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

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

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

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

dumbcow
 2 years ago
Best 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 \[Area = 2\int\limits_{1}^{1}\sqrt{1x^{2}} dx = \pi\]

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

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

dumbcow
 2 years ago
Best ResponseYou've already chosen the best response.0in 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

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

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