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Math Help - Please Help: find the area of the region

  1. #1
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    Please Help: find the area of the region

    Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

    y=6cos(x), y=(2sec(x))^2, x=-pi/4, x=pi/4

    I keep getting .4853 but is incorrect

    Thanks
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by waite3 View Post
    Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

    y=6cos(x), y=(2sec(x))^2, x=-pi/4, x=pi/4

    I keep getting .4853 but is incorrect

    Thanks
    \int_{-\pi/4}^{\pi/4}6\cos(x)-2 \sec^{2}xdx=6\sin(x)-2\tan(x)|_{-\pi/4}^{\pi/4}

    =6[\sin(\pi /4)-\sin(-\pi /4)]-2[\tan(\pi /4)-tan(-\pi /4)]

    6[ \frac {\sqrt{2}}{2}- \frac{- \sqrt{2}}{2}]-2[1-(-1)]=6\sqrt{2}-4 \approx 4.485
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  3. #3
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    Quote Originally Posted by waite3 View Post
    Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

    y=6cos(x), y=(2sec(x))^2, x=-pi/4, x=pi/4

    I keep getting .4853 but is incorrect

    Thanks
    See the image below for the intersection region. (This entire area fits within the required bounds, so there are no cut-offs.

    The intersection points of the two curves may be found by
    6~cos(x) = 4~sec^2(x)

    cos^3(x) = \frac{2}{3}

    cos(x) = \sqrt[3]{\frac{2}{3}}

    Keeping in mind that cosine is an even function of x:
    x = \pm cos^{-1} \left ( \sqrt[3]{\frac{2}{3}} \right )

    x \approx \pm 0.508285

    We obviously integrate over x.

    A = \int_{-0.508285}^{0.508285}( 6~cos(x) - 4~sec^2(x) )~dx

    = (6~sin(0.508285) - 4~tan(0.508285) ) - ( 6~sin(-0.508285) - 4~tan(-0.508285) )  \approx  1.38328
    Attached Thumbnails Attached Thumbnails Please Help: find the area of the region-area.jpg  
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  4. #4
    Behold, the power of SARDINES!
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    I'm blind

    y=6cos(x), y=(2sec(x))^2, x=-pi/4, x=pi/4
    doh!!
    I did 2\sec^2(x)not (2\sec(x))^2

    That was just plain lazy
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  5. #5
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    Help please!

    I tried the 1.3833 answer but was incorrect and the teacher said im missing the sqrt 2 so any help would be greatly appreciated. thanks
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