Why does the sign of the second items is negative












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Consider
$$int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy= int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$
My question is why the sign of the second item is negative?






calculus integration







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    -1












    $begingroup$


    Consider
    $$int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy= int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$
    My question is why the sign of the second item is negative?






    calculus integration







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$


      Consider
      $$int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy= int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$
      My question is why the sign of the second item is negative?






      calculus integration







      share|cite|improve this question











      $endgroup$




      Consider
      $$int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy= int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$
      My question is why the sign of the second item is negative?






      calculus integration




      calculus integration






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      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 22 '18 at 7:47









      Robert Z

      98.3k1067139




      98.3k1067139










      asked Dec 22 '18 at 6:42









      BloodpolyhydronBloodpolyhydron

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          2 Answers
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          $begingroup$

          In the integral on the $LHS$, in the region $pile xle2pi,-1le yle0,y$ ranges from $0tosin(x)$ for any value of $x$. In the corresponding term on the $RHS$, that is $displaystyleint_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx,y$ ranges from $-1to0$, which is the opposite direction. So we add a negative sign:$$-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx=int^{-1}_0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$






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            $begingroup$

            Here is the step-by-step solution:
            $$begin{align}int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy+int_{pi}^{2pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy=\
            &=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy-int_{pi}^{2pi}dxint_{color{blue}{sin(x)}}^0 f(x,y)dy=\
            &=int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx.end{align}$$
            Refer to the graph:



            enter image description here






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              2 Answers
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              2 Answers
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              0












              $begingroup$

              In the integral on the $LHS$, in the region $pile xle2pi,-1le yle0,y$ ranges from $0tosin(x)$ for any value of $x$. In the corresponding term on the $RHS$, that is $displaystyleint_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx,y$ ranges from $-1to0$, which is the opposite direction. So we add a negative sign:$$-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx=int^{-1}_0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                In the integral on the $LHS$, in the region $pile xle2pi,-1le yle0,y$ ranges from $0tosin(x)$ for any value of $x$. In the corresponding term on the $RHS$, that is $displaystyleint_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx,y$ ranges from $-1to0$, which is the opposite direction. So we add a negative sign:$$-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx=int^{-1}_0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  In the integral on the $LHS$, in the region $pile xle2pi,-1le yle0,y$ ranges from $0tosin(x)$ for any value of $x$. In the corresponding term on the $RHS$, that is $displaystyleint_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx,y$ ranges from $-1to0$, which is the opposite direction. So we add a negative sign:$$-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx=int^{-1}_0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$






                  share|cite|improve this answer









                  $endgroup$



                  In the integral on the $LHS$, in the region $pile xle2pi,-1le yle0,y$ ranges from $0tosin(x)$ for any value of $x$. In the corresponding term on the $RHS$, that is $displaystyleint_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx,y$ ranges from $-1to0$, which is the opposite direction. So we add a negative sign:$$-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx=int^{-1}_0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$







                  share|cite|improve this answer












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                  share|cite|improve this answer










                  answered Dec 22 '18 at 7:48









                  Shubham JohriShubham Johri

                  5,177717




                  5,177717























                      0












                      $begingroup$

                      Here is the step-by-step solution:
                      $$begin{align}int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy+int_{pi}^{2pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy=\
                      &=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy-int_{pi}^{2pi}dxint_{color{blue}{sin(x)}}^0 f(x,y)dy=\
                      &=int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx.end{align}$$
                      Refer to the graph:



                      enter image description here






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Here is the step-by-step solution:
                        $$begin{align}int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy+int_{pi}^{2pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy=\
                        &=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy-int_{pi}^{2pi}dxint_{color{blue}{sin(x)}}^0 f(x,y)dy=\
                        &=int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx.end{align}$$
                        Refer to the graph:



                        enter image description here






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Here is the step-by-step solution:
                          $$begin{align}int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy+int_{pi}^{2pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy=\
                          &=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy-int_{pi}^{2pi}dxint_{color{blue}{sin(x)}}^0 f(x,y)dy=\
                          &=int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx.end{align}$$
                          Refer to the graph:



                          enter image description here






                          share|cite|improve this answer









                          $endgroup$



                          Here is the step-by-step solution:
                          $$begin{align}int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy+int_{pi}^{2pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy=\
                          &=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy-int_{pi}^{2pi}dxint_{color{blue}{sin(x)}}^0 f(x,y)dy=\
                          &=int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx.end{align}$$
                          Refer to the graph:



                          enter image description here







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 22 '18 at 8:45









                          farruhotafarruhota

                          20.4k2739




                          20.4k2739






























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