Results 1 to 6 of 6

Math Help - Area of region lying inside r(t) = sin(2t) and r(t) = cos(2t)

  1. #1
    Junior Member
    Joined
    Jul 2008
    Posts
    74

    Area of region lying inside r(t) = sin(2t) and r(t) = cos(2t)

    Dear all,

    I am having trouble with the following exercise and would like to ask if the step underlined in red could be please explained.

    I understand everything up to that point but don't understand why only  \sin 2\theta is considered in the integral. What about  \cos 2\theta?

    Also, what does the  8 \cdot 2\int^{\pi/8}_{0} \frac{1}{2} \sin^2 2\theta d\theta mean? Does it represent the fact that the wedge (highlighted in light green below in my graph) is multiplied by 16 times?

    Thank you very much.

    ---

    Question:
    Find the area of the region that lies inside both  r = \sin 2\theta and  r = \cos 2\theta .

    Given solution:

    My graph:


    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    note the attached graph ...

    r = \sin(2t) in red from 0 to \frac{\pi}{8}

    r = \cos(2t) in blue from \frac{\pi}{8} to \frac{\pi}{4}

    area swept out by r = \sin(2t) from 0 to \frac{\pi}{8} equals the area swept out by \cos(2t) from \frac{\pi}{8} to \frac{\pi}{4}

    so, taking advantage of the symmetry of the graph, 16 times one of those single areas is the desired area defined by the problem.

    \displaystyle 16 \int_0^{\frac{\pi}{8}} \frac{\sin^2(2t)}{2} \, dt = 16 \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} \frac{\cos^2(2t)}{2} \, dt

    either one of the above will calculate the total area.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by skeeter View Post
    note the attached graph ...

    r = \sin(2t) in red from 0 to \frac{\pi}{8}

    r = \cos(2t) in blue from \frac{\pi}{8} to \frac{\pi}{4}

    area swept out by r = \sin(2t) from 0 to \frac{\pi}{8} equals the area swept out by \cos(2t) from \frac{\pi}{8} to \frac{\pi}{4}

    so, taking advantage of the symmetry of the graph, 16 times one of those single areas is the desired area defined by the problem.

    \displaystyle 16 \int_0^{\frac{\pi}{8}} \frac{\sin^2(2t)}{2} \, dt = 16 \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} \frac{\cos^2(2t)}{2} \, dt

    either one of the above will calculate the total area.


    Pretty nice, but I think the OP's problem, and mine as well, is to justify why we integrate \frac{\sin^22t}{2}} to

    obtain that area, since that function doesn't appear at all in the given data...

    Tonio
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by tonio View Post
    Pretty nice, but I think the OP's problem, and mine as well, is to justify why we integrate \frac{\sin^22t}{2}} to

    obtain that area, since that function doesn't appear at all in the given data...

    Tonio
    Since this was not a simple/beginner type area problem, I made the assumption that the OP was familiar with the basics of finding the area enclosed by a polar curve ...

    \displaystyle A = \int_{\theta_1}^{\theta_2} \frac{[r(\theta)]^2}{2} \, d\theta
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jul 2008
    Posts
    74
    Thank you for your response, skeeter.

    I can see by inspection that

    Quote Originally Posted by skeeter View Post
    ...

    the area swept out by r = \sin(2t) from 0 to \frac{\pi}{8} equals the area swept out by \cos(2t) from \frac{\pi}{8} to \frac{\pi}{4}

    so, taking advantage of the symmetry of the graph, 16 times one of those single areas is the desired area defined by the problem ...
    .

    However, if I were not sure about the symmetry of your graph, what computation could I perform to verify this symmetry? Must I compute \displaystyle \int_{0}^{\frac{\pi}{8}} \sin(2t) dt and \displaystyle \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} \cos(2t) dt or is there some easier method?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by scherz0 View Post
    Thank you for your response, skeeter.

    I can see by inspection that

    .

    However, if I were not sure about the symmetry of your graph, what computation could I perform to verify this symmetry? Must I compute \displaystyle \int_{0}^{\frac{\pi}{8}} \sin(2t) dt and \displaystyle \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} \cos(2t) dt or is there some easier method?
    \sin(2t) and \cos(2t) are not the integrands ...

    \frac{\sin^2(2t)}{2} and \frac{\cos^2(2t)}{2} are the integrands for the area.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. point inside an area
    Posted in the Geometry Forum
    Replies: 5
    Last Post: March 28th 2011, 09:09 AM
  2. Surface area of cone lying inside a cylinder
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 24th 2010, 09:25 AM
  3. Area inside cardioid and also inside circle??
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 19th 2010, 05:57 PM
  4. Replies: 2
    Last Post: September 6th 2008, 02:20 AM
  5. Replies: 13
    Last Post: June 23rd 2008, 09:43 AM

Search Tags


/mathhelpforum @mathhelpforum