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agentx5 Group Title

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...

  • 2 years ago
  • 2 years ago

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  1. agentx5 Group Title
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    |dw:1343424196418:dw|

    • 2 years ago
  2. agentx5 Group Title
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    Hideous graph I know but it's the best I could do relatively quickly free-handed with a mouse.

    • 2 years ago
  3. mukushla Group Title
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    very Compatible topic What the heck does this question mean? ....:)

    • 2 years ago
  4. agentx5 Group Title
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    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.

    • 2 years ago
  5. mukushla Group Title
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    |dw:1343424564173:dw|

    • 2 years ago
  6. agentx5 Group Title
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    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.

    • 2 years ago
  7. agentx5 Group Title
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    "Find the area of the region that lies inside both curves. r = sin 2θ, r = cos 2θ" That's it. Literally.

    • 2 years ago
  8. mukushla Group Title
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    me too...i have any idea...almost forgot these polar stuff.

    • 2 years ago
  9. dpaInc Group Title
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    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|

    • 2 years ago
  10. agentx5 Group Title
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    I'm trying to figure out when they intersect I guess, right? sin 2θ = cos 2θ So every \(\frac{\pi}{8}\) radians?

    • 2 years ago
  11. mukushla Group Title
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    yes... i think we should start with that...

    • 2 years ago
  12. agentx5 Group Title
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    Integrate for one from 0 to \(\frac{\pi}{8}\) and then multiply by 8?

    • 2 years ago
  13. agentx5 Group Title
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    I mean I'm out of ideas, there's no "outer" and "inner" function here, they're radially equivalent for symmetry.

    • 2 years ago
  14. dpaInc Group Title
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    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....:(

    • 2 years ago
  15. mukushla Group Title
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    |dw:1343425260518:dw|

    • 2 years ago
  16. agentx5 Group Title
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    \[8 \int\limits_0^{\pi/8} \frac{1}{2} (\sin^2(2 x)-\cos^2(2 x)) \ \ \ dx = -\frac{1}{8}\] ???

    • 2 years ago
  17. agentx5 Group Title
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    ack or -1

    • 2 years ago
  18. dpaInc Group Title
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    cos(2theta) is the top graph sin(2theta) is the bottom graph

    • 2 years ago
  19. agentx5 Group Title
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    top - bottom?

    • 2 years ago
  20. mukushla Group Title
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    II for \( \sin 2\theta \) from 0 to \(\frac{\pi}{8} \) and I for \( \cos 2\theta \) from \(\frac{\pi}{8} \) to ....?

    • 2 years ago
  21. mukushla Group Title
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    ?= \( \frac{\pi}{4} \) ?

    • 2 years ago
  22. mukushla Group Title
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    i think u should evaluate area of I and II separatly...

    • 2 years ago
  23. agentx5 Group Title
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    Well I mean that as your limits for 0 to \(\frac{\pi}{16}\)

    • 2 years ago
  24. mukushla Group Title
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    |dw:1343425734006:dw|

    • 2 years ago
  25. mukushla Group Title
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    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

    • 2 years ago
  26. agentx5 Group Title
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    so from... \(\frac{\pi}{8}\) to... \(\frac{\pi}{8}\) + \(\frac{\pi}{8}\) Basically?

    • 2 years ago
  27. mukushla Group Title
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    from pi/8 to ( where \( \cos 2 \theta=0 \) )

    • 2 years ago
  28. mukushla Group Title
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    |dw:1343426180407:dw|

    • 2 years ago
  29. mukushla Group Title
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    so i think u should set up ur integrals like this: A=Total Area=8\(( A_I + A_{II} )\)

    • 2 years ago
  30. mukushla Group Title
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    \[A=\frac{8}{2} \left( \int\limits_{0}^{\frac{\pi}{8}} \sin^2 2 \theta \ d \theta+\int\limits_{\frac{\pi}{8}}^{\frac{\pi}{4}} \cos^2 2 \theta \ d \theta \right)\]

    • 2 years ago
  31. dpaInc Group Title
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    |dw:1343505800216:dw|

    • 2 years ago
  32. dpaInc Group Title
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    So the area of the entire shaded region is obtained by : \[\huge A=16\int_\frac{\pi}{8}^\frac{\pi}{4} \frac{1}{2}cos(2\theta)d\theta \] =\(\huge \frac{\pi - 2}{2} \)

    • 2 years ago
  33. agentx5 Group Title
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    Yay @dpaInc ! That was correct, and that method worked for the second part too

    • 2 years ago
  34. dpaInc Group Title
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    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....

    • 2 years ago
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