Double integral change of variables to polar coordinates.












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$begingroup$


Sketch the region represented in the following double integral:
$I =∬ln(x^2+y^2)dxdy$



In the region:



$x:ycotβ$ to $sqrt(a^2+y^2)$



$y:0$ to $asinβ$



where a > 0 and 0 < β < π/2.



By changing to polar coordinates or otherwise, show that
I = $a^2β(ln a − 1/2)$.



So far I have found that the sketch yields a triangle inscribed in the circle of radius $a$ in the first quadrant. I am having trouble converting to polar coordinates.










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    0












    $begingroup$


    Sketch the region represented in the following double integral:
    $I =∬ln(x^2+y^2)dxdy$



    In the region:



    $x:ycotβ$ to $sqrt(a^2+y^2)$



    $y:0$ to $asinβ$



    where a > 0 and 0 < β < π/2.



    By changing to polar coordinates or otherwise, show that
    I = $a^2β(ln a − 1/2)$.



    So far I have found that the sketch yields a triangle inscribed in the circle of radius $a$ in the first quadrant. I am having trouble converting to polar coordinates.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Sketch the region represented in the following double integral:
      $I =∬ln(x^2+y^2)dxdy$



      In the region:



      $x:ycotβ$ to $sqrt(a^2+y^2)$



      $y:0$ to $asinβ$



      where a > 0 and 0 < β < π/2.



      By changing to polar coordinates or otherwise, show that
      I = $a^2β(ln a − 1/2)$.



      So far I have found that the sketch yields a triangle inscribed in the circle of radius $a$ in the first quadrant. I am having trouble converting to polar coordinates.










      share|cite|improve this question









      $endgroup$




      Sketch the region represented in the following double integral:
      $I =∬ln(x^2+y^2)dxdy$



      In the region:



      $x:ycotβ$ to $sqrt(a^2+y^2)$



      $y:0$ to $asinβ$



      where a > 0 and 0 < β < π/2.



      By changing to polar coordinates or otherwise, show that
      I = $a^2β(ln a − 1/2)$.



      So far I have found that the sketch yields a triangle inscribed in the circle of radius $a$ in the first quadrant. I am having trouble converting to polar coordinates.







      calculus integration multivariable-calculus






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      asked Feb 3 at 11:29









      matmat

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          $begingroup$

          Hint: In the polar coordinate system $x^2+y^2 Rightarrow r^2$ and $dxdy Rightarrow rcdot drdtheta$. Also, we have that $y=rcdot sin{theta}$ and $x=rcdot cos{theta}$.






          share|cite|improve this answer









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          • $begingroup$
            My problem is finding the limits for $r$ and theta.
            $endgroup$
            – mat
            Feb 3 at 12:09












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          1 Answer
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          0












          $begingroup$

          Hint: In the polar coordinate system $x^2+y^2 Rightarrow r^2$ and $dxdy Rightarrow rcdot drdtheta$. Also, we have that $y=rcdot sin{theta}$ and $x=rcdot cos{theta}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            My problem is finding the limits for $r$ and theta.
            $endgroup$
            – mat
            Feb 3 at 12:09
















          0












          $begingroup$

          Hint: In the polar coordinate system $x^2+y^2 Rightarrow r^2$ and $dxdy Rightarrow rcdot drdtheta$. Also, we have that $y=rcdot sin{theta}$ and $x=rcdot cos{theta}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            My problem is finding the limits for $r$ and theta.
            $endgroup$
            – mat
            Feb 3 at 12:09














          0












          0








          0





          $begingroup$

          Hint: In the polar coordinate system $x^2+y^2 Rightarrow r^2$ and $dxdy Rightarrow rcdot drdtheta$. Also, we have that $y=rcdot sin{theta}$ and $x=rcdot cos{theta}$.






          share|cite|improve this answer









          $endgroup$



          Hint: In the polar coordinate system $x^2+y^2 Rightarrow r^2$ and $dxdy Rightarrow rcdot drdtheta$. Also, we have that $y=rcdot sin{theta}$ and $x=rcdot cos{theta}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 3 at 11:59









          Peter ForemanPeter Foreman

          7,5371320




          7,5371320












          • $begingroup$
            My problem is finding the limits for $r$ and theta.
            $endgroup$
            – mat
            Feb 3 at 12:09


















          • $begingroup$
            My problem is finding the limits for $r$ and theta.
            $endgroup$
            – mat
            Feb 3 at 12:09
















          $begingroup$
          My problem is finding the limits for $r$ and theta.
          $endgroup$
          – mat
          Feb 3 at 12:09




          $begingroup$
          My problem is finding the limits for $r$ and theta.
          $endgroup$
          – mat
          Feb 3 at 12:09


















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