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Math Help - Area Shared by Polar Curves

  1. #1
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    Area Shared by Polar Curves

    Hi guys!!

    Today I took my calc test and I'm curious to see if I did the last one right.

    Set up the integral or integrals to find the area shared by the polar curves given by:

     r=\sin2\theta

     r=\frac{1}{2}

    0\leq\theta\leq\frac{\pi}{2}

    I found the points of intersection:

    \sin2\theta=\frac{1}{2}

    \theta=\frac{\pi}{12},\frac{5\pi}{12}

    From there I can't remember exactly what I did, but if someone could just set up the integrals, I'm sure I can remember if that's what I did when I see it. I know there are probably a few different ways, but I would appreciate any input! I can't wait a week because I've been stressing out about this problem! And, if I got it wrong, I'd rather cry at home...(I'm kidding!...well, not really....
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by elizsimca View Post
    Hi guys!!

    Today I took my calc test and I'm curious to see if I did the last one right.

    Set up the integral or integrals to find the area shared by the polar curves given by:

     r=\sin2\theta

     r=\frac{1}{2}

    0\leq\theta\leq\frac{\pi}{2}

    I found the points of intersection:

    \sin2\theta=\frac{1}{2}

    \theta=\frac{\pi}{12},\frac{5\pi}{12}

    From there I can't remember exactly what I did, but if someone could just set up the integrals, I'm sure I can remember if that's what I did when I see it. I know there are probably a few different ways, but I would appreciate any input! I can't wait a week because I've been stressing out about this problem! And, if I got it wrong, I'd rather cry at home...(I'm kidding!...well, not really....
    I got A=\frac{\sqrt{3}}{2}square_units?
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  3. #3
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    I didn't actually have to solve it, I just had to set up the integral (because we only have 50 minutes for tests and we don't have enough time to actually do the problem.) Could you show me how you set up that integral if it's not too much trouble

    Thanks a ton!!
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by elizsimca View Post
    I didn't actually have to solve it, I just had to set up the integral (because we only have 50 minutes for tests and we don't have enough time to actually do the problem.) Could you show me how you set up that integral if it's not too much trouble

    Thanks a ton!!
    \frac{1}{2}\int_{\frac{\pi}{12}}^{\frac{5\pi}{12}}  sin^2(2\theta)d\theta
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  5. #5
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    Hello, elizsimca!

    Set up the integral to find the area shared by the polar curves given by:
    . . r\:=\:\sin2\theta,\;\;r\:=\:\frac{1}{2},\;\;\left[0,\,\frac{\pi}{2}\right]

    I found the points of intersection: . \sin2\theta\:=\:\frac{1}{2} \quad\Rightarrow\quad \theta\:=\:\frac{\pi}{12},\;\;\frac{5\pi}{12} . . . . Good!
    The first curve is a 4-leaf rose curve.
    The petals at one unit long and are aligned along the 45 lines.

    The second is a circle centered at the origin with radius .

    We are in Quadrant 1.
    We want the area of that petal which is inside the circle.
    Code:
          |
          |               *
          |        *    /
          * *         /
          |   .*    /     *
          | .:::::*
          |*::::/::*   *
          |:::/:::::*
          |:/:::*   *
        - * - - - - * - - - -
          |
    I'm afraid it takes three integrals . . .

    A \;=\;\frac{1}{2}\int^{\frac{\pi}{12}}_0 \sin^2\!2\theta\,d\theta + \frac{1}{2}\int^{\frac{5\pi}{12}}_{\frac{\pi}{12}}  \left[\left(\frac{1}{2}\right)^2 - \sin^2\!2\theta\right]\,d\theta + \frac{1}{2}\int^{\frac{\pi}{2}}_{\frac{5\pi}{12}}\  sin^2\!2\theta\,d\theta

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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    \frac{1}{2}\int_{\frac{\pi}{12}}^{\frac{5\pi}{12}}  sin^2(2\theta)d\theta

    Would that give us the whole area of the leaf from pi/12 to 5pi/12? Then how would you subtract off the outer portion of the leaf to just have the area of the "leaf" bounded above by r=1/2?
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by elizsimca View Post
    Would that give us the whole area of the leaf from pi/12 to 5pi/12? Then how would you subtract off the outer portion of the leaf to just have the area of the "leaf" bounded above by r=1/2?
    Oops! missed that part...haha...sorry...well Soroban already answered it...sorry
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  8. #8
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    Quote Originally Posted by Soroban View Post
    Hello, elizsimca!

    The first curve is a 4-leaf rose curve.
    The petals at one unit long and are aligned along the 45 lines.

    The second is a circle centered at the origin with radius .

    We are in Quadrant 1.
    We want the area of that petal which is inside the circle.
    Code:
          |
          |               *
          |        *    /
          * *         /
          |   .*    /     *
          | .:::::*
          |*::::/::*   *
          |:::/:::::*
          |:/:::*   *
        - * - - - - * - - - -
          |
    I'm afraid it takes three integrals . . .

    A \;=\;\frac{1}{2}\int^{\frac{\pi}{12}}_0 \sin^2\!2\theta\,d\theta + \frac{1}{2}\int^{\frac{5\pi}{12}}_{\frac{\pi}{12}}  \left[\left(\frac{1}{2}\right)^2 - \sin^2\!2\theta\right]\,d\theta + \frac{1}{2}\int^{\frac{\pi}{2}}_{\frac{5\pi}{12}}\  sin^2\!2\theta\,d\theta
    Nice picture!

    I think I did somethimg similar on my test, but I'm afraid that wasn't it. Luckily, my teacher iis a believer in partial credit, I think I got 2 out of the 3 integrals right.

    Thanks MathStud and Soroban! Now I'm going to go cry
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