anonymous
  • anonymous
Find the areas of the region shared by the circles of the polar equations:
Mathematics
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anonymous
  • anonymous
Find the areas of the region shared by the circles of the polar equations:
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
\[r = 2 \cos \theta\] \[r = 2 \sin \theta\]
EarthCitizen
  • EarthCitizen
\[\theta=?\]
anonymous
  • anonymous
What do you mean? The upper and lower bounds?

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anonymous
  • anonymous
I found them to be 0 and pi if that's what you mean.
EarthCitizen
  • EarthCitizen
yp, are they given ?
anonymous
  • anonymous
Nope.
anonymous
  • anonymous
I just used the calculator.
anonymous
  • anonymous
It's the same for all the questions in my book.
EarthCitizen
  • EarthCitizen
alryt which is the inner curve and the outer curve or gr8r and smaller curve?
anonymous
  • anonymous
2 sin theta is tangent to the x-axis and isis positive with the y-axis separating it in 2 while 2 cos theta is chopped in half by the x-axis to the left of sin theta. It is tangent to the origin.
EarthCitizen
  • EarthCitizen
r2 and r1 is ?
anonymous
  • anonymous
The sin theta is on top.
anonymous
  • anonymous
The equations are on top.
anonymous
  • anonymous
But the questions is to find he area shared by the two. I tried to find the area of one circle and then the area without the area of the shared. I planned to substract them to get the shared area but one of the integrations led toa negative number or a zero. So it didn't happen.
anonymous
  • anonymous
I was hoping if there were any other ways, especially with only one equation. But the other method would be great as long as I don't get negative or zero.
EarthCitizen
  • EarthCitizen
|dw:1327830143156:dw|
anonymous
  • anonymous
|dw:1327830380231:dw|
anonymous
  • anonymous
The top is the sin
EarthCitizen
  • EarthCitizen
\[A=\int\limits_{\alpha}^{\beta}(1/2(r _{2}^{2}-r _{1}^{2})d \theta\]
EarthCitizen
  • EarthCitizen
\[A=\int\limits_{0}^{\pi}1/2((2\cos \theta)^{2}-(2\sin \theta)^{2})d \theta\] \[=\int\limits_{0}^{\pi}1/2(4\cos ^{2}\theta-4\sin ^{2}\theta)d \theta\] =\[=\int\limits_{0}^{\pi}4/2(\cos ^{2}\theta-(1-\cos ^{2}\theta)d \theta\] =\[=2\int\limits_{0}^{\pi}(2\cos ^{2}\theta-1)d \theta\] \[=\int\limits_{0}^{\pi}(4\cos2\theta)d \theta\]
anonymous
  • anonymous
That's the problem. Integrating this makes the answer 0.
EarthCitizen
  • EarthCitizen
\[=\int\limits_{0}^{\pi}(2\cos2\theta)d \theta\]
anonymous
  • anonymous
It's 4 sin 2 theta. Plugging pi and 0 would make it zero.
anonymous
  • anonymous
sin 2 pi and sin 0 are both 0.
EarthCitizen
  • EarthCitizen
\[(\sin2\pi)-0 +c\]
EarthCitizen
  • EarthCitizen
yh it's not symmetrical ? it's could be from the intrsc pts
EarthCitizen
  • EarthCitizen
alpha=pi/6

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