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420chan is Getting Overhauled - Changelog/Bug Report/Request Thread (Updated July 10)
haalp Ignore Report Reply
Angus Worthingshaw - Fri, 05 Dec 2014 16:44:51 EST ID:wdBv6x7g No.14503
File: 1417815891494.jpg -(55091B / 53.80KB, 500x471) Thumbnail displayed, click image for full size. 55091
i need to find the area of the polar curve 2(a)sin(theta)

i keep getting 2(pi)(a)^2 but the book says its just (pi)(a)^2

i used the bounds 0 to 2pi which could be wrong but i dont see why.
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Angus Worthingshaw - Fri, 05 Dec 2014 16:48:52 EST ID:wdBv6x7g No.14504 Ignore Report Reply
>>14503
ugh fuck im an idiot guys ill the bounds are 0 to pi for this curve
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Nigel Depperham - Sat, 06 Dec 2014 00:41:32 EST ID:XpKKr8Wn No.14505 Ignore Report Reply
the fuck is (a) supposed to be? what are you integrating with respect to?
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Betsy Drocklekudging - Sat, 06 Dec 2014 23:47:00 EST ID:qz3c7Bt+ No.14506 Ignore Report Reply
1417927620903.gif -(3740B / 3.65KB, 338x155) Thumbnail displayed, click image for full size.
DICKS EVERYWHERE
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Phineas Mellyshit - Wed, 17 Dec 2014 04:30:13 EST ID:xatex5Q1 No.14536 Ignore Report Reply
The shape you are finding the area of is under the polar curve. You only integrate over the interval of theta that it takes the polar curve to draw the shape. After that its just tracing itself.


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