Trying to solve this triple integral: $iiint (x-1)(y-1) ,dx,dy,dz$












3












$begingroup$


Here's the question
$$iiint (x-1)(y-1) ,dx,dy,dz.$$
I am asked to evaluate this integral over the region $$D:=left { (x,y,z) inmathbb{R}^3 :x^2+y^2 leq z leq 2x+2y+2 right }.$$
There are the bounds of integration in set D (the variable $z$ is isolated) and well I tried to find the solution of this integral :



begin{align*}
&iint_{Pr_{(y,x)}(D)}int_{x^2+y^2}^{2x+2y+2}(x-1)(y-1) ,dx, dy, dz \
=&iint_{Pr_{(y,x)}(D)}int_{x^2+y^2}^{2x+2y+2}(xy-x-y+1) ,dx, dy, dz,
end{align*}
and integrate only with respect to $z.$ I have that:
begin{align*}
int_{x^2+y^2}^{2x+2y+2}(xy-x-y+1) ,dz&=(xy-x-y+1)*(2x+2y+2-(x^2+y^2)) \
&=3x^2y+3xy^2-2x^3y-2xy^3-3x^2+x^3-2xy-3y^2+y^3+2.
end{align*}



It looks like this way is too long. The second thing that came to mind when I saw the set $D$ was to apply cylindrical coordinates, but this doesn't make easier the left member of the set $D.$
What can I do or what have I done wrong up until now?



Any support for this question would be appreciated.






integration multivariable-calculus definite-integrals spherical-coordinates







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Here's the question
    $$iiint (x-1)(y-1) ,dx,dy,dz.$$
    I am asked to evaluate this integral over the region $$D:=left { (x,y,z) inmathbb{R}^3 :x^2+y^2 leq z leq 2x+2y+2 right }.$$
    There are the bounds of integration in set D (the variable $z$ is isolated) and well I tried to find the solution of this integral :



    begin{align*}
    &iint_{Pr_{(y,x)}(D)}int_{x^2+y^2}^{2x+2y+2}(x-1)(y-1) ,dx, dy, dz \
    =&iint_{Pr_{(y,x)}(D)}int_{x^2+y^2}^{2x+2y+2}(xy-x-y+1) ,dx, dy, dz,
    end{align*}
    and integrate only with respect to $z.$ I have that:
    begin{align*}
    int_{x^2+y^2}^{2x+2y+2}(xy-x-y+1) ,dz&=(xy-x-y+1)*(2x+2y+2-(x^2+y^2)) \
    &=3x^2y+3xy^2-2x^3y-2xy^3-3x^2+x^3-2xy-3y^2+y^3+2.
    end{align*}



    It looks like this way is too long. The second thing that came to mind when I saw the set $D$ was to apply cylindrical coordinates, but this doesn't make easier the left member of the set $D.$
    What can I do or what have I done wrong up until now?



    Any support for this question would be appreciated.






    integration multivariable-calculus definite-integrals spherical-coordinates







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Here's the question
      $$iiint (x-1)(y-1) ,dx,dy,dz.$$
      I am asked to evaluate this integral over the region $$D:=left { (x,y,z) inmathbb{R}^3 :x^2+y^2 leq z leq 2x+2y+2 right }.$$
      There are the bounds of integration in set D (the variable $z$ is isolated) and well I tried to find the solution of this integral :



      begin{align*}
      &iint_{Pr_{(y,x)}(D)}int_{x^2+y^2}^{2x+2y+2}(x-1)(y-1) ,dx, dy, dz \
      =&iint_{Pr_{(y,x)}(D)}int_{x^2+y^2}^{2x+2y+2}(xy-x-y+1) ,dx, dy, dz,
      end{align*}
      and integrate only with respect to $z.$ I have that:
      begin{align*}
      int_{x^2+y^2}^{2x+2y+2}(xy-x-y+1) ,dz&=(xy-x-y+1)*(2x+2y+2-(x^2+y^2)) \
      &=3x^2y+3xy^2-2x^3y-2xy^3-3x^2+x^3-2xy-3y^2+y^3+2.
      end{align*}



      It looks like this way is too long. The second thing that came to mind when I saw the set $D$ was to apply cylindrical coordinates, but this doesn't make easier the left member of the set $D.$
      What can I do or what have I done wrong up until now?



      Any support for this question would be appreciated.






      integration multivariable-calculus definite-integrals spherical-coordinates







      share|cite|improve this question











      $endgroup$




      Here's the question
      $$iiint (x-1)(y-1) ,dx,dy,dz.$$
      I am asked to evaluate this integral over the region $$D:=left { (x,y,z) inmathbb{R}^3 :x^2+y^2 leq z leq 2x+2y+2 right }.$$
      There are the bounds of integration in set D (the variable $z$ is isolated) and well I tried to find the solution of this integral :



      begin{align*}
      &iint_{Pr_{(y,x)}(D)}int_{x^2+y^2}^{2x+2y+2}(x-1)(y-1) ,dx, dy, dz \
      =&iint_{Pr_{(y,x)}(D)}int_{x^2+y^2}^{2x+2y+2}(xy-x-y+1) ,dx, dy, dz,
      end{align*}
      and integrate only with respect to $z.$ I have that:
      begin{align*}
      int_{x^2+y^2}^{2x+2y+2}(xy-x-y+1) ,dz&=(xy-x-y+1)*(2x+2y+2-(x^2+y^2)) \
      &=3x^2y+3xy^2-2x^3y-2xy^3-3x^2+x^3-2xy-3y^2+y^3+2.
      end{align*}



      It looks like this way is too long. The second thing that came to mind when I saw the set $D$ was to apply cylindrical coordinates, but this doesn't make easier the left member of the set $D.$
      What can I do or what have I done wrong up until now?



      Any support for this question would be appreciated.






      integration multivariable-calculus definite-integrals spherical-coordinates




      integration multivariable-calculus definite-integrals spherical-coordinates






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 17 at 19:14









      Adrian Keister

      5,28571933




      5,28571933










      asked Jan 17 at 18:54









      andrewandrew

      698




      698






















          1 Answer
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          $begingroup$

          Hint. Note that the intersection of paraboloid $z=x^2+y^2$ and the plane $z=2x+2y+2$ projected onto $xy$-plane is given by the circle $x^2+y^2=2x+2y+2$ that is
          $$(x-1)^2+(y-1)^2=2^2.$$
          Hence the given triple integral becomes
          $$iint_{(x-1)^2+(y-1)^2leq 2^2}int_{x^2+y^2}^{2x+2y+2}(x-1)(y-1) dx dy dz,$$
          and after integrating with respect to $z$ we get
          $$iint_{(x-1)^2+(y-1)^2leq 2^2}(x-1)(y-1)(4-(x-1)^2-(y-1)^2) dx dy.$$
          Now use the polar coordinates centered at $(1,1)$. Can you take it from here?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes i understood
            $endgroup$
            – andrew
            Jan 17 at 19:34






          • 1




            $begingroup$
            @andrew OK. BTW I am happy that you appreciated my previous comment.
            $endgroup$
            – Robert Z
            Jan 17 at 19:41











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          2












          $begingroup$

          Hint. Note that the intersection of paraboloid $z=x^2+y^2$ and the plane $z=2x+2y+2$ projected onto $xy$-plane is given by the circle $x^2+y^2=2x+2y+2$ that is
          $$(x-1)^2+(y-1)^2=2^2.$$
          Hence the given triple integral becomes
          $$iint_{(x-1)^2+(y-1)^2leq 2^2}int_{x^2+y^2}^{2x+2y+2}(x-1)(y-1) dx dy dz,$$
          and after integrating with respect to $z$ we get
          $$iint_{(x-1)^2+(y-1)^2leq 2^2}(x-1)(y-1)(4-(x-1)^2-(y-1)^2) dx dy.$$
          Now use the polar coordinates centered at $(1,1)$. Can you take it from here?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes i understood
            $endgroup$
            – andrew
            Jan 17 at 19:34






          • 1




            $begingroup$
            @andrew OK. BTW I am happy that you appreciated my previous comment.
            $endgroup$
            – Robert Z
            Jan 17 at 19:41
















          2












          $begingroup$

          Hint. Note that the intersection of paraboloid $z=x^2+y^2$ and the plane $z=2x+2y+2$ projected onto $xy$-plane is given by the circle $x^2+y^2=2x+2y+2$ that is
          $$(x-1)^2+(y-1)^2=2^2.$$
          Hence the given triple integral becomes
          $$iint_{(x-1)^2+(y-1)^2leq 2^2}int_{x^2+y^2}^{2x+2y+2}(x-1)(y-1) dx dy dz,$$
          and after integrating with respect to $z$ we get
          $$iint_{(x-1)^2+(y-1)^2leq 2^2}(x-1)(y-1)(4-(x-1)^2-(y-1)^2) dx dy.$$
          Now use the polar coordinates centered at $(1,1)$. Can you take it from here?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yes i understood
            $endgroup$
            – andrew
            Jan 17 at 19:34






          • 1




            $begingroup$
            @andrew OK. BTW I am happy that you appreciated my previous comment.
            $endgroup$
            – Robert Z
            Jan 17 at 19:41














          2












          2








          2





          $begingroup$

          Hint. Note that the intersection of paraboloid $z=x^2+y^2$ and the plane $z=2x+2y+2$ projected onto $xy$-plane is given by the circle $x^2+y^2=2x+2y+2$ that is
          $$(x-1)^2+(y-1)^2=2^2.$$
          Hence the given triple integral becomes
          $$iint_{(x-1)^2+(y-1)^2leq 2^2}int_{x^2+y^2}^{2x+2y+2}(x-1)(y-1) dx dy dz,$$
          and after integrating with respect to $z$ we get
          $$iint_{(x-1)^2+(y-1)^2leq 2^2}(x-1)(y-1)(4-(x-1)^2-(y-1)^2) dx dy.$$
          Now use the polar coordinates centered at $(1,1)$. Can you take it from here?






          share|cite|improve this answer











          $endgroup$



          Hint. Note that the intersection of paraboloid $z=x^2+y^2$ and the plane $z=2x+2y+2$ projected onto $xy$-plane is given by the circle $x^2+y^2=2x+2y+2$ that is
          $$(x-1)^2+(y-1)^2=2^2.$$
          Hence the given triple integral becomes
          $$iint_{(x-1)^2+(y-1)^2leq 2^2}int_{x^2+y^2}^{2x+2y+2}(x-1)(y-1) dx dy dz,$$
          and after integrating with respect to $z$ we get
          $$iint_{(x-1)^2+(y-1)^2leq 2^2}(x-1)(y-1)(4-(x-1)^2-(y-1)^2) dx dy.$$
          Now use the polar coordinates centered at $(1,1)$. Can you take it from here?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 17 at 19:17

























          answered Jan 17 at 19:11









          Robert ZRobert Z

          97.1k1066137




          97.1k1066137












          • $begingroup$
            Yes i understood
            $endgroup$
            – andrew
            Jan 17 at 19:34






          • 1




            $begingroup$
            @andrew OK. BTW I am happy that you appreciated my previous comment.
            $endgroup$
            – Robert Z
            Jan 17 at 19:41


















          • $begingroup$
            Yes i understood
            $endgroup$
            – andrew
            Jan 17 at 19:34






          • 1




            $begingroup$
            @andrew OK. BTW I am happy that you appreciated my previous comment.
            $endgroup$
            – Robert Z
            Jan 17 at 19:41
















          $begingroup$
          Yes i understood
          $endgroup$
          – andrew
          Jan 17 at 19:34




          $begingroup$
          Yes i understood
          $endgroup$
          – andrew
          Jan 17 at 19:34




          1




          1




          $begingroup$
          @andrew OK. BTW I am happy that you appreciated my previous comment.
          $endgroup$
          – Robert Z
          Jan 17 at 19:41




          $begingroup$
          @andrew OK. BTW I am happy that you appreciated my previous comment.
          $endgroup$
          – Robert Z
          Jan 17 at 19:41


















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