Calculating the area of ​the plane $x+z=1$ portion that is enclosed by $x^2+y^2=2$ with $z≤0$












1












$begingroup$


the exercise is the text in the title. I'm studying surface integrals.
To start I thougt to make a change to cartecian coordinates with z as a function of $x$ and $y$, that is, $z=1-x$. With this I make the following parameterization: $O(x,y)=[x,y, x-1]$
Then I calculated the norm: $N=2^{1/2}$ of the normal vector: $[1,0,-1]$ of $O(x,y)$.
This is all I did because I do not understand how to project the surface on the xy plane or how to find between the limits of x and y.
In addition, how the $z≤0$ condition influences the projection.
Another question I have is: is the area of ​​the plane enclosed in the cylinder the same as the area of ​​the portion of the cylinder intersected by the plane?






calculus integration multivariable-calculus







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  • $begingroup$
    Is $2^{frac12}$ a vector?
    $endgroup$
    – Chris Custer
    Dec 14 '18 at 1:00










  • $begingroup$
    Sorry, $N$ is the norm of the vector $(1,0, -1)$
    $endgroup$
    – Ayesca
    Dec 14 '18 at 1:03
















1












$begingroup$


the exercise is the text in the title. I'm studying surface integrals.
To start I thougt to make a change to cartecian coordinates with z as a function of $x$ and $y$, that is, $z=1-x$. With this I make the following parameterization: $O(x,y)=[x,y, x-1]$
Then I calculated the norm: $N=2^{1/2}$ of the normal vector: $[1,0,-1]$ of $O(x,y)$.
This is all I did because I do not understand how to project the surface on the xy plane or how to find between the limits of x and y.
In addition, how the $z≤0$ condition influences the projection.
Another question I have is: is the area of ​​the plane enclosed in the cylinder the same as the area of ​​the portion of the cylinder intersected by the plane?






calculus integration multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is $2^{frac12}$ a vector?
    $endgroup$
    – Chris Custer
    Dec 14 '18 at 1:00










  • $begingroup$
    Sorry, $N$ is the norm of the vector $(1,0, -1)$
    $endgroup$
    – Ayesca
    Dec 14 '18 at 1:03














1












1








1





$begingroup$


the exercise is the text in the title. I'm studying surface integrals.
To start I thougt to make a change to cartecian coordinates with z as a function of $x$ and $y$, that is, $z=1-x$. With this I make the following parameterization: $O(x,y)=[x,y, x-1]$
Then I calculated the norm: $N=2^{1/2}$ of the normal vector: $[1,0,-1]$ of $O(x,y)$.
This is all I did because I do not understand how to project the surface on the xy plane or how to find between the limits of x and y.
In addition, how the $z≤0$ condition influences the projection.
Another question I have is: is the area of ​​the plane enclosed in the cylinder the same as the area of ​​the portion of the cylinder intersected by the plane?






calculus integration multivariable-calculus







share|cite|improve this question











$endgroup$




the exercise is the text in the title. I'm studying surface integrals.
To start I thougt to make a change to cartecian coordinates with z as a function of $x$ and $y$, that is, $z=1-x$. With this I make the following parameterization: $O(x,y)=[x,y, x-1]$
Then I calculated the norm: $N=2^{1/2}$ of the normal vector: $[1,0,-1]$ of $O(x,y)$.
This is all I did because I do not understand how to project the surface on the xy plane or how to find between the limits of x and y.
In addition, how the $z≤0$ condition influences the projection.
Another question I have is: is the area of ​​the plane enclosed in the cylinder the same as the area of ​​the portion of the cylinder intersected by the plane?






calculus integration multivariable-calculus




calculus integration multivariable-calculus






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share|cite|improve this question













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edited Dec 14 '18 at 1:14







Ayesca

















asked Dec 14 '18 at 0:54









AyescaAyesca

1999




1999












  • $begingroup$
    Is $2^{frac12}$ a vector?
    $endgroup$
    – Chris Custer
    Dec 14 '18 at 1:00










  • $begingroup$
    Sorry, $N$ is the norm of the vector $(1,0, -1)$
    $endgroup$
    – Ayesca
    Dec 14 '18 at 1:03


















  • $begingroup$
    Is $2^{frac12}$ a vector?
    $endgroup$
    – Chris Custer
    Dec 14 '18 at 1:00










  • $begingroup$
    Sorry, $N$ is the norm of the vector $(1,0, -1)$
    $endgroup$
    – Ayesca
    Dec 14 '18 at 1:03
















$begingroup$
Is $2^{frac12}$ a vector?
$endgroup$
– Chris Custer
Dec 14 '18 at 1:00




$begingroup$
Is $2^{frac12}$ a vector?
$endgroup$
– Chris Custer
Dec 14 '18 at 1:00












$begingroup$
Sorry, $N$ is the norm of the vector $(1,0, -1)$
$endgroup$
– Ayesca
Dec 14 '18 at 1:03




$begingroup$
Sorry, $N$ is the norm of the vector $(1,0, -1)$
$endgroup$
– Ayesca
Dec 14 '18 at 1:03










2 Answers
2






active

oldest

votes


















1












$begingroup$

I don't think the other answer is correct, the constraint that $zleq 0$ changes things quite a bit.



When you project upwards, this means you are integrating over
$$
{ (x,y):;x^2+y^2leq 2}cap { (x,y):xgeq 1}
$$
Since the combination $zleq 0$ and $x+z=1$ yields $xgeq 1$.



So as you found, the norm of the unit vector is $sqrt{2}=||(1,0,1||$ and your integral is
$$
int_{1}^{2}int_{-sqrt{2-x^2}}^{sqrt{2-x^2}} sqrt{2};mathrm dy mathrm dx
$$
Or much more easily in polar coordinates
$$
sqrt{2}int_1^2 int_{-pi/4}^{pi/4}r;mathrm drmathrm dtheta\
=frac{3pi sqrt{2}}{4}
$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    It is probably easier not to convert coordinates. To answer your second question: yes the portion enclosed by the cylinder is the same as the area of the intersection.



    The intersection between cylinder and the plane is an ellipse. Its semi-minor axis ($a$) is the same as the radius ($R$) of the cylinder. Its semi-major axis ($b$) is $sqrt2R$ since the plane is at a $45^o$ angle with the base of the cylinder. So we have:



    $areaspace of space enclosedspace ellipse=pi ab = pi R(sqrt2R)=sqrt2pi R^2=sqrt2pi times2=2sqrt2pi.$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      I don't think the other answer is correct, the constraint that $zleq 0$ changes things quite a bit.



      When you project upwards, this means you are integrating over
      $$
      { (x,y):;x^2+y^2leq 2}cap { (x,y):xgeq 1}
      $$
      Since the combination $zleq 0$ and $x+z=1$ yields $xgeq 1$.



      So as you found, the norm of the unit vector is $sqrt{2}=||(1,0,1||$ and your integral is
      $$
      int_{1}^{2}int_{-sqrt{2-x^2}}^{sqrt{2-x^2}} sqrt{2};mathrm dy mathrm dx
      $$
      Or much more easily in polar coordinates
      $$
      sqrt{2}int_1^2 int_{-pi/4}^{pi/4}r;mathrm drmathrm dtheta\
      =frac{3pi sqrt{2}}{4}
      $$






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        I don't think the other answer is correct, the constraint that $zleq 0$ changes things quite a bit.



        When you project upwards, this means you are integrating over
        $$
        { (x,y):;x^2+y^2leq 2}cap { (x,y):xgeq 1}
        $$
        Since the combination $zleq 0$ and $x+z=1$ yields $xgeq 1$.



        So as you found, the norm of the unit vector is $sqrt{2}=||(1,0,1||$ and your integral is
        $$
        int_{1}^{2}int_{-sqrt{2-x^2}}^{sqrt{2-x^2}} sqrt{2};mathrm dy mathrm dx
        $$
        Or much more easily in polar coordinates
        $$
        sqrt{2}int_1^2 int_{-pi/4}^{pi/4}r;mathrm drmathrm dtheta\
        =frac{3pi sqrt{2}}{4}
        $$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          I don't think the other answer is correct, the constraint that $zleq 0$ changes things quite a bit.



          When you project upwards, this means you are integrating over
          $$
          { (x,y):;x^2+y^2leq 2}cap { (x,y):xgeq 1}
          $$
          Since the combination $zleq 0$ and $x+z=1$ yields $xgeq 1$.



          So as you found, the norm of the unit vector is $sqrt{2}=||(1,0,1||$ and your integral is
          $$
          int_{1}^{2}int_{-sqrt{2-x^2}}^{sqrt{2-x^2}} sqrt{2};mathrm dy mathrm dx
          $$
          Or much more easily in polar coordinates
          $$
          sqrt{2}int_1^2 int_{-pi/4}^{pi/4}r;mathrm drmathrm dtheta\
          =frac{3pi sqrt{2}}{4}
          $$






          share|cite|improve this answer











          $endgroup$



          I don't think the other answer is correct, the constraint that $zleq 0$ changes things quite a bit.



          When you project upwards, this means you are integrating over
          $$
          { (x,y):;x^2+y^2leq 2}cap { (x,y):xgeq 1}
          $$
          Since the combination $zleq 0$ and $x+z=1$ yields $xgeq 1$.



          So as you found, the norm of the unit vector is $sqrt{2}=||(1,0,1||$ and your integral is
          $$
          int_{1}^{2}int_{-sqrt{2-x^2}}^{sqrt{2-x^2}} sqrt{2};mathrm dy mathrm dx
          $$
          Or much more easily in polar coordinates
          $$
          sqrt{2}int_1^2 int_{-pi/4}^{pi/4}r;mathrm drmathrm dtheta\
          =frac{3pi sqrt{2}}{4}
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 2:07

























          answered Dec 14 '18 at 1:52









          qbertqbert

          22.1k32561




          22.1k32561























              1












              $begingroup$

              It is probably easier not to convert coordinates. To answer your second question: yes the portion enclosed by the cylinder is the same as the area of the intersection.



              The intersection between cylinder and the plane is an ellipse. Its semi-minor axis ($a$) is the same as the radius ($R$) of the cylinder. Its semi-major axis ($b$) is $sqrt2R$ since the plane is at a $45^o$ angle with the base of the cylinder. So we have:



              $areaspace of space enclosedspace ellipse=pi ab = pi R(sqrt2R)=sqrt2pi R^2=sqrt2pi times2=2sqrt2pi.$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                It is probably easier not to convert coordinates. To answer your second question: yes the portion enclosed by the cylinder is the same as the area of the intersection.



                The intersection between cylinder and the plane is an ellipse. Its semi-minor axis ($a$) is the same as the radius ($R$) of the cylinder. Its semi-major axis ($b$) is $sqrt2R$ since the plane is at a $45^o$ angle with the base of the cylinder. So we have:



                $areaspace of space enclosedspace ellipse=pi ab = pi R(sqrt2R)=sqrt2pi R^2=sqrt2pi times2=2sqrt2pi.$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  It is probably easier not to convert coordinates. To answer your second question: yes the portion enclosed by the cylinder is the same as the area of the intersection.



                  The intersection between cylinder and the plane is an ellipse. Its semi-minor axis ($a$) is the same as the radius ($R$) of the cylinder. Its semi-major axis ($b$) is $sqrt2R$ since the plane is at a $45^o$ angle with the base of the cylinder. So we have:



                  $areaspace of space enclosedspace ellipse=pi ab = pi R(sqrt2R)=sqrt2pi R^2=sqrt2pi times2=2sqrt2pi.$






                  share|cite|improve this answer









                  $endgroup$



                  It is probably easier not to convert coordinates. To answer your second question: yes the portion enclosed by the cylinder is the same as the area of the intersection.



                  The intersection between cylinder and the plane is an ellipse. Its semi-minor axis ($a$) is the same as the radius ($R$) of the cylinder. Its semi-major axis ($b$) is $sqrt2R$ since the plane is at a $45^o$ angle with the base of the cylinder. So we have:



                  $areaspace of space enclosedspace ellipse=pi ab = pi R(sqrt2R)=sqrt2pi R^2=sqrt2pi times2=2sqrt2pi.$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 14 '18 at 1:23









                  IndominusIndominus

                  1,041316




                  1,041316






























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