Find the volume of intersection between cylinders
$begingroup$
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
$endgroup$
|
show 6 more comments
$begingroup$
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
$endgroup$
$begingroup$
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
|
show 6 more comments
$begingroup$
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
$endgroup$
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
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
|
show 6 more comments
$begingroup$
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
|
show 6 more comments
2 Answers
2
active
oldest
votes
$begingroup$
$$V = 16 intlimits_0^{pi/4} intlimits_0^1 s sqrt{1 - s^2 cos^2 theta} ds dtheta = 8(2 - sqrt{2}) $$
$endgroup$
$begingroup$
hey can you explain what is wrong with my answer since the number is slightly bigger
$endgroup$
– Mather
Dec 24 '18 at 21:54
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
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$begingroup$
$$V = 16 intlimits_0^{pi/4} intlimits_0^1 s sqrt{1 - s^2 cos^2 theta} ds dtheta = 8(2 - sqrt{2}) $$
$endgroup$
$begingroup$
hey can you explain what is wrong with my answer since the number is slightly bigger
$endgroup$
– Mather
Dec 24 '18 at 21:54
add a comment |
$begingroup$
$$V = 16 intlimits_0^{pi/4} intlimits_0^1 s sqrt{1 - s^2 cos^2 theta} ds dtheta = 8(2 - sqrt{2}) $$
$endgroup$
$begingroup$
hey can you explain what is wrong with my answer since the number is slightly bigger
$endgroup$
– Mather
Dec 24 '18 at 21:54
add a comment |
$begingroup$
$$V = 16 intlimits_0^{pi/4} intlimits_0^1 s sqrt{1 - s^2 cos^2 theta} ds dtheta = 8(2 - sqrt{2}) $$
$endgroup$
$$V = 16 intlimits_0^{pi/4} intlimits_0^1 s sqrt{1 - s^2 cos^2 theta} ds dtheta = 8(2 - sqrt{2}) $$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Dec 24 '18 at 22:06
answered Dec 24 '18 at 22:00
heropupheropup
63.9k762102
63.9k762102
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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