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anonymous
 4 years ago
Find the areas of the region shared by the circles of the polar equations:
anonymous
 4 years ago
Find the areas of the region shared by the circles of the polar equations:

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[r = 2 \cos \theta\] \[r = 2 \sin \theta\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What do you mean? The upper and lower bounds?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I found them to be 0 and pi if that's what you mean.

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1yp, are they given ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just used the calculator.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's the same for all the questions in my book.

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1alryt which is the inner curve and the outer curve or gr8r and smaller curve?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02 sin theta is tangent to the xaxis and isis positive with the yaxis separating it in 2 while 2 cos theta is chopped in half by the xaxis to the left of sin theta. It is tangent to the origin.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The sin theta is on top.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The equations are on top.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But 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
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327830143156:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327830380231:dw

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1\[A=\int\limits_{\alpha}^{\beta}(1/2(r _{2}^{2}r _{1}^{2})d \theta\]

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1\[A=\int\limits_{0}^{\pi}1/2((2\cos \theta)^{2}(2\sin \theta)^{2})d \theta\] \[=\int\limits_{0}^{\pi}1/2(4\cos ^{2}\theta4\sin ^{2}\theta)d \theta\] =\[=\int\limits_{0}^{\pi}4/2(\cos ^{2}\theta(1\cos ^{2}\theta)d \theta\] =\[=2\int\limits_{0}^{\pi}(2\cos ^{2}\theta1)d \theta\] \[=\int\limits_{0}^{\pi}(4\cos2\theta)d \theta\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's the problem. Integrating this makes the answer 0.

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1\[=\int\limits_{0}^{\pi}(2\cos2\theta)d \theta\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's 4 sin 2 theta. Plugging pi and 0 would make it zero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sin 2 pi and sin 0 are both 0.

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1\[(\sin2\pi)0 +c\]

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1yh it's not symmetrical ? it's could be from the intrsc pts
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