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Math Help - Areas in the Plane (Calculus)

  1. #1
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    Areas in the Plane (Calculus)

    Can someone help me with these problems and explain each step. Any help is appreciated.


    Find the area of the regions enclosed by the lines and curves.

    x = (tany)^2 and x = -(tany)^2, -Π/4 ≤y ≤ Π/4
    Last edited by turtle; February 4th 2007 at 09:29 PM.
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  2. #2
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    Quote Originally Posted by turtle View Post
    Can someone help me with these problems and explain each step. Any help is appreciated.
    Find the area of the regions enclosed by the lines and curves.
    x = (tany)^2 and x = -(tany)^2, -Π/4 ≤y ≤ Π/4
    Hello,

    the enclosed area is calculated by the difference of functions:

    x_1=\tan^2(y) and x_2=-\tan^2(y)

    x_1-x_2=2\tan^2(y)

    The enclosed area is:

    \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{2\tan^2(y)  dy} = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{2\cdot \frac{\sin^2(y)}{\cos^2(y)}  dy} =2\cdot \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{ \frac{1-\cos^2(y)}{\cos^2(y)}  dy}=2\cdot \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{ \left(\frac{1}{\cos^2(y)}-1 \right) dy}=
    \left[2 \cdot \tan(y) - 2y \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}=\left(2-\frac{\pi}{2}\right)-\left( -2-\frac{-\pi}{2} \right)=4-\pi

    I've attached a diagram of this area.

    EB
    Attached Thumbnails Attached Thumbnails Areas in the Plane (Calculus)-zweitangquad_flaeche.gif  
    Last edited by earboth; February 5th 2007 at 02:10 AM. Reason: incomplete answer
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