## 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

1. agentx5

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2. agentx5

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

3. mukushla

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

4. agentx5

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.

5. mukushla

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6. agentx5

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.

7. agentx5

"Find the area of the region that lies inside both curves. r = sin 2θ, r = cos 2θ" That's it. Literally.

8. mukushla

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

9. dpaInc

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|

10. agentx5

I'm trying to figure out when they intersect I guess, right? sin 2θ = cos 2θ So every $$\frac{\pi}{8}$$ radians?

11. mukushla

12. agentx5

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

13. agentx5

I mean I'm out of ideas, there's no "outer" and "inner" function here, they're radially equivalent for symmetry.

14. dpaInc

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....:(

15. mukushla

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16. agentx5

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

17. agentx5

ack or -1

18. dpaInc

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

19. agentx5

top - bottom?

20. mukushla

II for $$\sin 2\theta$$ from 0 to $$\frac{\pi}{8}$$ and I for $$\cos 2\theta$$ from $$\frac{\pi}{8}$$ to ....?

21. mukushla

?= $$\frac{\pi}{4}$$ ?

22. mukushla

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

23. agentx5

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

24. mukushla

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25. mukushla

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

26. agentx5

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

27. mukushla

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

28. mukushla

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29. mukushla

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

30. mukushla

$A=\frac{8}{2} \left( \int\limits_{0}^{\frac{\pi}{8}} \sin^2 2 \theta \ d \theta+\int\limits_{\frac{\pi}{8}}^{\frac{\pi}{4}} \cos^2 2 \theta \ d \theta \right)$

31. dpaInc

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32. dpaInc

So the area of the entire shaded region is obtained by : $\huge A=16\int_\frac{\pi}{8}^\frac{\pi}{4} \frac{1}{2}cos(2\theta)d\theta$ =$$\huge \frac{\pi - 2}{2}$$

33. agentx5

Yay @dpaInc ! That was correct, and that method worked for the second part too

34. dpaInc

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....