Here's the question you clicked on:
agentx5
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...
Hideous graph I know but it's the best I could do relatively quickly free-handed with a mouse.
very Compatible topic What the heck does this question mean? ....:)
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.
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.
"Find the area of the region that lies inside both curves. r = sin 2θ, r = cos 2θ" That's it. Literally.
me too...i have any idea...almost forgot these polar stuff.
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|
I'm trying to figure out when they intersect I guess, right? sin 2θ = cos 2θ So every \(\frac{\pi}{8}\) radians?
yes... i think we should start with that...
Integrate for one from 0 to \(\frac{\pi}{8}\) and then multiply by 8?
I mean I'm out of ideas, there's no "outer" and "inner" function here, they're radially equivalent for symmetry.
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....:(
\[8 \int\limits_0^{\pi/8} \frac{1}{2} (\sin^2(2 x)-\cos^2(2 x)) \ \ \ dx = -\frac{1}{8}\] ???
cos(2theta) is the top graph sin(2theta) is the bottom graph
II for \( \sin 2\theta \) from 0 to \(\frac{\pi}{8} \) and I for \( \cos 2\theta \) from \(\frac{\pi}{8} \) to ....?
?= \( \frac{\pi}{4} \) ?
i think u should evaluate area of I and II separatly...
Well I mean that as your limits for 0 to \(\frac{\pi}{16}\)
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
so from... \(\frac{\pi}{8}\) to... \(\frac{\pi}{8}\) + \(\frac{\pi}{8}\) Basically?
from pi/8 to ( where \( \cos 2 \theta=0 \) )
so i think u should set up ur integrals like this: A=Total Area=8\(( A_I + A_{II} )\)
\[A=\frac{8}{2} \left( \int\limits_{0}^{\frac{\pi}{8}} \sin^2 2 \theta \ d \theta+\int\limits_{\frac{\pi}{8}}^{\frac{\pi}{4}} \cos^2 2 \theta \ d \theta \right)\]
So the area of the entire shaded region is obtained by : \[\huge A=16\int_\frac{\pi}{8}^\frac{\pi}{4} \frac{1}{2}cos(2\theta)d\theta \] =\(\huge \frac{\pi - 2}{2} \)
Yay @dpaInc ! That was correct, and that method worked for the second part too
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....