## 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 Group Title

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2. agentx5 Group Title

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

3. mukushla Group Title

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

4. agentx5 Group Title

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 Group Title

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6. agentx5 Group Title

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 Group Title

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

8. mukushla Group Title

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

9. dpaInc Group Title

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 Group Title

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

11. mukushla Group Title

12. agentx5 Group Title

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

13. agentx5 Group Title

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

14. dpaInc Group Title

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 Group Title

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16. agentx5 Group Title

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

17. agentx5 Group Title

ack or -1

18. dpaInc Group Title

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

19. agentx5 Group Title

top - bottom?

20. mukushla Group Title

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

21. mukushla Group Title

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

22. mukushla Group Title

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

23. agentx5 Group Title

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

24. mukushla Group Title

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25. mukushla Group Title

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 Group Title

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

27. mukushla Group Title

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

28. mukushla Group Title

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29. mukushla Group Title

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

30. mukushla Group Title

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

31. dpaInc Group Title

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32. dpaInc Group Title

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 Group Title

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

34. dpaInc Group Title

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....