^{2} = 1\] \[y = \sqrt{1x^{2}}\] this is top half of circle...integrate from 1 to 1 \[Area = 2\int\limits_{1}^{1}\sqrt{1x^{2}} dx = \pi\]
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
Yes, but how do i integrate \[\sqrt{1x^2}\]
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
using trig substitution \[x = \sin u\] \[dx = \cos u\]
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
in the end you will just get pi =pi if you want to use this to get numerical approximation for pi, then integrate by computing area under curve using trapezoid rule or simpsons rule or something like that
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
Yes, but why? And how come we can do that?
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
why can we use trig sub? you can substitute anything you want to make the integral more manageable
 2 years ago
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