Evaluating $iint_{R} |xy| dx dy$ over the region $R$ enclosed by $x^{2} + 4y^{2} ≥1 , x^{2} + y^{2}≤1$.












0












$begingroup$


I am given the following integral -



$iint_{R} |xy| dx dy$ over the region $R$ enclosed by $x^{2} + 4y^{2} ≥1 , x^{2} + y^{2}≤1.$



I know how do we solve such problems, but I have a bit confusion.
The area is enclosed by an ellipse and circle. I calculated the value of integral in first quadrant by assuming $|xy|$ simply $xy$, then I multiplied it by 4 and got my answer right. However, I want to know how does this integral, when my function is $|xy|$, is different from the integral taking function as $xy$. Why I am getting same results in both cases.






calculus







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  • 1




    $begingroup$
    Consider simpler example: compare $int_{-1}^1 x dx$ and $int_{-1}^1 |x| dx$.
    $endgroup$
    – farruhota
    Jan 2 at 11:16
















0












$begingroup$


I am given the following integral -



$iint_{R} |xy| dx dy$ over the region $R$ enclosed by $x^{2} + 4y^{2} ≥1 , x^{2} + y^{2}≤1.$



I know how do we solve such problems, but I have a bit confusion.
The area is enclosed by an ellipse and circle. I calculated the value of integral in first quadrant by assuming $|xy|$ simply $xy$, then I multiplied it by 4 and got my answer right. However, I want to know how does this integral, when my function is $|xy|$, is different from the integral taking function as $xy$. Why I am getting same results in both cases.






calculus







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Consider simpler example: compare $int_{-1}^1 x dx$ and $int_{-1}^1 |x| dx$.
    $endgroup$
    – farruhota
    Jan 2 at 11:16














0












0








0





$begingroup$


I am given the following integral -



$iint_{R} |xy| dx dy$ over the region $R$ enclosed by $x^{2} + 4y^{2} ≥1 , x^{2} + y^{2}≤1.$



I know how do we solve such problems, but I have a bit confusion.
The area is enclosed by an ellipse and circle. I calculated the value of integral in first quadrant by assuming $|xy|$ simply $xy$, then I multiplied it by 4 and got my answer right. However, I want to know how does this integral, when my function is $|xy|$, is different from the integral taking function as $xy$. Why I am getting same results in both cases.






calculus







share|cite|improve this question









$endgroup$




I am given the following integral -



$iint_{R} |xy| dx dy$ over the region $R$ enclosed by $x^{2} + 4y^{2} ≥1 , x^{2} + y^{2}≤1.$



I know how do we solve such problems, but I have a bit confusion.
The area is enclosed by an ellipse and circle. I calculated the value of integral in first quadrant by assuming $|xy|$ simply $xy$, then I multiplied it by 4 and got my answer right. However, I want to know how does this integral, when my function is $|xy|$, is different from the integral taking function as $xy$. Why I am getting same results in both cases.






calculus




calculus






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asked Jan 2 at 8:21









MathsaddictMathsaddict

3669




3669








  • 1




    $begingroup$
    Consider simpler example: compare $int_{-1}^1 x dx$ and $int_{-1}^1 |x| dx$.
    $endgroup$
    – farruhota
    Jan 2 at 11:16














  • 1




    $begingroup$
    Consider simpler example: compare $int_{-1}^1 x dx$ and $int_{-1}^1 |x| dx$.
    $endgroup$
    – farruhota
    Jan 2 at 11:16








1




1




$begingroup$
Consider simpler example: compare $int_{-1}^1 x dx$ and $int_{-1}^1 |x| dx$.
$endgroup$
– farruhota
Jan 2 at 11:16




$begingroup$
Consider simpler example: compare $int_{-1}^1 x dx$ and $int_{-1}^1 |x| dx$.
$endgroup$
– farruhota
Jan 2 at 11:16










2 Answers
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$begingroup$

If there is no absolute value, by symmetry, the integral in the first quadrant is equal to the integral of the third quadrant (as $xy = (-x)(-y)$). However, the integral in the second and the fourth is going to be its negative.



Hence, they will cancel out and the answer for the problem without the absolute value sign is $0$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    In the first quadrant it is the same, as x and y are both positive and so the absolute value doesn't do anything. This is true in the third quadrant as well as x and y are both negative and so xy is positive. But in the second and fourth quadrants $|xy| = -xy$, so it matters if the absolute value signs are there. If you integrate xy over this area than the integral is the same in the first and third quadrants, and negative that value in the second and fourth quadrants, giving an answer of 0.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

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      2












      $begingroup$

      If there is no absolute value, by symmetry, the integral in the first quadrant is equal to the integral of the third quadrant (as $xy = (-x)(-y)$). However, the integral in the second and the fourth is going to be its negative.



      Hence, they will cancel out and the answer for the problem without the absolute value sign is $0$.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        If there is no absolute value, by symmetry, the integral in the first quadrant is equal to the integral of the third quadrant (as $xy = (-x)(-y)$). However, the integral in the second and the fourth is going to be its negative.



        Hence, they will cancel out and the answer for the problem without the absolute value sign is $0$.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          If there is no absolute value, by symmetry, the integral in the first quadrant is equal to the integral of the third quadrant (as $xy = (-x)(-y)$). However, the integral in the second and the fourth is going to be its negative.



          Hence, they will cancel out and the answer for the problem without the absolute value sign is $0$.






          share|cite|improve this answer











          $endgroup$



          If there is no absolute value, by symmetry, the integral in the first quadrant is equal to the integral of the third quadrant (as $xy = (-x)(-y)$). However, the integral in the second and the fourth is going to be its negative.



          Hence, they will cancel out and the answer for the problem without the absolute value sign is $0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 2 at 8:35









          Sauhard Sharma

          953318




          953318










          answered Jan 2 at 8:32









          Siong Thye GohSiong Thye Goh

          104k1468120




          104k1468120























              1












              $begingroup$

              In the first quadrant it is the same, as x and y are both positive and so the absolute value doesn't do anything. This is true in the third quadrant as well as x and y are both negative and so xy is positive. But in the second and fourth quadrants $|xy| = -xy$, so it matters if the absolute value signs are there. If you integrate xy over this area than the integral is the same in the first and third quadrants, and negative that value in the second and fourth quadrants, giving an answer of 0.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                In the first quadrant it is the same, as x and y are both positive and so the absolute value doesn't do anything. This is true in the third quadrant as well as x and y are both negative and so xy is positive. But in the second and fourth quadrants $|xy| = -xy$, so it matters if the absolute value signs are there. If you integrate xy over this area than the integral is the same in the first and third quadrants, and negative that value in the second and fourth quadrants, giving an answer of 0.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  In the first quadrant it is the same, as x and y are both positive and so the absolute value doesn't do anything. This is true in the third quadrant as well as x and y are both negative and so xy is positive. But in the second and fourth quadrants $|xy| = -xy$, so it matters if the absolute value signs are there. If you integrate xy over this area than the integral is the same in the first and third quadrants, and negative that value in the second and fourth quadrants, giving an answer of 0.






                  share|cite|improve this answer









                  $endgroup$



                  In the first quadrant it is the same, as x and y are both positive and so the absolute value doesn't do anything. This is true in the third quadrant as well as x and y are both negative and so xy is positive. But in the second and fourth quadrants $|xy| = -xy$, so it matters if the absolute value signs are there. If you integrate xy over this area than the integral is the same in the first and third quadrants, and negative that value in the second and fourth quadrants, giving an answer of 0.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 2 at 8:33









                  Erik ParkinsonErik Parkinson

                  1,17519




                  1,17519






























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