Finding area in polar coordinates

raincheck
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Homework Statement


"Find the area of the region described: The region that is enclosed by the rose r=4cos3[theta]"

Homework Equations



A= [integral] (1/2)r^2 d[theta]

The Attempt at a Solution



I'll use Q as [theta]..

I'm not really sure, but I set up (1/2) [integral] (16(cos^2)3Q) dQ

.. then I thought we were supposed to use the identity (cos^2)Q = (1/2)(1+cos2Q), but every time I substitute this in and integrate, I get 8[pi] instead of 4[pi], the correct answer. What am I doing wrong?

Thank you so much :]
 
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I get 8pi as well.
 
hmmm, I asked someone else and they got 8pi too...


Well I had one more question... if I were finding the area between
r = sqrt[cos2(theta)] and
r = 2cos(theta)

do I basically do the same thing as finding the area between two regular curves? Each time I try it, I come up with zeros and I feel like that can't be right :[
 
That's a really weird graph intersection. Where did you get that question from?
 
You don't say anything about what limits of integration you used. \theta going from 0 to what traces that figure exactly once?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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