Integration limits in a triple integral












1












$begingroup$


I'm having trouble describing a region in space.



Let $B$ the region bounded by the planes $x=0$, $y=0$, $z=0$, $x+y=1$, $z=x+y$. One way of calculating the volume of B is:



begin{equation}
V(B)=int_{0}^{1} int_{0}^{1-x} int_{0}^{x+y}dzdydx=frac{1}{3}
end{equation}



I'm trying to change the order to integration to $int int int dxdydz$, but I can't get my integration limits to work (I get different values for the volume).



I thought that $(0<z<1)$, and $(z-y<x<1-y)$, but I can't find the limits on the Y axis. I believe it sould be something like:



begin{equation}
int_0^1int_?^?int_{z-y}^{1-y}dxdydz
end{equation}






integration multivariable-calculus definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why would you want to switch to that order of integration?
    $endgroup$
    – StackTD
    Dec 5 '18 at 12:11






  • 1




    $begingroup$
    The book doesn't ask the reader to change the order of integration, but I'm just curious. Also, I'm studying, since I find region description to be kind of hard.
    $endgroup$
    – IchVerloren
    Dec 5 '18 at 12:35










  • $begingroup$
    It's very fiddly, that, as you'll have to find the intersection of the two slanted planes & divide the region up into parts having certain of their edges along that line of intersection.
    $endgroup$
    – AmbretteOrrisey
    Dec 5 '18 at 14:23










  • $begingroup$
    The plane z=x+y cuts off a pyramid from the unit box having vertices 000, 001, 101, 011 (ie containing the corner 001); and the plane x+y=1 cuts the unit box into two triangular prisms aligned along the z-axis & sharing vertices 010, 100, 011, 101.
    $endgroup$
    – AmbretteOrrisey
    Dec 5 '18 at 14:34












  • $begingroup$
    It's relatively easy to just see it geometrically: the plane x+y=1 cuts the cube equally into two, so that you're left with half it's volume; and then cutting-off that corner by the plane z=x+y takes-off a volume ⅙of the unit cube from that; so you're left with ⅓. But then ... how to formulate that as the limits of a triple-integral ... !?
    $endgroup$
    – AmbretteOrrisey
    Dec 5 '18 at 14:45


















1












$begingroup$


I'm having trouble describing a region in space.



Let $B$ the region bounded by the planes $x=0$, $y=0$, $z=0$, $x+y=1$, $z=x+y$. One way of calculating the volume of B is:



begin{equation}
V(B)=int_{0}^{1} int_{0}^{1-x} int_{0}^{x+y}dzdydx=frac{1}{3}
end{equation}



I'm trying to change the order to integration to $int int int dxdydz$, but I can't get my integration limits to work (I get different values for the volume).



I thought that $(0<z<1)$, and $(z-y<x<1-y)$, but I can't find the limits on the Y axis. I believe it sould be something like:



begin{equation}
int_0^1int_?^?int_{z-y}^{1-y}dxdydz
end{equation}






integration multivariable-calculus definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why would you want to switch to that order of integration?
    $endgroup$
    – StackTD
    Dec 5 '18 at 12:11






  • 1




    $begingroup$
    The book doesn't ask the reader to change the order of integration, but I'm just curious. Also, I'm studying, since I find region description to be kind of hard.
    $endgroup$
    – IchVerloren
    Dec 5 '18 at 12:35










  • $begingroup$
    It's very fiddly, that, as you'll have to find the intersection of the two slanted planes & divide the region up into parts having certain of their edges along that line of intersection.
    $endgroup$
    – AmbretteOrrisey
    Dec 5 '18 at 14:23










  • $begingroup$
    The plane z=x+y cuts off a pyramid from the unit box having vertices 000, 001, 101, 011 (ie containing the corner 001); and the plane x+y=1 cuts the unit box into two triangular prisms aligned along the z-axis & sharing vertices 010, 100, 011, 101.
    $endgroup$
    – AmbretteOrrisey
    Dec 5 '18 at 14:34












  • $begingroup$
    It's relatively easy to just see it geometrically: the plane x+y=1 cuts the cube equally into two, so that you're left with half it's volume; and then cutting-off that corner by the plane z=x+y takes-off a volume ⅙of the unit cube from that; so you're left with ⅓. But then ... how to formulate that as the limits of a triple-integral ... !?
    $endgroup$
    – AmbretteOrrisey
    Dec 5 '18 at 14:45
















1












1








1


1



$begingroup$


I'm having trouble describing a region in space.



Let $B$ the region bounded by the planes $x=0$, $y=0$, $z=0$, $x+y=1$, $z=x+y$. One way of calculating the volume of B is:



begin{equation}
V(B)=int_{0}^{1} int_{0}^{1-x} int_{0}^{x+y}dzdydx=frac{1}{3}
end{equation}



I'm trying to change the order to integration to $int int int dxdydz$, but I can't get my integration limits to work (I get different values for the volume).



I thought that $(0<z<1)$, and $(z-y<x<1-y)$, but I can't find the limits on the Y axis. I believe it sould be something like:



begin{equation}
int_0^1int_?^?int_{z-y}^{1-y}dxdydz
end{equation}






integration multivariable-calculus definite-integrals







share|cite|improve this question











$endgroup$




I'm having trouble describing a region in space.



Let $B$ the region bounded by the planes $x=0$, $y=0$, $z=0$, $x+y=1$, $z=x+y$. One way of calculating the volume of B is:



begin{equation}
V(B)=int_{0}^{1} int_{0}^{1-x} int_{0}^{x+y}dzdydx=frac{1}{3}
end{equation}



I'm trying to change the order to integration to $int int int dxdydz$, but I can't get my integration limits to work (I get different values for the volume).



I thought that $(0<z<1)$, and $(z-y<x<1-y)$, but I can't find the limits on the Y axis. I believe it sould be something like:



begin{equation}
int_0^1int_?^?int_{z-y}^{1-y}dxdydz
end{equation}






integration multivariable-calculus definite-integrals




integration multivariable-calculus definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 11:07







IchVerloren

















asked Dec 5 '18 at 10:30









IchVerlorenIchVerloren

969




969












  • $begingroup$
    Why would you want to switch to that order of integration?
    $endgroup$
    – StackTD
    Dec 5 '18 at 12:11






  • 1




    $begingroup$
    The book doesn't ask the reader to change the order of integration, but I'm just curious. Also, I'm studying, since I find region description to be kind of hard.
    $endgroup$
    – IchVerloren
    Dec 5 '18 at 12:35










  • $begingroup$
    It's very fiddly, that, as you'll have to find the intersection of the two slanted planes & divide the region up into parts having certain of their edges along that line of intersection.
    $endgroup$
    – AmbretteOrrisey
    Dec 5 '18 at 14:23










  • $begingroup$
    The plane z=x+y cuts off a pyramid from the unit box having vertices 000, 001, 101, 011 (ie containing the corner 001); and the plane x+y=1 cuts the unit box into two triangular prisms aligned along the z-axis & sharing vertices 010, 100, 011, 101.
    $endgroup$
    – AmbretteOrrisey
    Dec 5 '18 at 14:34












  • $begingroup$
    It's relatively easy to just see it geometrically: the plane x+y=1 cuts the cube equally into two, so that you're left with half it's volume; and then cutting-off that corner by the plane z=x+y takes-off a volume ⅙of the unit cube from that; so you're left with ⅓. But then ... how to formulate that as the limits of a triple-integral ... !?
    $endgroup$
    – AmbretteOrrisey
    Dec 5 '18 at 14:45




















  • $begingroup$
    Why would you want to switch to that order of integration?
    $endgroup$
    – StackTD
    Dec 5 '18 at 12:11






  • 1




    $begingroup$
    The book doesn't ask the reader to change the order of integration, but I'm just curious. Also, I'm studying, since I find region description to be kind of hard.
    $endgroup$
    – IchVerloren
    Dec 5 '18 at 12:35










  • $begingroup$
    It's very fiddly, that, as you'll have to find the intersection of the two slanted planes & divide the region up into parts having certain of their edges along that line of intersection.
    $endgroup$
    – AmbretteOrrisey
    Dec 5 '18 at 14:23










  • $begingroup$
    The plane z=x+y cuts off a pyramid from the unit box having vertices 000, 001, 101, 011 (ie containing the corner 001); and the plane x+y=1 cuts the unit box into two triangular prisms aligned along the z-axis & sharing vertices 010, 100, 011, 101.
    $endgroup$
    – AmbretteOrrisey
    Dec 5 '18 at 14:34












  • $begingroup$
    It's relatively easy to just see it geometrically: the plane x+y=1 cuts the cube equally into two, so that you're left with half it's volume; and then cutting-off that corner by the plane z=x+y takes-off a volume ⅙of the unit cube from that; so you're left with ⅓. But then ... how to formulate that as the limits of a triple-integral ... !?
    $endgroup$
    – AmbretteOrrisey
    Dec 5 '18 at 14:45


















$begingroup$
Why would you want to switch to that order of integration?
$endgroup$
– StackTD
Dec 5 '18 at 12:11




$begingroup$
Why would you want to switch to that order of integration?
$endgroup$
– StackTD
Dec 5 '18 at 12:11




1




1




$begingroup$
The book doesn't ask the reader to change the order of integration, but I'm just curious. Also, I'm studying, since I find region description to be kind of hard.
$endgroup$
– IchVerloren
Dec 5 '18 at 12:35




$begingroup$
The book doesn't ask the reader to change the order of integration, but I'm just curious. Also, I'm studying, since I find region description to be kind of hard.
$endgroup$
– IchVerloren
Dec 5 '18 at 12:35












$begingroup$
It's very fiddly, that, as you'll have to find the intersection of the two slanted planes & divide the region up into parts having certain of their edges along that line of intersection.
$endgroup$
– AmbretteOrrisey
Dec 5 '18 at 14:23




$begingroup$
It's very fiddly, that, as you'll have to find the intersection of the two slanted planes & divide the region up into parts having certain of their edges along that line of intersection.
$endgroup$
– AmbretteOrrisey
Dec 5 '18 at 14:23












$begingroup$
The plane z=x+y cuts off a pyramid from the unit box having vertices 000, 001, 101, 011 (ie containing the corner 001); and the plane x+y=1 cuts the unit box into two triangular prisms aligned along the z-axis & sharing vertices 010, 100, 011, 101.
$endgroup$
– AmbretteOrrisey
Dec 5 '18 at 14:34






$begingroup$
The plane z=x+y cuts off a pyramid from the unit box having vertices 000, 001, 101, 011 (ie containing the corner 001); and the plane x+y=1 cuts the unit box into two triangular prisms aligned along the z-axis & sharing vertices 010, 100, 011, 101.
$endgroup$
– AmbretteOrrisey
Dec 5 '18 at 14:34














$begingroup$
It's relatively easy to just see it geometrically: the plane x+y=1 cuts the cube equally into two, so that you're left with half it's volume; and then cutting-off that corner by the plane z=x+y takes-off a volume ⅙of the unit cube from that; so you're left with ⅓. But then ... how to formulate that as the limits of a triple-integral ... !?
$endgroup$
– AmbretteOrrisey
Dec 5 '18 at 14:45






$begingroup$
It's relatively easy to just see it geometrically: the plane x+y=1 cuts the cube equally into two, so that you're left with half it's volume; and then cutting-off that corner by the plane z=x+y takes-off a volume ⅙of the unit cube from that; so you're left with ⅓. But then ... how to formulate that as the limits of a triple-integral ... !?
$endgroup$
– AmbretteOrrisey
Dec 5 '18 at 14:45












2 Answers
2






active

oldest

votes


















1












$begingroup$

Try to work it out from the hints before reading the spoilers.



The first hint is to make a sketch of the cross-section of a plane $z=c$ with the volume of interest and $c$ some constant for which $0 leq c leq 1$. What is the shape of this cross-section?




Answer : a trapezoid, bounded by $x geq 0, x+ygeq z, y geq 0, 1 geq x+y$



Can you now find the integration limits?




Did you think of splitting the area in two parts to make it easier?




The simplest way would be to split the trapezoid in a parallelogram and a triangle separated by the line $y=c$. Can you solve the problem for both these areas?




$$
int_0^1 d z left{ int_0^z d y int_{z-y}^{1-x} d x + int_z^1 d y int_0^{1-y} d x right} = frac{1}{3}
$$
Obviously not the easiest way to evaluate the integral, but not too difficult either.










share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    That suggestion I made to 'cheat': it's not necessary - the 'fog' has lifted, and I can see now that $$int_{x=0}^1int_{y=0}^{1-x}int_{z=0}^{x+y}dz.dy.dx$$ does actually comprise the whole of the volume specified. If you see it through, you get $$int_{x=0}^1int_{y=0}^{1-x}(x+y)dy.dx$$$$=$$$$int_{x=0}^1{1over2}(2x(1-x)+(1-x)^2)dx$$$$=$$$${1over2}int_{x=0}^1(1-x^2)dx$$$$=$$$${1over3} .$$ It is of course necessary to do it this way if you are integrating some function of $x$ $y$ & $z$ over the region - which is generally the case when triple integrals arise in natural course ... the geometric shortcut that I broached in the comments can only be used for finding the volume ... except perhaps in the occasional problem having extraordinary symmetry.



    I think my 'cheat' is still relevant though. Because of the shape & orientation, a more 'fitting' set of coordinates for the integral would be $$r=x+y$$&$$s=x-y ,$$& $z$ kept as it is, whence the integral would be$$int_{r=0}^1int_{s=-r}^rint_{z=0}^r operatorname{f}({r+sover2}, {r-sover2},z).dz.{dsoversqrt{2}}.{droversqrt{2}} ,$$ with the factors of $1/sqrt{2}$ at the differentials $dr$ & $ds$ so that they represent distance in this space. It might also fit the function being integrated over that region better also, as if the region has that shape & orientation, there's an appreciable likelyhood that the function also has some convenient symmetry with respect to that orientation.






    share|cite|improve this answer











    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026913%2fintegration-limits-in-a-triple-integral%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Try to work it out from the hints before reading the spoilers.



      The first hint is to make a sketch of the cross-section of a plane $z=c$ with the volume of interest and $c$ some constant for which $0 leq c leq 1$. What is the shape of this cross-section?




      Answer : a trapezoid, bounded by $x geq 0, x+ygeq z, y geq 0, 1 geq x+y$



      Can you now find the integration limits?




      Did you think of splitting the area in two parts to make it easier?




      The simplest way would be to split the trapezoid in a parallelogram and a triangle separated by the line $y=c$. Can you solve the problem for both these areas?




      $$
      int_0^1 d z left{ int_0^z d y int_{z-y}^{1-x} d x + int_z^1 d y int_0^{1-y} d x right} = frac{1}{3}
      $$
      Obviously not the easiest way to evaluate the integral, but not too difficult either.










      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Try to work it out from the hints before reading the spoilers.



        The first hint is to make a sketch of the cross-section of a plane $z=c$ with the volume of interest and $c$ some constant for which $0 leq c leq 1$. What is the shape of this cross-section?




        Answer : a trapezoid, bounded by $x geq 0, x+ygeq z, y geq 0, 1 geq x+y$



        Can you now find the integration limits?




        Did you think of splitting the area in two parts to make it easier?




        The simplest way would be to split the trapezoid in a parallelogram and a triangle separated by the line $y=c$. Can you solve the problem for both these areas?




        $$
        int_0^1 d z left{ int_0^z d y int_{z-y}^{1-x} d x + int_z^1 d y int_0^{1-y} d x right} = frac{1}{3}
        $$
        Obviously not the easiest way to evaluate the integral, but not too difficult either.










        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Try to work it out from the hints before reading the spoilers.



          The first hint is to make a sketch of the cross-section of a plane $z=c$ with the volume of interest and $c$ some constant for which $0 leq c leq 1$. What is the shape of this cross-section?




          Answer : a trapezoid, bounded by $x geq 0, x+ygeq z, y geq 0, 1 geq x+y$



          Can you now find the integration limits?




          Did you think of splitting the area in two parts to make it easier?




          The simplest way would be to split the trapezoid in a parallelogram and a triangle separated by the line $y=c$. Can you solve the problem for both these areas?




          $$
          int_0^1 d z left{ int_0^z d y int_{z-y}^{1-x} d x + int_z^1 d y int_0^{1-y} d x right} = frac{1}{3}
          $$
          Obviously not the easiest way to evaluate the integral, but not too difficult either.










          share|cite|improve this answer









          $endgroup$



          Try to work it out from the hints before reading the spoilers.



          The first hint is to make a sketch of the cross-section of a plane $z=c$ with the volume of interest and $c$ some constant for which $0 leq c leq 1$. What is the shape of this cross-section?




          Answer : a trapezoid, bounded by $x geq 0, x+ygeq z, y geq 0, 1 geq x+y$



          Can you now find the integration limits?




          Did you think of splitting the area in two parts to make it easier?




          The simplest way would be to split the trapezoid in a parallelogram and a triangle separated by the line $y=c$. Can you solve the problem for both these areas?




          $$
          int_0^1 d z left{ int_0^z d y int_{z-y}^{1-x} d x + int_z^1 d y int_0^{1-y} d x right} = frac{1}{3}
          $$
          Obviously not the easiest way to evaluate the integral, but not too difficult either.











          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 14:40









          Ronald BlaakRonald Blaak

          2,394310




          2,394310























              0












              $begingroup$

              That suggestion I made to 'cheat': it's not necessary - the 'fog' has lifted, and I can see now that $$int_{x=0}^1int_{y=0}^{1-x}int_{z=0}^{x+y}dz.dy.dx$$ does actually comprise the whole of the volume specified. If you see it through, you get $$int_{x=0}^1int_{y=0}^{1-x}(x+y)dy.dx$$$$=$$$$int_{x=0}^1{1over2}(2x(1-x)+(1-x)^2)dx$$$$=$$$${1over2}int_{x=0}^1(1-x^2)dx$$$$=$$$${1over3} .$$ It is of course necessary to do it this way if you are integrating some function of $x$ $y$ & $z$ over the region - which is generally the case when triple integrals arise in natural course ... the geometric shortcut that I broached in the comments can only be used for finding the volume ... except perhaps in the occasional problem having extraordinary symmetry.



              I think my 'cheat' is still relevant though. Because of the shape & orientation, a more 'fitting' set of coordinates for the integral would be $$r=x+y$$&$$s=x-y ,$$& $z$ kept as it is, whence the integral would be$$int_{r=0}^1int_{s=-r}^rint_{z=0}^r operatorname{f}({r+sover2}, {r-sover2},z).dz.{dsoversqrt{2}}.{droversqrt{2}} ,$$ with the factors of $1/sqrt{2}$ at the differentials $dr$ & $ds$ so that they represent distance in this space. It might also fit the function being integrated over that region better also, as if the region has that shape & orientation, there's an appreciable likelyhood that the function also has some convenient symmetry with respect to that orientation.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                That suggestion I made to 'cheat': it's not necessary - the 'fog' has lifted, and I can see now that $$int_{x=0}^1int_{y=0}^{1-x}int_{z=0}^{x+y}dz.dy.dx$$ does actually comprise the whole of the volume specified. If you see it through, you get $$int_{x=0}^1int_{y=0}^{1-x}(x+y)dy.dx$$$$=$$$$int_{x=0}^1{1over2}(2x(1-x)+(1-x)^2)dx$$$$=$$$${1over2}int_{x=0}^1(1-x^2)dx$$$$=$$$${1over3} .$$ It is of course necessary to do it this way if you are integrating some function of $x$ $y$ & $z$ over the region - which is generally the case when triple integrals arise in natural course ... the geometric shortcut that I broached in the comments can only be used for finding the volume ... except perhaps in the occasional problem having extraordinary symmetry.



                I think my 'cheat' is still relevant though. Because of the shape & orientation, a more 'fitting' set of coordinates for the integral would be $$r=x+y$$&$$s=x-y ,$$& $z$ kept as it is, whence the integral would be$$int_{r=0}^1int_{s=-r}^rint_{z=0}^r operatorname{f}({r+sover2}, {r-sover2},z).dz.{dsoversqrt{2}}.{droversqrt{2}} ,$$ with the factors of $1/sqrt{2}$ at the differentials $dr$ & $ds$ so that they represent distance in this space. It might also fit the function being integrated over that region better also, as if the region has that shape & orientation, there's an appreciable likelyhood that the function also has some convenient symmetry with respect to that orientation.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  That suggestion I made to 'cheat': it's not necessary - the 'fog' has lifted, and I can see now that $$int_{x=0}^1int_{y=0}^{1-x}int_{z=0}^{x+y}dz.dy.dx$$ does actually comprise the whole of the volume specified. If you see it through, you get $$int_{x=0}^1int_{y=0}^{1-x}(x+y)dy.dx$$$$=$$$$int_{x=0}^1{1over2}(2x(1-x)+(1-x)^2)dx$$$$=$$$${1over2}int_{x=0}^1(1-x^2)dx$$$$=$$$${1over3} .$$ It is of course necessary to do it this way if you are integrating some function of $x$ $y$ & $z$ over the region - which is generally the case when triple integrals arise in natural course ... the geometric shortcut that I broached in the comments can only be used for finding the volume ... except perhaps in the occasional problem having extraordinary symmetry.



                  I think my 'cheat' is still relevant though. Because of the shape & orientation, a more 'fitting' set of coordinates for the integral would be $$r=x+y$$&$$s=x-y ,$$& $z$ kept as it is, whence the integral would be$$int_{r=0}^1int_{s=-r}^rint_{z=0}^r operatorname{f}({r+sover2}, {r-sover2},z).dz.{dsoversqrt{2}}.{droversqrt{2}} ,$$ with the factors of $1/sqrt{2}$ at the differentials $dr$ & $ds$ so that they represent distance in this space. It might also fit the function being integrated over that region better also, as if the region has that shape & orientation, there's an appreciable likelyhood that the function also has some convenient symmetry with respect to that orientation.






                  share|cite|improve this answer











                  $endgroup$



                  That suggestion I made to 'cheat': it's not necessary - the 'fog' has lifted, and I can see now that $$int_{x=0}^1int_{y=0}^{1-x}int_{z=0}^{x+y}dz.dy.dx$$ does actually comprise the whole of the volume specified. If you see it through, you get $$int_{x=0}^1int_{y=0}^{1-x}(x+y)dy.dx$$$$=$$$$int_{x=0}^1{1over2}(2x(1-x)+(1-x)^2)dx$$$$=$$$${1over2}int_{x=0}^1(1-x^2)dx$$$$=$$$${1over3} .$$ It is of course necessary to do it this way if you are integrating some function of $x$ $y$ & $z$ over the region - which is generally the case when triple integrals arise in natural course ... the geometric shortcut that I broached in the comments can only be used for finding the volume ... except perhaps in the occasional problem having extraordinary symmetry.



                  I think my 'cheat' is still relevant though. Because of the shape & orientation, a more 'fitting' set of coordinates for the integral would be $$r=x+y$$&$$s=x-y ,$$& $z$ kept as it is, whence the integral would be$$int_{r=0}^1int_{s=-r}^rint_{z=0}^r operatorname{f}({r+sover2}, {r-sover2},z).dz.{dsoversqrt{2}}.{droversqrt{2}} ,$$ with the factors of $1/sqrt{2}$ at the differentials $dr$ & $ds$ so that they represent distance in this space. It might also fit the function being integrated over that region better also, as if the region has that shape & orientation, there's an appreciable likelyhood that the function also has some convenient symmetry with respect to that orientation.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 6 '18 at 5:17

























                  answered Dec 5 '18 at 18:12









                  AmbretteOrriseyAmbretteOrrisey

                  54210




                  54210






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026913%2fintegration-limits-in-a-triple-integral%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Le Mesnil-Réaume

                      Image
                      Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
                      Read more

                      Bundesstraße 106

                      Image
                      Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                      Read more