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Math Help - Find the area of the region enclosed

  1. #1
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    Find the area of the region enclosed

    Find the area of the region enclosed between and from to .

    can anyone help with this one.
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  2. #2
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    <br />
dA = f(x) - g(x) <=> A = \int {f(x) - g(x)}\ where f(x) is the top-function

    So, figure out what function is above the other and get the anti derivative, then put in your limits
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    so the anti derivative is -3cos(x)-2sin(x)
    what do i do after i plug in the limits
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  4. #4
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    Quote Originally Posted by dat1611 View Post
    Find the area of the region enclosed between and from to .

    can anyone help with this one.
    first find the intersection of the two functions in the given interval ...

    3\sin{x} = 2\cos{x}

    \tan{x} = \frac{2}{3}

    x = \arctan\left(\frac{2}{3}\right)


    finding the area will consist of two integrals because the "top" function changes ...

    let c = \arctan\left(\frac{2}{3}\right)

    A = \int_0^c 2\cos{x} - 3\sin{x} \, dx + \int_c^{\frac{7\pi}{10}} 3\sin{x} - 2\cos{x} \, dx<br />

    integrate and use the Fundamental Theorem of Calculus to find the area value.

    note that this area problem is probably better suited for completion with a calculator.
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by dat1611 View Post
    Find the area of the region enclosed between and from to .

    can anyone help with this one.
    So you have {y_1} = 3\sin x,{\text{ }}{y_2} = 2\cos x,{\text{ }}0 \leqslant x \leqslant \frac{{7\pi }}{{10}},{\text{ }}S = ?

    {y_1} = {y_2} \Leftrightarrow 3\sin x = 2\cos x \Leftrightarrow \tan x = \frac{2}{3} \Leftrightarrow x = \arctan \frac{2}{3} + \pi k{\text{ }}\left( {k \in \mathbb{Z}} \right).

    S = {S_1} + {S_2}.

    {S_1} = \int\limits_0^{\arctan \frac{2}{3}} {\left( {2\cos x - 3\sin x} \right)dx}  = \left. {\left( {2\sin x + 3\cos x} \right)} \right|_0^{\arctan \frac{2}{3}} =

    = 2\sin \left( {\arctan \frac{2}{3}} \right) + 3\cos \left( {\arctan \frac{2}{3}} \right) - 3 =

    = 2 \cdot \frac{{{2 \mathord{\left/{\vphantom {2 3}} \right.\kern-\nulldelimiterspace} 3}}}{{\sqrt {1 + {{\left( {{2 \mathord{\left/<br />
 {\vphantom {2 3}} \right.\kern-\nulldelimiterspace} 3}} \right)}^2}} }} + 3 \cdot \frac{1}{{\sqrt {1 + {{\left( {{2 \mathord{\left/{\vphantom {2 3}} \right.\kern-\nulldelimiterspace} 3}} \right)}^2}} }} - 3 =

    = \frac{4}{{\sqrt {13} }} + \frac{9}{{\sqrt {13} }} - 3 = \frac{{13}}{{\sqrt {13} }} - 3 = \sqrt {13}  - 3.

    {S_1} = \int\limits_{\arctan \frac{2}{3}}^{{{7\pi } \mathord{\left/{\vphantom {{7\pi } {10}}} \right.\kern-\nulldelimiterspace} {10}}} {\left( {3\sin x - 2\cos x} \right)dx}  =  - \left. {\left( {3\cos x + 2\sin x} \right)} \right|_{\arctan \frac{2}{3}}^{{{7\pi } \mathord{\left/{\vphantom {{7\pi } {10}}} \right.\kern-\nulldelimiterspace} {10}}} =

    =  - \left( {3\cos \frac{{7\pi }}{{10}} + 2\sin \frac{{7\pi }}{{10}} - \sqrt {13} } \right) = \sqrt {13}  - 3\cos \left( {\pi  - \frac{{3\pi }}{{10}}} \right) - 2\sin \left( {\pi  - \frac{{3\pi }}{{10}}} \right) =

    = \sqrt {13}  + 3\cos \frac{{3\pi }}{{10}} - 2\sin \frac{{3\pi }}{{10}} = \sqrt {13}  + 3 \cdot \frac{{\sqrt {5 - \sqrt 5 } }}{{2\sqrt 2 }} - 2 \cdot \frac{{\sqrt 5  + 1}}{4} =

    = \sqrt {13}  + \frac{{3\sqrt 2 \sqrt {5 - \sqrt 5 } }}{4} - \frac{{\sqrt 5  + 1}}{2} = \sqrt {13}  + \frac{3}{4}\sqrt {10 - 2\sqrt 5 }  - \frac{1}{2}\left( {\sqrt 5  + 1} \right).

    Finally

    S = \sqrt {13}  - 3 + \sqrt {13}  + \frac{3}{4}\sqrt {10 - 2\sqrt 5 }  - \frac{1}{2}\left( {\sqrt 5  + 1} \right) =

    = \frac{1}{2}\left( {4\sqrt {13}  - \sqrt 5 } \right) + \frac{3}{4}\sqrt {10 - 2\sqrt 5 }  - \frac{7}{2} \approx {\text{4}}{\text{.35642}}

    See this picture

    Last edited by DeMath; August 5th 2009 at 02:45 PM.
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