With the help of a suitable transformation and Fubini I want to determine the integral ∫ V x 3 y d λ 2 ( x , y ) ,where V is the open subset of R + 2 bounded by the following curves: x 2 + y 2 = 4 x 2 − y 2 = 2 x 2 − y 2 = 1 I know how to do that. The only problem is finding V . Is it V = { ( x , y ) ∈ R 2 : 1 &lt; x 2 − y 2 &lt; 2 , 0 &lt; x 2 + y 2 &lt; 4 }Because then I set x 2 + y 2 = v x 2 − y 2 = u and get ( x , y ) = ( 1 / 2 ( v − u ) , 1 / 2 ( u + v ) ) So either my transformation is wrong or the limits of the intervalls I chose.Thanks for any kind of help.

Question

With the help of a suitable transformation and Fubini I want to determine the integral      ∫          V            x          3        y  d      λ          2        (  x  ,  y  )  ,where   V is the open subset of             R              +              2       bounded by the following curves:                                    x                      2                          +                  y                      2                          =        4                                            x                      2                          −                  y                      2                          =        2                                            x                      2                          −                  y                      2                          =        1            I know how to do that. The only problem is finding   V  . Is it  V  =  {  (  x  ,  y  )  ∈            R        2    :  1  &amp;lt;      x          2        −      y          2        &amp;lt;  2  ,  0  &amp;lt;      x          2        +      y          2        &amp;lt;  4  }Because then I set                                    x                      2                          +                  y                      2                          =        v                                            x                      2                          −                  y                      2                          =        u            and get  (  x  ,  y  )  =      (          1              /            2      (      v      −      u        )    ,          1              /            2      (      u      +      v      )        )  So either my transformation is wrong or the limits of the intervalls I chose.Thanks for any kind of help.

timmeraared · Accepted Answer

We have  V  =      {    (    x    ,    y    )    ∈                  R            +              2              :    x    ,    y    &amp;gt;    0    ,          x      2        +          y      2        &amp;lt;    4    ,    1    &amp;lt;          x      2        −          y      2        &amp;lt;    2    }    .Let us consider the new variables  u  =      x    2    +      y    2    ,    v  =      x    2    −      y    2  and the corresponding figure  (  u  ,  v  )  =  Ψ  (  x  ,  y  )  :=      (          x      2        +          y      2        ,          x      2        −          y      2        )    .Obviously we have  U  :=  Ψ  (  V  )  =      {    (    u    ,    v    )    ∈                  R            +              2              :    1    &amp;lt;    v    &amp;lt;    2    ,    v    &amp;lt;    u    &amp;lt;    4    }    .The mapping   Ψ  :  V  →  U is continuously differentiable and the inverse mapping   Φ  =      Ψ          −      1       exists and looks like this:  (  x  ,  y  )  =  Φ  (  u  ,  v  )  =      (                            u          +          v                2              ,                            u          −          v                2              )    .The mapping   Φ  :  U  →  V is also continuously differentiable and hence a diffeomorphism. We have      Φ          ′        =      (                                        ∂                          u                                            Φ                          1                                                            ∂                          v                                            Φ                          1                                                                        ∂                          u                                            Φ                          2                                                            ∂                          v                                            Φ                          2                                            )    =      1          2              2                  (                                                      1                              u                +                v                                                                                        1                              u                +                v                                                                                                    1                              u                −                v                                                              −                                    1                              u                −                v                                                          )  and  det  (      Φ          ′        )  =  −      1    u    ⋅      1                  u        2            −              v        2              .Using the transformation theorem and Fubini&#039;s theorem, we obtain                                            ∫                      V                                    x                      3                          y                d                  λ                      2                          (        x        ,        y        )        =                  ∫                      U                                    x                      3                          y                          |                det        (                  Φ                      ′                          )                  |                        d                  λ                      2                          (        u        ,        v        )        =                  ∫                      U                                    1          4                                      (                                                            u                  +                  v                                2                                      )                                3                                                              u              −              v                        2                                    1                                    u              2                        −                          v              2                                              d                  λ                      2                          (        u        ,        v        )                                          =                  ∫          1          2                          (                      ∫            v            4                                                                                u                  −                  v                                                                      (                    u                    +                    v                    )                                                        3                    2                                                                              16                                                      u                    2                                    −                                      v                    2                                                                                d          u          )                d        v        =                  ∫                      1                                2                                                [            −                                                            3                                      v                    2                                    −                  8                  v                  −                  16                                32                                      ]                                v                                4                          d        v        =                              [            −                                                            v                                      (                                          v                      2                                        −                    4                    v                    −                    16                    )                                                  32                                      ]                    1                      2                          =                              21            32

With the help of a suitable transformation and Fubini I want to determine the integral &#

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