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How to calculate the volume of intersection of sphere and cylinder
I have to calculate the volume of intersection of a sphere and a cylinder.
The cylinder's radius is $r$ and the center point is $(r,0,0)$.
The sphere's center point is $(0,0,0)$ and the radius $2r$.
I calculated the estimated volume with Monte Carlo methods but now I have to calculate the real volume with a formula.
Is there any kind of explicit equation for this volume?
I will appreciate any kind of help!
integration volume spheres monte-carlo
edited Jan 10 '17 at 19:24
The Count
2,29961431
asked Jan 10 '17 at 19:05
VanBubuu
1
add a comment |
I have to calculate the volume of intersection of a sphere and a cylinder.
The cylinder's radius is $r$ and the center point is $(r,0,0)$.
The sphere's center point is $(0,0,0)$ and the radius $2r$.
I calculated the estimated volume with Monte Carlo methods but now I have to calculate the real volume with a formula.
Is there any kind of explicit equation for this volume?
I will appreciate any kind of help!
integration volume spheres monte-carlo
edited Jan 10 '17 at 19:24
The Count
2,29961431
asked Jan 10 '17 at 19:05
VanBubuu
1
Please show that estimated volume calculation and any and all other relevant info, please!
– The Count
Jan 10 '17 at 19:09
ok=0; for i=1:n x = rand(1,3); x = x*4*R-2*R; if((x(1))^2+(x(2))^2+x(3)^2-(2*R)^2 <= 0 && ((x(1)-R)/R)^2+(x(2)/R)^2-1 <= 0) ok=ok+1; end end montecarlo_V = ok/n * (2*R)^3;
– VanBubuu
Jan 10 '17 at 20:29
I used this to calculate the estimated, now I have to calculate the theoretical one.
– VanBubuu
Jan 10 '17 at 20:29
add a comment |
I have to calculate the volume of intersection of a sphere and a cylinder.
The cylinder's radius is $r$ and the center point is $(r,0,0)$.
The sphere's center point is $(0,0,0)$ and the radius $2r$.
I calculated the estimated volume with Monte Carlo methods but now I have to calculate the real volume with a formula.
Is there any kind of explicit equation for this volume?
I will appreciate any kind of help!
integration volume spheres monte-carlo
edited Jan 10 '17 at 19:24
The Count
2,29961431
asked Jan 10 '17 at 19:05
VanBubuu
1
I have to calculate the volume of intersection of a sphere and a cylinder.
The cylinder's radius is $r$ and the center point is $(r,0,0)$.
The sphere's center point is $(0,0,0)$ and the radius $2r$.
I calculated the estimated volume with Monte Carlo methods but now I have to calculate the real volume with a formula.
Is there any kind of explicit equation for this volume?
I will appreciate any kind of help!
integration volume spheres monte-carlo
integration volume spheres monte-carlo
edited Jan 10 '17 at 19:24
The Count
2,29961431
asked Jan 10 '17 at 19:05
VanBubuu
1
edited Jan 10 '17 at 19:24
The Count
2,29961431
asked Jan 10 '17 at 19:05
VanBubuu
1
edited Jan 10 '17 at 19:24
The Count
2,29961431
edited Jan 10 '17 at 19:24
The Count
2,29961431
edited Jan 10 '17 at 19:24
The Count
2,29961431
2,29961431
asked Jan 10 '17 at 19:05
VanBubuu
1
asked Jan 10 '17 at 19:05
VanBubuu
1
asked Jan 10 '17 at 19:05
VanBubuu
1
1
Please show that estimated volume calculation and any and all other relevant info, please!
– The Count
Jan 10 '17 at 19:09
ok=0; for i=1:n x = rand(1,3); x = x*4*R-2*R; if((x(1))^2+(x(2))^2+x(3)^2-(2*R)^2 <= 0 && ((x(1)-R)/R)^2+(x(2)/R)^2-1 <= 0) ok=ok+1; end end montecarlo_V = ok/n * (2*R)^3;
– VanBubuu
Jan 10 '17 at 20:29
I used this to calculate the estimated, now I have to calculate the theoretical one.
– VanBubuu
Jan 10 '17 at 20:29
add a comment |
Please show that estimated volume calculation and any and all other relevant info, please!
– The Count
Jan 10 '17 at 19:09
ok=0; for i=1:n x = rand(1,3); x = x*4*R-2*R; if((x(1))^2+(x(2))^2+x(3)^2-(2*R)^2 <= 0 && ((x(1)-R)/R)^2+(x(2)/R)^2-1 <= 0) ok=ok+1; end end montecarlo_V = ok/n * (2*R)^3;
– VanBubuu
Jan 10 '17 at 20:29
I used this to calculate the estimated, now I have to calculate the theoretical one.
– VanBubuu
Jan 10 '17 at 20:29
Please show that estimated volume calculation and any and all other relevant info, please!
– The Count
Jan 10 '17 at 19:09
Please show that estimated volume calculation and any and all other relevant info, please!
– The Count
Jan 10 '17 at 19:09
ok=0; for i=1:n x = rand(1,3); x = x*4*R-2*R; if((x(1))^2+(x(2))^2+x(3)^2-(2*R)^2 <= 0 && ((x(1)-R)/R)^2+(x(2)/R)^2-1 <= 0) ok=ok+1; end end montecarlo_V = ok/n * (2*R)^3;
– VanBubuu
Jan 10 '17 at 20:29
ok=0; for i=1:n x = rand(1,3); x = x*4*R-2*R; if((x(1))^2+(x(2))^2+x(3)^2-(2*R)^2 <= 0 && ((x(1)-R)/R)^2+(x(2)/R)^2-1 <= 0) ok=ok+1; end end montecarlo_V = ok/n * (2*R)^3;
– VanBubuu
Jan 10 '17 at 20:29
I used this to calculate the estimated, now I have to calculate the theoretical one.
– VanBubuu
Jan 10 '17 at 20:29
I used this to calculate the estimated, now I have to calculate the theoretical one.
– VanBubuu
Jan 10 '17 at 20:29
add a comment |
2 Answers
2
active
oldest
votes
Yes, draw a figure, and you will realize that the volume is given by
$$
2iint_D sqrt{(2r)^2-x^2-y^2},dx,dy
$$
where
$$
D={(x,y)inmathbb R^2~|~(x-r)^2+y^2leq r^2}.
$$
I leave it to you to calculate the integral.
answered Jan 10 '17 at 19:11
mickep
18.5k12250
Thank you for the answer. Can you give me some hints, links how can I calculate this? I'm not very good in math :(
– VanBubuu
Jan 10 '17 at 20:30
add a comment |
Let the sphere be $x^2+y^2+z^2 = 4a^2$
And the cylinder be $(x - a)^2+y^2 = a^2$
Using polar coordinates we have
$$x = r costheta cosphi$$
$$y = r sintheta cosphi$$
$$z = r sinphi$$
Then the equation of the sphere is r = 2a
And the cylinder $r cosphi = 2a costheta$
Solving the 2 equations we have the curve of intersection $cosphi = costheta$ or $phi = theta$
This means that for a fixed $theta$ we have $phi$ varies from 0 to $theta$ and r varies from 0 to 2a and $theta$ from 0 to $pi$ / 2 by considering only the first quadrant.
By transforming to polar coordinates we have the volume integrant to be
$$frac{D(x, y, z)}{D(theta, phi, r)} = r^2cosphi$$
Hence the total volume is
$$4int_0^frac{pi}{2}int_0^{2a}int_0^theta r^2cos{phi}d{phi}drd{theta}$$
$$= 4int_0^frac{pi}{2}int_0^{2a}r^2sin{theta}drd{theta}$$
$$= 4int_0^frac{pi}{2}frac{8a^3}{3}sin{theta}d{theta}$$
$$= frac{32a^3}{3}$$
answered Nov 21 '18 at 1:47
KY Tang
644
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes, draw a figure, and you will realize that the volume is given by
$$
2iint_D sqrt{(2r)^2-x^2-y^2},dx,dy
$$
where
$$
D={(x,y)inmathbb R^2~|~(x-r)^2+y^2leq r^2}.
$$
I leave it to you to calculate the integral.
answered Jan 10 '17 at 19:11
mickep
18.5k12250
Thank you for the answer. Can you give me some hints, links how can I calculate this? I'm not very good in math :(
– VanBubuu
Jan 10 '17 at 20:30
add a comment |
Yes, draw a figure, and you will realize that the volume is given by
$$
2iint_D sqrt{(2r)^2-x^2-y^2},dx,dy
$$
where
$$
D={(x,y)inmathbb R^2~|~(x-r)^2+y^2leq r^2}.
$$
I leave it to you to calculate the integral.
answered Jan 10 '17 at 19:11
mickep
18.5k12250
Thank you for the answer. Can you give me some hints, links how can I calculate this? I'm not very good in math :(
– VanBubuu
Jan 10 '17 at 20:30
add a comment |
Yes, draw a figure, and you will realize that the volume is given by
$$
2iint_D sqrt{(2r)^2-x^2-y^2},dx,dy
$$
where
$$
D={(x,y)inmathbb R^2~|~(x-r)^2+y^2leq r^2}.
$$
I leave it to you to calculate the integral.
answered Jan 10 '17 at 19:11
mickep
18.5k12250
Yes, draw a figure, and you will realize that the volume is given by
$$
2iint_D sqrt{(2r)^2-x^2-y^2},dx,dy
$$
where
$$
D={(x,y)inmathbb R^2~|~(x-r)^2+y^2leq r^2}.
$$
I leave it to you to calculate the integral.
answered Jan 10 '17 at 19:11
mickep
18.5k12250
answered Jan 10 '17 at 19:11
mickep
18.5k12250
answered Jan 10 '17 at 19:11
mickep
18.5k12250
answered Jan 10 '17 at 19:11
mickep
18.5k12250
18.5k12250
Thank you for the answer. Can you give me some hints, links how can I calculate this? I'm not very good in math :(
– VanBubuu
Jan 10 '17 at 20:30
add a comment |
Thank you for the answer. Can you give me some hints, links how can I calculate this? I'm not very good in math :(
– VanBubuu
Jan 10 '17 at 20:30
Thank you for the answer. Can you give me some hints, links how can I calculate this? I'm not very good in math :(
– VanBubuu
Jan 10 '17 at 20:30
Thank you for the answer. Can you give me some hints, links how can I calculate this? I'm not very good in math :(
– VanBubuu
Jan 10 '17 at 20:30
add a comment |
Let the sphere be $x^2+y^2+z^2 = 4a^2$
And the cylinder be $(x - a)^2+y^2 = a^2$
Using polar coordinates we have
$$x = r costheta cosphi$$
$$y = r sintheta cosphi$$
$$z = r sinphi$$
Then the equation of the sphere is r = 2a
And the cylinder $r cosphi = 2a costheta$
Solving the 2 equations we have the curve of intersection $cosphi = costheta$ or $phi = theta$
This means that for a fixed $theta$ we have $phi$ varies from 0 to $theta$ and r varies from 0 to 2a and $theta$ from 0 to $pi$ / 2 by considering only the first quadrant.
By transforming to polar coordinates we have the volume integrant to be
$$frac{D(x, y, z)}{D(theta, phi, r)} = r^2cosphi$$
Hence the total volume is
$$4int_0^frac{pi}{2}int_0^{2a}int_0^theta r^2cos{phi}d{phi}drd{theta}$$
$$= 4int_0^frac{pi}{2}int_0^{2a}r^2sin{theta}drd{theta}$$
$$= 4int_0^frac{pi}{2}frac{8a^3}{3}sin{theta}d{theta}$$
$$= frac{32a^3}{3}$$
answered Nov 21 '18 at 1:47
KY Tang
644
add a comment |
Let the sphere be $x^2+y^2+z^2 = 4a^2$
And the cylinder be $(x - a)^2+y^2 = a^2$
Using polar coordinates we have
$$x = r costheta cosphi$$
$$y = r sintheta cosphi$$
$$z = r sinphi$$
Then the equation of the sphere is r = 2a
And the cylinder $r cosphi = 2a costheta$
Solving the 2 equations we have the curve of intersection $cosphi = costheta$ or $phi = theta$
This means that for a fixed $theta$ we have $phi$ varies from 0 to $theta$ and r varies from 0 to 2a and $theta$ from 0 to $pi$ / 2 by considering only the first quadrant.
By transforming to polar coordinates we have the volume integrant to be
$$frac{D(x, y, z)}{D(theta, phi, r)} = r^2cosphi$$
Hence the total volume is
$$4int_0^frac{pi}{2}int_0^{2a}int_0^theta r^2cos{phi}d{phi}drd{theta}$$
$$= 4int_0^frac{pi}{2}int_0^{2a}r^2sin{theta}drd{theta}$$
$$= 4int_0^frac{pi}{2}frac{8a^3}{3}sin{theta}d{theta}$$
$$= frac{32a^3}{3}$$
answered Nov 21 '18 at 1:47
KY Tang
644
add a comment |
Let the sphere be $x^2+y^2+z^2 = 4a^2$
And the cylinder be $(x - a)^2+y^2 = a^2$
Using polar coordinates we have
$$x = r costheta cosphi$$
$$y = r sintheta cosphi$$
$$z = r sinphi$$
Then the equation of the sphere is r = 2a
And the cylinder $r cosphi = 2a costheta$
Solving the 2 equations we have the curve of intersection $cosphi = costheta$ or $phi = theta$
This means that for a fixed $theta$ we have $phi$ varies from 0 to $theta$ and r varies from 0 to 2a and $theta$ from 0 to $pi$ / 2 by considering only the first quadrant.
By transforming to polar coordinates we have the volume integrant to be
$$frac{D(x, y, z)}{D(theta, phi, r)} = r^2cosphi$$
Hence the total volume is
$$4int_0^frac{pi}{2}int_0^{2a}int_0^theta r^2cos{phi}d{phi}drd{theta}$$
$$= 4int_0^frac{pi}{2}int_0^{2a}r^2sin{theta}drd{theta}$$
$$= 4int_0^frac{pi}{2}frac{8a^3}{3}sin{theta}d{theta}$$
$$= frac{32a^3}{3}$$
answered Nov 21 '18 at 1:47
KY Tang
644
Let the sphere be $x^2+y^2+z^2 = 4a^2$
And the cylinder be $(x - a)^2+y^2 = a^2$
Using polar coordinates we have
$$x = r costheta cosphi$$
$$y = r sintheta cosphi$$
$$z = r sinphi$$
Then the equation of the sphere is r = 2a
And the cylinder $r cosphi = 2a costheta$
Solving the 2 equations we have the curve of intersection $cosphi = costheta$ or $phi = theta$
This means that for a fixed $theta$ we have $phi$ varies from 0 to $theta$ and r varies from 0 to 2a and $theta$ from 0 to $pi$ / 2 by considering only the first quadrant.
By transforming to polar coordinates we have the volume integrant to be
$$frac{D(x, y, z)}{D(theta, phi, r)} = r^2cosphi$$
Hence the total volume is
$$4int_0^frac{pi}{2}int_0^{2a}int_0^theta r^2cos{phi}d{phi}drd{theta}$$
$$= 4int_0^frac{pi}{2}int_0^{2a}r^2sin{theta}drd{theta}$$
$$= 4int_0^frac{pi}{2}frac{8a^3}{3}sin{theta}d{theta}$$
$$= frac{32a^3}{3}$$
answered Nov 21 '18 at 1:47
KY Tang
644
answered Nov 21 '18 at 1:47
KY Tang
644
answered Nov 21 '18 at 1:47
KY Tang
644
answered Nov 21 '18 at 1:47
KY Tang
644
644
add a comment |
add a comment |
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Please show that estimated volume calculation and any and all other relevant info, please!
– The Count
Jan 10 '17 at 19:09
ok=0; for i=1:n x = rand(1,3); x = x*4*R-2*R; if((x(1))^2+(x(2))^2+x(3)^2-(2*R)^2 <= 0 && ((x(1)-R)/R)^2+(x(2)/R)^2-1 <= 0) ok=ok+1; end end montecarlo_V = ok/n * (2*R)^3;
– VanBubuu
Jan 10 '17 at 20:29
I used this to calculate the estimated, now I have to calculate the theoretical one.
– VanBubuu
Jan 10 '17 at 20:29