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anonymous

  • 4 years ago

Find the areas of the region shared by the circles of the polar equations:

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  1. anonymous
    • 4 years ago
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    \[r = 2 \cos \theta\] \[r = 2 \sin \theta\]

  2. EarthCitizen
    • 4 years ago
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    \[\theta=?\]

  3. anonymous
    • 4 years ago
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    What do you mean? The upper and lower bounds?

  4. anonymous
    • 4 years ago
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    I found them to be 0 and pi if that's what you mean.

  5. EarthCitizen
    • 4 years ago
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    yp, are they given ?

  6. anonymous
    • 4 years ago
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    Nope.

  7. anonymous
    • 4 years ago
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    I just used the calculator.

  8. anonymous
    • 4 years ago
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    It's the same for all the questions in my book.

  9. EarthCitizen
    • 4 years ago
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    alryt which is the inner curve and the outer curve or gr8r and smaller curve?

  10. anonymous
    • 4 years ago
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    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.

  11. EarthCitizen
    • 4 years ago
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    r2 and r1 is ?

  12. anonymous
    • 4 years ago
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    The sin theta is on top.

  13. anonymous
    • 4 years ago
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    The equations are on top.

  14. anonymous
    • 4 years ago
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    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.

  15. anonymous
    • 4 years ago
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    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.

  16. EarthCitizen
    • 4 years ago
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    |dw:1327830143156:dw|

  17. anonymous
    • 4 years ago
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    |dw:1327830380231:dw|

  18. anonymous
    • 4 years ago
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    The top is the sin

  19. EarthCitizen
    • 4 years ago
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    \[A=\int\limits_{\alpha}^{\beta}(1/2(r _{2}^{2}-r _{1}^{2})d \theta\]

  20. EarthCitizen
    • 4 years ago
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    \[A=\int\limits_{0}^{\pi}1/2((2\cos \theta)^{2}-(2\sin \theta)^{2})d \theta\] \[=\int\limits_{0}^{\pi}1/2(4\cos ^{2}\theta-4\sin ^{2}\theta)d \theta\] =\[=\int\limits_{0}^{\pi}4/2(\cos ^{2}\theta-(1-\cos ^{2}\theta)d \theta\] =\[=2\int\limits_{0}^{\pi}(2\cos ^{2}\theta-1)d \theta\] \[=\int\limits_{0}^{\pi}(4\cos2\theta)d \theta\]

  21. anonymous
    • 4 years ago
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    That's the problem. Integrating this makes the answer 0.

  22. EarthCitizen
    • 4 years ago
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    \[=\int\limits_{0}^{\pi}(2\cos2\theta)d \theta\]

  23. anonymous
    • 4 years ago
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    It's 4 sin 2 theta. Plugging pi and 0 would make it zero.

  24. anonymous
    • 4 years ago
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    sin 2 pi and sin 0 are both 0.

  25. EarthCitizen
    • 4 years ago
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    \[(\sin2\pi)-0 +c\]

  26. EarthCitizen
    • 4 years ago
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    yh it's not symmetrical ? it's could be from the intrsc pts

  27. EarthCitizen
    • 4 years ago
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    alpha=pi/6

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