Compute $int int(a^2-x^2) dx dy$ taken over half the circle $x^2+y^2=a^2$ in the positive quadrant.











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Since it is given as positive quadrant, I took the limits as follows,
w.r.t. x from 0 to a & w.r.t y from 0 to root of $$(a^2-x^2)$$.
I proceeded with that & finally substituted $x=asin(theta)$. The answer which I got was $(3(a^4)π)/16$. But the actual answer is $(3a^4)/8$. I don't know where I am making a mistake.






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    Since it is given as positive quadrant, I took the limits as follows,
    w.r.t. x from 0 to a & w.r.t y from 0 to root of $$(a^2-x^2)$$.
    I proceeded with that & finally substituted $x=asin(theta)$. The answer which I got was $(3(a^4)π)/16$. But the actual answer is $(3a^4)/8$. I don't know where I am making a mistake.






    integration







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      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Since it is given as positive quadrant, I took the limits as follows,
      w.r.t. x from 0 to a & w.r.t y from 0 to root of $$(a^2-x^2)$$.
      I proceeded with that & finally substituted $x=asin(theta)$. The answer which I got was $(3(a^4)π)/16$. But the actual answer is $(3a^4)/8$. I don't know where I am making a mistake.






      integration







      share|cite|improve this question















      Since it is given as positive quadrant, I took the limits as follows,
      w.r.t. x from 0 to a & w.r.t y from 0 to root of $$(a^2-x^2)$$.
      I proceeded with that & finally substituted $x=asin(theta)$. The answer which I got was $(3(a^4)π)/16$. But the actual answer is $(3a^4)/8$. I don't know where I am making a mistake.






      integration




      integration






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      edited 12 hours ago









      David G. Stork

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      asked 13 hours ago









      Renuka

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          The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.






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            $$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$






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              The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.






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                The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.






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                  The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.






                  share|cite|improve this answer














                  The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.







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                  edited 13 hours ago

























                  answered 13 hours ago









                  Kavi Rama Murthy

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                      $$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$






                      share|cite|improve this answer



























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                        $$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$






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                          $$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$






                          share|cite|improve this answer














                          $$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$







                          share|cite|improve this answer














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                          edited 12 hours ago

























                          answered 13 hours ago









                          David G. Stork

                          8,90521232




                          8,90521232






























                               

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