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What the heck does this question mean? They're just two 4leaf clovers shifted 45\(^o\) (aka. \(\frac{\pi}{4}\)) relative to each other...
Find the area of the region that lies inside both curves.
r = sin 2θ, r = cos 2θ
Polar coordinates, and looking at the graphs I'm not sure what is meant here...
 one year ago
 one year ago
What the heck does this question mean? They're just two 4leaf clovers shifted 45\(^o\) (aka. \(\frac{\pi}{4}\)) relative to each other... Find the area of the region that lies inside both curves. r = sin 2θ, r = cos 2θ Polar coordinates, and looking at the graphs I'm not sure what is meant here...
 one year ago
 one year ago

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agentx5Best ResponseYou've already chosen the best response.0
Hideous graph I know but it's the best I could do relatively quickly freehanded with a mouse.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
very Compatible topic What the heck does this question mean? ....:)
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Eh? I'm asking, what are they looking for me to find the area of. Typically this "inbetween the curves" area type problems involved two circles or a circle and a carotid, or something else. I'm not really sure what "sections" of area they are looking for with the question worded like so for these very similar functions.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
dw:1343424564173:dw
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
That's what I was wondering, find the area of all 8 overlaps? O_o I seriously DON'T have any more information than what I posted at the start of this question. It's just in a section of challenge problems for polar stuff.
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
"Find the area of the region that lies inside both curves. r = sin 2θ, r = cos 2θ" That's it. Literally.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
me too...i have any idea...almost forgot these polar stuff.
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
you'll probably have to use symmetry and just find the area of just one of those "intersections" then multiply by 8dw:1343425073279:dw
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
I'm trying to figure out when they intersect I guess, right? sin 2θ = cos 2θ So every \(\frac{\pi}{8}\) radians?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
yes... i think we should start with that...
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Integrate for one from 0 to \(\frac{\pi}{8}\) and then multiply by 8?
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
I mean I'm out of ideas, there's no "outer" and "inner" function here, they're radially equivalent for symmetry.
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
no... don't multiply by eight now.. just find the area of that intersection first... yeah... and me too, i'm very rusty with polar equations....:(
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
dw:1343425260518:dw
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
\[8 \int\limits_0^{\pi/8} \frac{1}{2} (\sin^2(2 x)\cos^2(2 x)) \ \ \ dx = \frac{1}{8}\] ???
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
cos(2theta) is the top graph sin(2theta) is the bottom graph
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
II for \( \sin 2\theta \) from 0 to \(\frac{\pi}{8} \) and I for \( \cos 2\theta \) from \(\frac{\pi}{8} \) to ....?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
?= \( \frac{\pi}{4} \) ?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
i think u should evaluate area of I and II separatly...
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Well I mean that as your limits for 0 to \(\frac{\pi}{16}\)
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
dw:1343425734006:dw
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
according to what dpalnc said cos(2theta) is the top graph (I) bounds from pi/8 to pi/4 and sin(2theta) is the bottom graph (II) bounds from 0 to pi/8
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
so from... \(\frac{\pi}{8}\) to... \(\frac{\pi}{8}\) + \(\frac{\pi}{8}\) Basically?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
from pi/8 to ( where \( \cos 2 \theta=0 \) )
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
dw:1343426180407:dw
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
so i think u should set up ur integrals like this: A=Total Area=8\(( A_I + A_{II} )\)
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
\[A=\frac{8}{2} \left( \int\limits_{0}^{\frac{\pi}{8}} \sin^2 2 \theta \ d \theta+\int\limits_{\frac{\pi}{8}}^{\frac{\pi}{4}} \cos^2 2 \theta \ d \theta \right)\]
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
So the area of the entire shaded region is obtained by : \[\huge A=16\int_\frac{\pi}{8}^\frac{\pi}{4} \frac{1}{2}cos(2\theta)d\theta \] =\(\huge \frac{\pi  2}{2} \)
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Yay @dpaInc ! That was correct, and that method worked for the second part too
 one year ago

dpaIncBest ResponseYou've already chosen the best response.1
glad to hear it....:) i was really rusty with polar so i reviewed this a bit.... what i said earlier about "top" minus "bottom" function does not apply in polar coordinates...... it is "outer function" minus "inner function" when dealing with polar graphs....
 one year ago
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