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|dw:1343424196418:dw|

Hideous graph I know but it's the best I could do relatively quickly free-handed with a mouse.

very Compatible topic
What the heck does this question mean?
....:)

|dw:1343424564173:dw|

me too...i have any idea...almost forgot these polar stuff.

yes... i think we should start with that...

Integrate for one from 0 to \(\frac{\pi}{8}\) and then multiply by 8?

|dw:1343425260518:dw|

\[8 \int\limits_0^{\pi/8} \frac{1}{2} (\sin^2(2 x)-\cos^2(2 x)) \ \ \ dx = -\frac{1}{8}\]
???

ack or -1

cos(2theta) is the top graph
sin(2theta) is the bottom graph

top - bottom?

?= \( \frac{\pi}{4} \) ?

i think u should evaluate area of I and II separatly...

Well I mean that as your limits for 0 to \(\frac{\pi}{16}\)

|dw:1343425734006:dw|

so from... \(\frac{\pi}{8}\) to... \(\frac{\pi}{8}\) + \(\frac{\pi}{8}\) Basically?

from pi/8 to ( where \( \cos 2 \theta=0 \) )

|dw:1343426180407:dw|

so i think u should set up ur integrals like this:
A=Total Area=8\(( A_I + A_{II} )\)

|dw:1343505800216:dw|