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juantweaverBest ResponseYou've already chosen the best response.0
do you have an equation and what coordinates are you in?
 one year ago

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

apple_piBest ResponseYou've already chosen the best response.0
In doing so I am trying to find pi
 one year ago

dumbcowBest ResponseYou've already chosen the best response.0
haha prob not what you're looking for
 one year ago

dumbcowBest 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 \[Area = 2\int\limits_{1}^{1}\sqrt{1x^{2}} dx = \pi\]
 one year ago

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

dumbcowBest ResponseYou've already chosen the best response.0
using trig substitution \[x = \sin u\] \[dx = \cos u\]
 one year ago

dumbcowBest 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
 one year ago

apple_piBest ResponseYou've already chosen the best response.0
Yes, but why? And how come we can do that?
 one year ago

dumbcowBest 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
 one year ago
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