anonymous
  • anonymous
Solve the double integral by first converting to polar coordinates (for no apparent reason)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1370935037671:dw|
dan815
  • dan815
loll
dan815
  • dan815
its the upper half of this circle

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More answers

whpalmer4
  • whpalmer4
Maybe do it while standing on your head, too :-)
dan815
  • dan815
|dw:1370935521183:dw|
anonymous
  • anonymous
How do you figure?
anonymous
  • anonymous
|dw:1370936303132:dw|
anonymous
  • anonymous
I can see that but I can't see how to convert the limits
anonymous
  • anonymous
|dw:1370936354561:dw|
dan815
  • dan815
|dw:1370936313408:dw|
dan815
  • dan815
|dw:1370936381535:dw|
dan815
  • dan815
|dw:1370936402456:dw|
anonymous
  • anonymous
You got me. Damn you are awesome
dan815
  • dan815
bahahAHAHAHAha
anonymous
  • anonymous
Gold metal to you
dan815
  • dan815
THANKYOU THANKYOU BUDDDY
anonymous
  • anonymous
How do I know that y=sqrt(2x-x^2) though)
dan815
  • dan815
thats our bound
dan815
  • dan815
|dw:1370936504905:dw|
anonymous
  • anonymous
Ok. My experience is being enhanced because I've been acid tripping on this song all day http://www.youtube.com/watch?v=OciYCYiR2X8
anonymous
  • anonymous
|dw:1370936611378:dw|
anonymous
  • anonymous
How did you get theta limits of 0 to pi/2?
dan815
  • dan815
|dw:1370936645740:dw|
anonymous
  • anonymous
Wouldn't it be -pi/2 to pi/2 then?
dan815
  • dan815
nono that is +/- sqrt2x-x^2
dan815
  • dan815
only +sqrt is the top of that circle
dan815
  • dan815
if ur questions integration was this |dw:1370936773349:dw|
dan815
  • dan815
if it was like this then u gotta do the whole circle
anonymous
  • anonymous
I'm curious. if we left it in regular coordinates why would we have x going from 1 to 2 then?
dan815
  • dan815
ohh i didnt even notice that 1 to 2
dan815
  • dan815
i did for 0 to 2
dan815
  • dan815
|dw:1370936889553:dw|
dan815
  • dan815
this changes everything
dan815
  • dan815
|dw:1370936951250:dw|
anonymous
  • anonymous
gg
dan815
  • dan815
do it like u wud normally do except u gotta subtract the area of the triangle
dan815
  • dan815
|dw:1370936988709:dw|
dan815
  • dan815
THERE!
dan815
  • dan815
the first integral will find this area |dw:1370937120305:dw|
dan815
  • dan815
the 2nd integral subtracts this |dw:1370937136511:dw|
anonymous
  • anonymous
Would R go from sec0 to 2/cos0?
anonymous
  • anonymous
Would R go from \[\sec \theta \to 2/\cos \theta?\]
dan815
  • dan815
i used that triangle formula we figued out from last one
dan815
  • dan815
for which one?? u tryna to set up the triangle?
dan815
  • dan815
the triangle is r from 0 to 1/costheta
anonymous
  • anonymous
\[x=r*\cos \theta\]\[1=r*\cos \theta\]\[\frac{ 1 }{ \cos \theta }=r\]\[\sec \theta = r\]
dan815
  • dan815
yep
anonymous
  • anonymous
Remember the first limits are from 1 to 2
dan815
  • dan815
yeah
dan815
  • dan815
i fixed it for that
anonymous
  • anonymous
So R would start at sec0 and go to 2/cos0?
dan815
  • dan815
|dw:1370937299511:dw|
dan815
  • dan815
yeah
dan815
  • dan815
from sec to 2cos theta
dan815
  • dan815
thats pretty cool!
dan815
  • dan815
|dw:1370937390919:dw|
dan815
  • dan815
i never remember what sec and csc are...
dan815
  • dan815
|dw:1370937466320:dw|

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