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











    $endgroup$













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      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.







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                  edited Dec 6 '18 at 5:17

























                  answered Dec 5 '18 at 18:12









                  AmbretteOrriseyAmbretteOrrisey

                  54210




                  54210






























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