Compute ∬ D x 2 1 + x 2 + y 2 d x d ywhere D = { ( x , y ) | x 2 + y 2 ≤ 1 , x ≥ 0 }

Question

Compute      ∬    D              x      2              1      +              x        2            +              y        2              d  x  d  ywhere   D  =  {  (  x  ,  y  )      |        x    2    +      y    2    ≤  1  ,  x  ≥  0  }

exhumatql · Accepted Answer

∬          D                                      x            2                                1            +                          x              2                        +                          y              2                                              d        x                d        y                            =                  ∫          0          1                          ∫                      −                          π              2                                                          π              2                                                                          r              3                                      cos              2                        ⁡            (            θ            )                                1            +                          r              2                                              d        θ                d        r                                          =                  ∫          0          1                                      r            3                                1            +                          r              2                                              d        r                  ∫                      −                          π              2                                                          π              2                                                            cos            ⁡            (            2            θ            )            +            1                    2                        d        θ            Now, we have      ∫          −              π        2                            π        2                        cos      ⁡      (      2      θ      )      +      1        2      d  θ  =      π    2  and                              ∫          0          1                                      r            3                                1            +                          r              2                                              d        r                            =                  1          2                          ∫          0          1                          u                      1            +            u                                  d        u                    (        u        =                  r          2                )                                          =                  1          2                          ∫          1          2                                      s            −            1                                s                                  d        s                    (        s        =        1        +        u        )                                          =                  1          2                          ∫          1          2                          s                                    1              2                                      −                  s                      −                          1              2                                              d        s                                          =                  2          3                (        2        −                  2                )            giving a final answer of             π      (      2      −              2            )        3  . There might be some mistake which I&#039;m not aware of which causes the answer not to be the one you claimed.

Compute ∬ D </msub> x 2 </msup> <msqrt>

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