anonymous
  • anonymous
What the heck does this question mean? They're just two 4-leaf 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...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1343424196418:dw|
anonymous
  • anonymous
Hideous graph I know but it's the best I could do relatively quickly free-handed with a mouse.
anonymous
  • anonymous
very Compatible topic What the heck does this question mean? ....:)

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anonymous
  • anonymous
Eh? I'm asking, what are they looking for me to find the area of. Typically this "in-between 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
  • anonymous
|dw:1343424564173:dw|
anonymous
  • anonymous
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.
anonymous
  • anonymous
"Find the area of the region that lies inside both curves. r = sin 2θ, r = cos 2θ" That's it. Literally.
anonymous
  • anonymous
me too...i have any idea...almost forgot these polar stuff.
anonymous
  • anonymous
you'll probably have to use symmetry and just find the area of just one of those "intersections" then multiply by 8|dw:1343425073279:dw|
anonymous
  • anonymous
I'm trying to figure out when they intersect I guess, right? sin 2θ = cos 2θ So every \(\frac{\pi}{8}\) radians?
anonymous
  • anonymous
yes... i think we should start with that...
anonymous
  • anonymous
Integrate for one from 0 to \(\frac{\pi}{8}\) and then multiply by 8?
anonymous
  • anonymous
I mean I'm out of ideas, there's no "outer" and "inner" function here, they're radially equivalent for symmetry.
anonymous
  • anonymous
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....:(
anonymous
  • anonymous
|dw:1343425260518:dw|
anonymous
  • anonymous
\[8 \int\limits_0^{\pi/8} \frac{1}{2} (\sin^2(2 x)-\cos^2(2 x)) \ \ \ dx = -\frac{1}{8}\] ???
anonymous
  • anonymous
ack or -1
anonymous
  • anonymous
cos(2theta) is the top graph sin(2theta) is the bottom graph
anonymous
  • anonymous
top - bottom?
anonymous
  • anonymous
II for \( \sin 2\theta \) from 0 to \(\frac{\pi}{8} \) and I for \( \cos 2\theta \) from \(\frac{\pi}{8} \) to ....?
anonymous
  • anonymous
?= \( \frac{\pi}{4} \) ?
anonymous
  • anonymous
i think u should evaluate area of I and II separatly...
anonymous
  • anonymous
Well I mean that as your limits for 0 to \(\frac{\pi}{16}\)
anonymous
  • anonymous
|dw:1343425734006:dw|
anonymous
  • anonymous
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
anonymous
  • anonymous
so from... \(\frac{\pi}{8}\) to... \(\frac{\pi}{8}\) + \(\frac{\pi}{8}\) Basically?
anonymous
  • anonymous
from pi/8 to ( where \( \cos 2 \theta=0 \) )
anonymous
  • anonymous
|dw:1343426180407:dw|
anonymous
  • anonymous
so i think u should set up ur integrals like this: A=Total Area=8\(( A_I + A_{II} )\)
anonymous
  • anonymous
\[A=\frac{8}{2} \left( \int\limits_{0}^{\frac{\pi}{8}} \sin^2 2 \theta \ d \theta+\int\limits_{\frac{\pi}{8}}^{\frac{\pi}{4}} \cos^2 2 \theta \ d \theta \right)\]
anonymous
  • anonymous
|dw:1343505800216:dw|
anonymous
  • anonymous
So the area of the entire shaded region is obtained by : \[\huge A=16\int_\frac{\pi}{8}^\frac{\pi}{4} \frac{1}{2}cos(2\theta)d\theta \] =\(\huge \frac{\pi - 2}{2} \)
anonymous
  • anonymous
Yay @dpaInc ! That was correct, and that method worked for the second part too
anonymous
  • anonymous
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....

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