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anonymous
 3 years 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...
anonymous
 3 years 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...

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1343424196418:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hideous graph I know but it's the best I could do relatively quickly freehanded with a mouse.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0very Compatible topic What the heck does this question mean? ....:)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Eh? 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1343424564173:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That'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.

anonymous
 3 years ago
Best 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0me too...i have any idea...almost forgot these polar stuff.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you'll probably have to use symmetry and just find the area of just one of those "intersections" then multiply by 8dw:1343425073279:dw

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes... i think we should start with that...

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no... 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....:(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1343425260518:dw

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0cos(2theta) is the top graph sin(2theta) is the bottom graph

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0II for \( \sin 2\theta \) from 0 to \(\frac{\pi}{8} \) and I for \( \cos 2\theta \) from \(\frac{\pi}{8} \) to ....?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0?= \( \frac{\pi}{4} \) ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think u should evaluate area of I and II separatly...

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1343425734006:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0according 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so from... \(\frac{\pi}{8}\) to... \(\frac{\pi}{8}\) + \(\frac{\pi}{8}\) Basically?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0from pi/8 to ( where \( \cos 2 \theta=0 \) )

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1343426180407:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i think u should set up ur integrals like this: A=Total Area=8\(( A_I + A_{II} )\)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1343505800216:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So the area of the entire shaded region is obtained by : \[\huge A=16\int_\frac{\pi}{8}^\frac{\pi}{4} \frac{1}{2}cos(2\theta)d\theta \] =\(\huge \frac{\pi  2}{2} \)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yay @dpaInc ! That was correct, and that method worked for the second part too

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0glad 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....
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