Find the volume of intersection between cylinders












0












$begingroup$



Find the volume of intersection of the cylinder

{$ x^2 +
y^2 leq 1 $} , {$ x^2 + z^2 leq 1$}, {$ y^2 + z^2 leq 1$}.




i am having tough time finding the volume how do i solve this kind of questions ? .



my trial :
i will move to the cylinder coordinates of the xy cylinder let :



$x^2 + y^2 = r^2 $



$ z = z $



$0leqtheta leq 2pi$



solving the inequalties i get :



$ 0 leq r^2 leq 1$



$ -sqrt{1-frac{r^2}{2}}leq z leq sqrt{1-frac{r^2}{2}}$



$0leqtheta leq 2pi$



the integral is :



$ int_{z=-sqrt{1-frac{r^2}{2}}}^{z=sqrt{1-frac{r^2}{2}}}int_{r=0}^{r=1}int_0^{2pi} dz dr d{theta}$ = $ frac{4pi}{sqrt{2}}frac{(2-r^2)^{frac{3}{2}}}{-3} |_{r=0}^{r=1}$






multivariable-calculus jacobian cylindrical-coordinates







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  • $begingroup$
    What does your integral look like? If you show your attempted solution you are more likely to get useful feedback.
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:16










  • $begingroup$
    i'll write that in a moment.
    $endgroup$
    – Mather
    Dec 24 '18 at 21:17










  • $begingroup$
    i have updated it
    $endgroup$
    – Mather
    Dec 24 '18 at 21:23










  • $begingroup$
    I'm not sure about your bounds for $z$, how exactly did you get those?
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:34










  • $begingroup$
    Either way it looks like you're missing the Jacobian, which is $r$ for cylindrical coordinates.
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:37
















0












$begingroup$



Find the volume of intersection of the cylinder

{$ x^2 +
y^2 leq 1 $} , {$ x^2 + z^2 leq 1$}, {$ y^2 + z^2 leq 1$}.




i am having tough time finding the volume how do i solve this kind of questions ? .



my trial :
i will move to the cylinder coordinates of the xy cylinder let :



$x^2 + y^2 = r^2 $



$ z = z $



$0leqtheta leq 2pi$



solving the inequalties i get :



$ 0 leq r^2 leq 1$



$ -sqrt{1-frac{r^2}{2}}leq z leq sqrt{1-frac{r^2}{2}}$



$0leqtheta leq 2pi$



the integral is :



$ int_{z=-sqrt{1-frac{r^2}{2}}}^{z=sqrt{1-frac{r^2}{2}}}int_{r=0}^{r=1}int_0^{2pi} dz dr d{theta}$ = $ frac{4pi}{sqrt{2}}frac{(2-r^2)^{frac{3}{2}}}{-3} |_{r=0}^{r=1}$






multivariable-calculus jacobian cylindrical-coordinates







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does your integral look like? If you show your attempted solution you are more likely to get useful feedback.
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:16










  • $begingroup$
    i'll write that in a moment.
    $endgroup$
    – Mather
    Dec 24 '18 at 21:17










  • $begingroup$
    i have updated it
    $endgroup$
    – Mather
    Dec 24 '18 at 21:23










  • $begingroup$
    I'm not sure about your bounds for $z$, how exactly did you get those?
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:34










  • $begingroup$
    Either way it looks like you're missing the Jacobian, which is $r$ for cylindrical coordinates.
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:37














0












0








0





$begingroup$



Find the volume of intersection of the cylinder

{$ x^2 +
y^2 leq 1 $} , {$ x^2 + z^2 leq 1$}, {$ y^2 + z^2 leq 1$}.




i am having tough time finding the volume how do i solve this kind of questions ? .



my trial :
i will move to the cylinder coordinates of the xy cylinder let :



$x^2 + y^2 = r^2 $



$ z = z $



$0leqtheta leq 2pi$



solving the inequalties i get :



$ 0 leq r^2 leq 1$



$ -sqrt{1-frac{r^2}{2}}leq z leq sqrt{1-frac{r^2}{2}}$



$0leqtheta leq 2pi$



the integral is :



$ int_{z=-sqrt{1-frac{r^2}{2}}}^{z=sqrt{1-frac{r^2}{2}}}int_{r=0}^{r=1}int_0^{2pi} dz dr d{theta}$ = $ frac{4pi}{sqrt{2}}frac{(2-r^2)^{frac{3}{2}}}{-3} |_{r=0}^{r=1}$






multivariable-calculus jacobian cylindrical-coordinates







share|cite|improve this question











$endgroup$





Find the volume of intersection of the cylinder

{$ x^2 +
y^2 leq 1 $} , {$ x^2 + z^2 leq 1$}, {$ y^2 + z^2 leq 1$}.




i am having tough time finding the volume how do i solve this kind of questions ? .



my trial :
i will move to the cylinder coordinates of the xy cylinder let :



$x^2 + y^2 = r^2 $



$ z = z $



$0leqtheta leq 2pi$



solving the inequalties i get :



$ 0 leq r^2 leq 1$



$ -sqrt{1-frac{r^2}{2}}leq z leq sqrt{1-frac{r^2}{2}}$



$0leqtheta leq 2pi$



the integral is :



$ int_{z=-sqrt{1-frac{r^2}{2}}}^{z=sqrt{1-frac{r^2}{2}}}int_{r=0}^{r=1}int_0^{2pi} dz dr d{theta}$ = $ frac{4pi}{sqrt{2}}frac{(2-r^2)^{frac{3}{2}}}{-3} |_{r=0}^{r=1}$






multivariable-calculus jacobian cylindrical-coordinates




multivariable-calculus jacobian cylindrical-coordinates






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 21:23







Mather

















asked Dec 24 '18 at 21:15









Mather Mather

3437




3437












  • $begingroup$
    What does your integral look like? If you show your attempted solution you are more likely to get useful feedback.
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:16










  • $begingroup$
    i'll write that in a moment.
    $endgroup$
    – Mather
    Dec 24 '18 at 21:17










  • $begingroup$
    i have updated it
    $endgroup$
    – Mather
    Dec 24 '18 at 21:23










  • $begingroup$
    I'm not sure about your bounds for $z$, how exactly did you get those?
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:34










  • $begingroup$
    Either way it looks like you're missing the Jacobian, which is $r$ for cylindrical coordinates.
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:37


















  • $begingroup$
    What does your integral look like? If you show your attempted solution you are more likely to get useful feedback.
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:16










  • $begingroup$
    i'll write that in a moment.
    $endgroup$
    – Mather
    Dec 24 '18 at 21:17










  • $begingroup$
    i have updated it
    $endgroup$
    – Mather
    Dec 24 '18 at 21:23










  • $begingroup$
    I'm not sure about your bounds for $z$, how exactly did you get those?
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:34










  • $begingroup$
    Either way it looks like you're missing the Jacobian, which is $r$ for cylindrical coordinates.
    $endgroup$
    – pwerth
    Dec 24 '18 at 21:37
















$begingroup$
What does your integral look like? If you show your attempted solution you are more likely to get useful feedback.
$endgroup$
– pwerth
Dec 24 '18 at 21:16




$begingroup$
What does your integral look like? If you show your attempted solution you are more likely to get useful feedback.
$endgroup$
– pwerth
Dec 24 '18 at 21:16












$begingroup$
i'll write that in a moment.
$endgroup$
– Mather
Dec 24 '18 at 21:17




$begingroup$
i'll write that in a moment.
$endgroup$
– Mather
Dec 24 '18 at 21:17












$begingroup$
i have updated it
$endgroup$
– Mather
Dec 24 '18 at 21:23




$begingroup$
i have updated it
$endgroup$
– Mather
Dec 24 '18 at 21:23












$begingroup$
I'm not sure about your bounds for $z$, how exactly did you get those?
$endgroup$
– pwerth
Dec 24 '18 at 21:34




$begingroup$
I'm not sure about your bounds for $z$, how exactly did you get those?
$endgroup$
– pwerth
Dec 24 '18 at 21:34












$begingroup$
Either way it looks like you're missing the Jacobian, which is $r$ for cylindrical coordinates.
$endgroup$
– pwerth
Dec 24 '18 at 21:37




$begingroup$
Either way it looks like you're missing the Jacobian, which is $r$ for cylindrical coordinates.
$endgroup$
– pwerth
Dec 24 '18 at 21:37










2 Answers
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enter image description here



$$V = 16 intlimits_0^{pi/4} intlimits_0^1 s sqrt{1 - s^2 cos^2 theta} ds dtheta = 8(2 - sqrt{2}) $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    hey can you explain what is wrong with my answer since the number is slightly bigger
    $endgroup$
    – Mather
    Dec 24 '18 at 21:54



















0












$begingroup$

Perhaps surprisingly, it is not necessary to use polar coordinates, nor even multiple integration or trigonometry, to find the volume.



What we need to do is consider the volume of a solid whose cross-sections perpendicular to the $x$-axis are rectangles whose width is $x$ and whose height is $sqrt{1-x^2} - 1/sqrt{2}$, over the interval $x in [0,1/sqrt{2}]$. This is simply $$int_{x=0}^{1/sqrt{2}} x left(sqrt{1 - x^2} - 1/sqrt{2}right) , dx = left[-frac{(1-x^2)^{3/2}}{3} - frac{x^2}{2sqrt{2}}right]_{x=0}^{1/sqrt{2}} = frac{8 - 5 sqrt{2}}{24}.$$ Then the desired volume is $48$ times this volume, plus the volume of the inscribed cube, which gives $$2sqrt{2} + 2(8-5sqrt{2}) = 8(2 - sqrt{2}).$$ I have left the reasoning as an exercise for the reader.



An alternative computation can be set up that includes the inscribed cube in the integral, simply by changing the height of the perpendicular cross section to $sqrt{1-x^2} - x$, giving $$frac{V}{48} = int_{x=0}^{1/sqrt{2}} x left(sqrt{1-x^2} - xright) , dx.$$ Again, the details are left for the reader.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    enter image description here



    $$V = 16 intlimits_0^{pi/4} intlimits_0^1 s sqrt{1 - s^2 cos^2 theta} ds dtheta = 8(2 - sqrt{2}) $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      hey can you explain what is wrong with my answer since the number is slightly bigger
      $endgroup$
      – Mather
      Dec 24 '18 at 21:54
















    0












    $begingroup$

    enter image description here



    $$V = 16 intlimits_0^{pi/4} intlimits_0^1 s sqrt{1 - s^2 cos^2 theta} ds dtheta = 8(2 - sqrt{2}) $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      hey can you explain what is wrong with my answer since the number is slightly bigger
      $endgroup$
      – Mather
      Dec 24 '18 at 21:54














    0












    0








    0





    $begingroup$

    enter image description here



    $$V = 16 intlimits_0^{pi/4} intlimits_0^1 s sqrt{1 - s^2 cos^2 theta} ds dtheta = 8(2 - sqrt{2}) $$






    share|cite|improve this answer









    $endgroup$



    enter image description here



    $$V = 16 intlimits_0^{pi/4} intlimits_0^1 s sqrt{1 - s^2 cos^2 theta} ds dtheta = 8(2 - sqrt{2}) $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 24 '18 at 21:43









    David G. StorkDavid G. Stork

    11k41432




    11k41432












    • $begingroup$
      hey can you explain what is wrong with my answer since the number is slightly bigger
      $endgroup$
      – Mather
      Dec 24 '18 at 21:54


















    • $begingroup$
      hey can you explain what is wrong with my answer since the number is slightly bigger
      $endgroup$
      – Mather
      Dec 24 '18 at 21:54
















    $begingroup$
    hey can you explain what is wrong with my answer since the number is slightly bigger
    $endgroup$
    – Mather
    Dec 24 '18 at 21:54




    $begingroup$
    hey can you explain what is wrong with my answer since the number is slightly bigger
    $endgroup$
    – Mather
    Dec 24 '18 at 21:54











    0












    $begingroup$

    Perhaps surprisingly, it is not necessary to use polar coordinates, nor even multiple integration or trigonometry, to find the volume.



    What we need to do is consider the volume of a solid whose cross-sections perpendicular to the $x$-axis are rectangles whose width is $x$ and whose height is $sqrt{1-x^2} - 1/sqrt{2}$, over the interval $x in [0,1/sqrt{2}]$. This is simply $$int_{x=0}^{1/sqrt{2}} x left(sqrt{1 - x^2} - 1/sqrt{2}right) , dx = left[-frac{(1-x^2)^{3/2}}{3} - frac{x^2}{2sqrt{2}}right]_{x=0}^{1/sqrt{2}} = frac{8 - 5 sqrt{2}}{24}.$$ Then the desired volume is $48$ times this volume, plus the volume of the inscribed cube, which gives $$2sqrt{2} + 2(8-5sqrt{2}) = 8(2 - sqrt{2}).$$ I have left the reasoning as an exercise for the reader.



    An alternative computation can be set up that includes the inscribed cube in the integral, simply by changing the height of the perpendicular cross section to $sqrt{1-x^2} - x$, giving $$frac{V}{48} = int_{x=0}^{1/sqrt{2}} x left(sqrt{1-x^2} - xright) , dx.$$ Again, the details are left for the reader.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Perhaps surprisingly, it is not necessary to use polar coordinates, nor even multiple integration or trigonometry, to find the volume.



      What we need to do is consider the volume of a solid whose cross-sections perpendicular to the $x$-axis are rectangles whose width is $x$ and whose height is $sqrt{1-x^2} - 1/sqrt{2}$, over the interval $x in [0,1/sqrt{2}]$. This is simply $$int_{x=0}^{1/sqrt{2}} x left(sqrt{1 - x^2} - 1/sqrt{2}right) , dx = left[-frac{(1-x^2)^{3/2}}{3} - frac{x^2}{2sqrt{2}}right]_{x=0}^{1/sqrt{2}} = frac{8 - 5 sqrt{2}}{24}.$$ Then the desired volume is $48$ times this volume, plus the volume of the inscribed cube, which gives $$2sqrt{2} + 2(8-5sqrt{2}) = 8(2 - sqrt{2}).$$ I have left the reasoning as an exercise for the reader.



      An alternative computation can be set up that includes the inscribed cube in the integral, simply by changing the height of the perpendicular cross section to $sqrt{1-x^2} - x$, giving $$frac{V}{48} = int_{x=0}^{1/sqrt{2}} x left(sqrt{1-x^2} - xright) , dx.$$ Again, the details are left for the reader.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Perhaps surprisingly, it is not necessary to use polar coordinates, nor even multiple integration or trigonometry, to find the volume.



        What we need to do is consider the volume of a solid whose cross-sections perpendicular to the $x$-axis are rectangles whose width is $x$ and whose height is $sqrt{1-x^2} - 1/sqrt{2}$, over the interval $x in [0,1/sqrt{2}]$. This is simply $$int_{x=0}^{1/sqrt{2}} x left(sqrt{1 - x^2} - 1/sqrt{2}right) , dx = left[-frac{(1-x^2)^{3/2}}{3} - frac{x^2}{2sqrt{2}}right]_{x=0}^{1/sqrt{2}} = frac{8 - 5 sqrt{2}}{24}.$$ Then the desired volume is $48$ times this volume, plus the volume of the inscribed cube, which gives $$2sqrt{2} + 2(8-5sqrt{2}) = 8(2 - sqrt{2}).$$ I have left the reasoning as an exercise for the reader.



        An alternative computation can be set up that includes the inscribed cube in the integral, simply by changing the height of the perpendicular cross section to $sqrt{1-x^2} - x$, giving $$frac{V}{48} = int_{x=0}^{1/sqrt{2}} x left(sqrt{1-x^2} - xright) , dx.$$ Again, the details are left for the reader.






        share|cite|improve this answer











        $endgroup$



        Perhaps surprisingly, it is not necessary to use polar coordinates, nor even multiple integration or trigonometry, to find the volume.



        What we need to do is consider the volume of a solid whose cross-sections perpendicular to the $x$-axis are rectangles whose width is $x$ and whose height is $sqrt{1-x^2} - 1/sqrt{2}$, over the interval $x in [0,1/sqrt{2}]$. This is simply $$int_{x=0}^{1/sqrt{2}} x left(sqrt{1 - x^2} - 1/sqrt{2}right) , dx = left[-frac{(1-x^2)^{3/2}}{3} - frac{x^2}{2sqrt{2}}right]_{x=0}^{1/sqrt{2}} = frac{8 - 5 sqrt{2}}{24}.$$ Then the desired volume is $48$ times this volume, plus the volume of the inscribed cube, which gives $$2sqrt{2} + 2(8-5sqrt{2}) = 8(2 - sqrt{2}).$$ I have left the reasoning as an exercise for the reader.



        An alternative computation can be set up that includes the inscribed cube in the integral, simply by changing the height of the perpendicular cross section to $sqrt{1-x^2} - x$, giving $$frac{V}{48} = int_{x=0}^{1/sqrt{2}} x left(sqrt{1-x^2} - xright) , dx.$$ Again, the details are left for the reader.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 24 '18 at 22:06

























        answered Dec 24 '18 at 22:00









        heropupheropup

        63.9k762102




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