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Math Help - double integral bounds question

  1. #1
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    double integral bounds question

    I am to evaluate the double integral over a region s of 1/(x^2 + y^2)^(1/2)

    S is the region where x^2 + y^2 is greater or equal to 4
    and where x^2 + y^2 is less than or equal to 25
    y is greater or equal to 0


    I see right away that my coordinates limits for r should be 2 to 5 and that wehave the function of 1/r times r dr dtheta

    I can not see why theta goes from 0 to pi.

    Once the bounds are set this is a ridiculously easy integral to take I know!!!!

    I realize this is a simple concept that I keep misunderstanding. Would someone please let me know how we know that theta is from 0 to pi? Thanks very much, Frostking
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Frostking View Post
    I am to evaluate the double integral over a region s of 1/(x^2 + y^2)^(1/2)

    S is the region where x^2 + y^2 is greater or equal to 4
    and where x^2 + y^2 is less than or equal to 25
    y is greater or equal to 0


    I see right away that my coordinates limits for r should be 2 to 5 and that wehave the function of 1/r times r dr dtheta

    I can not see why theta goes from 0 to pi.

    Once the bounds are set this is a ridiculously easy integral to take I know!!!!

    I realize this is a simple concept that I keep misunderstanding. Would someone please let me know how we know that theta is from 0 to pi? Thanks very much, Frostking
    Since you are told that y is greater than zero, you're considering all values above the x axis. Since the x-axis is a horizontal line, it has a degree measure of 180 or \pi radians. That is why 0\leqslant\theta\leqslant\pi.
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Frostking View Post
    I am to evaluate the double integral over a region s of 1/(x^2 + y^2)^(1/2)

    S is the region where x^2 + y^2 is greater or equal to 4
    and where x^2 + y^2 is less than or equal to 25
    y is greater or equal to 0

    I see right away that my coordinates limits for r should be 2 to 5 and that wehave the function of 1/r times r dr dtheta

    I can not see why theta goes from 0 to pi.

    Once the bounds are set this is a ridiculously easy integral to take I know!!!!

    I realize this is a simple concept that I keep misunderstanding. Would someone please let me know how we know that theta is from 0 to pi? Thanks very much, Frostking
    S = {S_2}\backslash {S_1}, where

    {S_1} = \left\{ {\left. {\left( {x,y} \right)} \right|{\text{ }}{x^2} + {y^2} \leqslant 4,{\text{ }}y \geqslant 0,{\text{ }} - 2 \leqslant x \leqslant 2} \right\} = \left\{ {\left. {\left( {x,y} \right)} \right|{\text{ 0}} \leqslant y \leqslant \sqrt {4 - {x^2}} ,{\text{ }} - 2 \leqslant x \leqslant 2} \right\}
    and
    {S_2} = \left\{ {\left. {\left( {x,y} \right)} \right|{\text{ }}{x^2} + {y^2} \leqslant 25,{\text{ }}y \geqslant 0,{\text{ }} - 5 \leqslant x \leqslant 5} \right\} = \left\{ {\left. {\left( {x,y} \right)} \right|{\text{ 0}} \leqslant y \leqslant \sqrt {25 - {x^2}} ,{\text{ }} - 5 \leqslant x \leqslant 5} \right\}.

    \iint\limits_S {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} = \iint\limits_{{S_2}} {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} - \iint\limits_{{S_1}} {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} = \int\limits_{ - 5}^5 {\int\limits_0^{\sqrt {25 - {x^2}} } {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} }  - \int\limits_{ - 2}^2 {\int\limits_0^{\sqrt {4 - {x^2}} } {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} } .

    Look a projection of the integration's area on the plane XY on this picture



    Your answer must be 3\pi.
    Last edited by DeMath; March 15th 2009 at 11:15 AM.
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  4. #4
    Senior Member DeMath's Avatar
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    You can provide a projection area of integration in the XY-plane and thus

    S = {S_1} \cup 2{S_2}, where

    {S_1} = \left\{ {\left. {\left( {x,y} \right)} \right|{\text{ }}{x^2} + {y^2} \geqslant 4,{\text{ }}{x^2} + {y^2} \leqslant 25,{\text{ }}y \geqslant 0,{\text{ }} - 2 \leqslant x \leqslant 2,} \right\} = ^{}\left\{ {\left. {\left( {x,y} \right)} \right|{\text{ }}\sqrt {4 - {x^2}}  \leqslant {\text{ }}y \leqslant \sqrt {25 - {x^2}} ,{\text{ }} - 2 \leqslant x \leqslant 2} \right\}
    and
    {S_2} = \left\{ {\left. {\left( {x,y} \right)} \right|{\text{ }}{x^2} + {y^2} \leqslant 25,{\text{ }}y \geqslant 0,{\text{ }}2 \leqslant x \leqslant 5,} \right\} = ^{\text{ }}\left\{ {\left. {\left( {x,y} \right)} \right|{\text{ 0}} \leqslant y \leqslant \sqrt {25 - {x^2}} ,{\text{ }}2 \leqslant x \leqslant 5} \right\}.

    Then we have

    \iint\limits_S {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} = \iint\limits_{{S_1}} {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} + 2\iint\limits_{{S_2}} {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} = \int\limits_{ - 2}^2 {\int\limits_{\sqrt {4 - {x^2}} }^{\sqrt {25 - {x^2}} } {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }}} }  + 2\int\limits_2^5 {\int\limits_0^{\sqrt {25 - {x^2}} } {\frac{{dydx}}{{\sqrt {{x^2} + {y^2}} }} =  \ldots  = 3\pi } } .

    Look this picture

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