Results 1 to 7 of 7

Math Help - Area between two curves

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    12

    Area between two curves

    Find the area of the regions by the lines and curves given below.

    y= 8cos(x) and y = sec^2(x) between  \frac{-\pi}{3} \leq  x \leq  \frac{\pi}{3}
    Last edited by casemeister; May 17th 2008 at 09:43 PM. Reason: \pi does not = pi
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    First notice that the curves both intersect at -\frac{\pi}{3} and \frac{\pi}{3} which will be our limits of the integrand. Now, if you sketch it out, you'll see that \cos x > \sec^{2} x over the interval \left(-\frac{\pi}{3}, \frac{\pi}{3}\right).

    So the area bounded is given by: \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left( 8\cos x - \sec^{2} x \right) dx

    Should be simple to integrate.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by casemeister View Post
    Find the area of the regions by the lines and curves given below.

    y= 8cos(x) and y = sec^2(x) between  \frac{-\pi}{3} \leq x \leq \frac{\pi}{3}
    Here is a graph

    Area between two curves-capture.jpg

    \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(8\cos(x)-\sec^{2}(x))dx=8\sin(x)-\tan(x)\bigg|_{-\frac{\pi}{3}}^{\frac{\pi}{3}}=6\sqrt{3}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2008
    Posts
    12
    See, thats what I need some help on, integrating it. Im guessing you use Substitution, but i've tried that and i dont get anywhere, i just keep substituting multiple times.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left( 8\cos x - \sec^{2} x \right)dx = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 8\cos x dx - \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \sec^{2} x dx

    You should know that: \int \cos x dx = \sin x + C
    And: \int sec^{2} x dx = \tan x + C
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by casemeister View Post
    See, thats what I need some help on, integrating it. Im guessing you use Substitution, but i've tried that and i dont get anywhere, i just keep substituting multiple times.
    \int{8\cos(x)-sec^{2}(x)dx}

    seperating we get \int{8\cos(x)dx}-\int{\sec^2(x)dx}

    Note that \frac{D[\tan(x)]}{dx}=\sec^2(x)

    So \int{8\cos(x)dx}-\int{\sec^2(x)dx}=8\sin(x)-\tan(x)+C
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Apr 2008
    Posts
    12
    Doh', I really should have listened to my professor and drill those derivatives into my head...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Area between Curves
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 15th 2011, 04:24 PM
  2. Area between two curves
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 11th 2010, 10:52 PM
  3. area between curves.
    Posted in the Calculus Forum
    Replies: 6
    Last Post: January 11th 2010, 12:39 PM
  4. Area Between Two Curves
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 10th 2009, 04:15 PM
  5. Area between two curves
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 27th 2009, 05:59 PM

Search Tags


/mathhelpforum @mathhelpforum