Results 1 to 8 of 8

Math Help - find the area

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    4

    find the area

    of the region inside r^2 = 4 sin 2theta and outside r = sqrt(2)

    I've sketched it out, but it just made me angrier. How would i go about setting up an equation for this type of problem?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member SengNee's Avatar
    Joined
    Jan 2008
    From
    Pangkor Island, Perak, Malaysia.
    Posts
    155
    Quote Originally Posted by tomobson View Post
    of the region inside r^2 = 4 sin 2theta and outside r = sqrt(2)

    I've sketched it out, but it just made me angrier. How would i go about setting up an equation for this type of problem?
    You should find their intersections 1st.
    Then find the area of each lines from one intersection to another.
    The difference between these lines will be your answer.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by tomobson View Post
    of the region inside r^2 = 4 sin 2theta and outside r = sqrt(2)

    I've sketched it out, but it just made me angrier. How would i go about setting up an equation for this type of problem?
    Graphing this shows me that it is symmetric around the origin so I'll just calculate it in the first quadrant and multiply by two.

    You first want to find where the two curves intersect. If r^2=4\sin(2\theta) and r^2=2, these two curves intersect at \theta=\frac{\pi}{12} and \theta=\frac{5\pi}{12}.

    The formula you want to use is \frac{1}{2}\int_{\theta_1}^{\theta_2}(r_1^2 - r_2^2)\,d\theta

    Plugging in gives you 2\cdot\frac{1}{2}\int_{\pi/12}^{5\pi/12}(4\sin(2\theta)-2)\,d\theta
    Spoiler:
    = \bigg[-2\cos(2\theta)-2\theta\bigg]_{\pi/12}^{5\pi/12} = \left[\left(\sqrt{3}-\frac{5\pi}{6}\right)-\left(-\sqrt{3}-\frac{\pi}{6}\right)\right] = 2\sqrt{3}-\frac{2\pi}{3}
    Last edited by redsoxfan325; April 28th 2009 at 09:35 AM. Reason: Changed to multiply by 2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member SengNee's Avatar
    Joined
    Jan 2008
    From
    Pangkor Island, Perak, Malaysia.
    Posts
    155
    Quote Originally Posted by redsoxfan325 View Post
    Graphing this shows me that it is symmetric so I'll just calculate it in the first quadrant and multiply by four.

    You first want to find where the two curves intersect. If r^2=4\sin(2\theta) and r^2=2, these two curves intersect at \theta=\frac{\pi}{12} and \theta=\frac{5\pi}{12}.

    The formula you want to use is \frac{1}{2}\int_{\theta_1}^{\theta_2}(r_1^2 - r_2^2)\,d\theta

    Plugging in gives you 4\cdot\frac{1}{2}\int_{\pi/12}^{5\pi/12}(4\sin(2\theta)-2)\,d\theta
    Spoiler:
    = 2\bigg[-2\cos(2\theta)-2\theta\bigg]_{\pi/12}^{5\pi/12} = 2\left[\left(\sqrt{3}-\frac{5\pi}{6}\right)-\left(-\sqrt{3}-\frac{\pi}{6}\right)\right] = 2\left(2\sqrt{3}-\frac{2\pi}{3}\right) = 4\sqrt{3}-\frac{4\pi}{3}
    Why
    \frac{1}{2}\int_{\theta_1}^{\theta_2}(r_1^2 - r_2^2)\,d\theta
    but not
    \frac{1}{2}\int_{\theta_1}^{\theta_2}(r_1 - r_2)\,d\theta
    ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    That's just the formula for polar curves. It comes from double integrals.

    \int\int\,dA = \int_{\theta_1}^{\theta_2}\int_{r_1(\theta)}^{r_2(  \theta)}r\,dr\,d\theta = \frac{1}{2}\int_{\theta_1}^{\theta_2}(r_2^2-r_1^2)\,d\theta
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2009
    Posts
    4
    i guess i graphed this out wrong, so i'm not seeing why i have to calculate four areas and not just two. isn't the first curve a lemniscate with a 45 degree tilt, and the second one just a circle, making it only two areas i have to calculate for the area?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by tomobson View Post
    i guess i graphed this out wrong, so i'm not seeing why i have to calculate four areas and not just two. isn't the first curve a lemniscate with a 45 degree tilt, and the second one just a circle, making it only two areas i have to calculate for the area?
    Yes, you're right. When I graphed the equation of the first curve, I forgot it was r^2=4\sin(2\theta). (I graphed r=4\sin(2\theta).) However, most of my calculations are still correct, I believe, except that instead of multiplying by 4 at the end, I should have multiplied by 2. I have edited my original post.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member SengNee's Avatar
    Joined
    Jan 2008
    From
    Pangkor Island, Perak, Malaysia.
    Posts
    155
    Quote Originally Posted by redsoxfan325 View Post
    That's just the formula for polar curves. It comes from double integrals.

    \int\int\,dA = \int_{\theta_1}^{\theta_2}\int_{r_1(\theta)}^{r_2(  \theta)}r\,dr\,d\theta = \frac{1}{2}\int_{\theta_1}^{\theta_2}(r_2^2-r_1^2)\,d\theta

    I see.
    I haven't learn it yet.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 5th 2010, 08:48 PM
  2. Find the area..
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 21st 2010, 08:56 PM
  3. find the area
    Posted in the Geometry Forum
    Replies: 3
    Last Post: May 4th 2009, 04:51 AM
  4. Find the area...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 4th 2008, 11:22 AM
  5. find area of shaded area (yellow)
    Posted in the Geometry Forum
    Replies: 8
    Last Post: September 4th 2007, 08:34 AM

Search Tags


/mathhelpforum @mathhelpforum