How to calculate the average projected area of a circle?
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As a follow up to Average projected area in higher dimension, I started to think about a variation of the original post. Let's visualize a circle with radius $1$ in $3D$ space and cast its shadow on the $x$-$z$ plane. The area of the shadow will change as we orient the circle in space. After we obtain all possible orientations of the circle, what will the average area of the shadow be? Just for clarification, average area is the sum of projected area divided by the number of possible orientations. For this problem, there are infinitely many orientations, so we have to apply calculus here. I think we have to set an integral over the circle's area, but I am not sure how exactly I should do that. Any help or related information would be truly appreciated.
calculus definite-integrals surfaces
asked Dec 9 '18 at 12:21
LarryLarry
2,31431028
$endgroup$
add a comment |
$begingroup$
As a follow up to Average projected area in higher dimension, I started to think about a variation of the original post. Let's visualize a circle with radius $1$ in $3D$ space and cast its shadow on the $x$-$z$ plane. The area of the shadow will change as we orient the circle in space. After we obtain all possible orientations of the circle, what will the average area of the shadow be? Just for clarification, average area is the sum of projected area divided by the number of possible orientations. For this problem, there are infinitely many orientations, so we have to apply calculus here. I think we have to set an integral over the circle's area, but I am not sure how exactly I should do that. Any help or related information would be truly appreciated.
calculus definite-integrals surfaces
asked Dec 9 '18 at 12:21
LarryLarry
2,31431028
$endgroup$
$begingroup$
I think this will depend on the position of the light origin as well. Your diagrams suggest that you're thinking of it as located far away (infinitely far), so that, when the circle lies parallel to the $x, z$ plane, the area of the shadow is equal to that of the circle. Am I right?
$endgroup$
– Patricio
Dec 9 '18 at 12:47
$begingroup$
Yes, you are right.
$endgroup$
– Larry
Dec 9 '18 at 12:49
1
$begingroup$
As a 3-d object, your circle is convex and has total surface area $2pi r^2$, so the average area of its projection along different orientations is $frac14$ of that, i.e $frac{pi}{2}{r^2}$. In general, when you randomly rotate your circle and its normal vector is making an angle with the $z$-axis, the area of its projection onto $xy$-plane equals to $pi r^2|costheta|$. So the average area of projection is $$pi r^2 times frac{int_0^{2pi}int_0^{pi} |costheta| sintheta dtheta dphi}{int_0^{2pi}int_0^pi sintheta dtheta dphi} = frac{pi}{2}r^2$$
$endgroup$
– achille hui
Dec 9 '18 at 13:21
$begingroup$
@achillehui: Thank you, this is really helpful. Could you explain which angle is $theta$ and which angle is $phi$? I am not very familiar with 3D vector.
$endgroup$
– Larry
Dec 9 '18 at 13:32
1
$begingroup$
I am using physicist's version of spherical coordinate system $$(r,theta,phi) mapsto (x,y,z) = (r sintheta cosphi,r sintheta sinphi, rcostheta)$$ where $r ge 0, theta in [0,pi]$ and $phi in [0,2pi)$. $theta$ is the angle between the point and +ve $z$-axis.
$endgroup$
– achille hui
Dec 9 '18 at 13:43
add a comment |
$begingroup$
As a follow up to Average projected area in higher dimension, I started to think about a variation of the original post. Let's visualize a circle with radius $1$ in $3D$ space and cast its shadow on the $x$-$z$ plane. The area of the shadow will change as we orient the circle in space. After we obtain all possible orientations of the circle, what will the average area of the shadow be? Just for clarification, average area is the sum of projected area divided by the number of possible orientations. For this problem, there are infinitely many orientations, so we have to apply calculus here. I think we have to set an integral over the circle's area, but I am not sure how exactly I should do that. Any help or related information would be truly appreciated.
calculus definite-integrals surfaces
asked Dec 9 '18 at 12:21
LarryLarry
2,31431028
$endgroup$
As a follow up to Average projected area in higher dimension, I started to think about a variation of the original post. Let's visualize a circle with radius $1$ in $3D$ space and cast its shadow on the $x$-$z$ plane. The area of the shadow will change as we orient the circle in space. After we obtain all possible orientations of the circle, what will the average area of the shadow be? Just for clarification, average area is the sum of projected area divided by the number of possible orientations. For this problem, there are infinitely many orientations, so we have to apply calculus here. I think we have to set an integral over the circle's area, but I am not sure how exactly I should do that. Any help or related information would be truly appreciated.
calculus definite-integrals surfaces
calculus definite-integrals surfaces
asked Dec 9 '18 at 12:21
LarryLarry
2,31431028
asked Dec 9 '18 at 12:21
LarryLarry
2,31431028
asked Dec 9 '18 at 12:21
LarryLarry
2,31431028
asked Dec 9 '18 at 12:21
LarryLarry
2,31431028
asked Dec 9 '18 at 12:21
LarryLarry
2,31431028
2,31431028
$begingroup$
I think this will depend on the position of the light origin as well. Your diagrams suggest that you're thinking of it as located far away (infinitely far), so that, when the circle lies parallel to the $x, z$ plane, the area of the shadow is equal to that of the circle. Am I right?
$endgroup$
– Patricio
Dec 9 '18 at 12:47
$begingroup$
Yes, you are right.
$endgroup$
– Larry
Dec 9 '18 at 12:49
1
$begingroup$
As a 3-d object, your circle is convex and has total surface area $2pi r^2$, so the average area of its projection along different orientations is $frac14$ of that, i.e $frac{pi}{2}{r^2}$. In general, when you randomly rotate your circle and its normal vector is making an angle with the $z$-axis, the area of its projection onto $xy$-plane equals to $pi r^2|costheta|$. So the average area of projection is $$pi r^2 times frac{int_0^{2pi}int_0^{pi} |costheta| sintheta dtheta dphi}{int_0^{2pi}int_0^pi sintheta dtheta dphi} = frac{pi}{2}r^2$$
$endgroup$
– achille hui
Dec 9 '18 at 13:21
$begingroup$
@achillehui: Thank you, this is really helpful. Could you explain which angle is $theta$ and which angle is $phi$? I am not very familiar with 3D vector.
$endgroup$
– Larry
Dec 9 '18 at 13:32
1
$begingroup$
I am using physicist's version of spherical coordinate system $$(r,theta,phi) mapsto (x,y,z) = (r sintheta cosphi,r sintheta sinphi, rcostheta)$$ where $r ge 0, theta in [0,pi]$ and $phi in [0,2pi)$. $theta$ is the angle between the point and +ve $z$-axis.
$endgroup$
– achille hui
Dec 9 '18 at 13:43
add a comment |
$begingroup$
I think this will depend on the position of the light origin as well. Your diagrams suggest that you're thinking of it as located far away (infinitely far), so that, when the circle lies parallel to the $x, z$ plane, the area of the shadow is equal to that of the circle. Am I right?
$endgroup$
– Patricio
Dec 9 '18 at 12:47
$begingroup$
Yes, you are right.
$endgroup$
– Larry
Dec 9 '18 at 12:49
1
$begingroup$
As a 3-d object, your circle is convex and has total surface area $2pi r^2$, so the average area of its projection along different orientations is $frac14$ of that, i.e $frac{pi}{2}{r^2}$. In general, when you randomly rotate your circle and its normal vector is making an angle with the $z$-axis, the area of its projection onto $xy$-plane equals to $pi r^2|costheta|$. So the average area of projection is $$pi r^2 times frac{int_0^{2pi}int_0^{pi} |costheta| sintheta dtheta dphi}{int_0^{2pi}int_0^pi sintheta dtheta dphi} = frac{pi}{2}r^2$$
$endgroup$
– achille hui
Dec 9 '18 at 13:21
$begingroup$
@achillehui: Thank you, this is really helpful. Could you explain which angle is $theta$ and which angle is $phi$? I am not very familiar with 3D vector.
$endgroup$
– Larry
Dec 9 '18 at 13:32
1
$begingroup$
I am using physicist's version of spherical coordinate system $$(r,theta,phi) mapsto (x,y,z) = (r sintheta cosphi,r sintheta sinphi, rcostheta)$$ where $r ge 0, theta in [0,pi]$ and $phi in [0,2pi)$. $theta$ is the angle between the point and +ve $z$-axis.
$endgroup$
– achille hui
Dec 9 '18 at 13:43
$begingroup$
I think this will depend on the position of the light origin as well. Your diagrams suggest that you're thinking of it as located far away (infinitely far), so that, when the circle lies parallel to the $x, z$ plane, the area of the shadow is equal to that of the circle. Am I right?
$endgroup$
– Patricio
Dec 9 '18 at 12:47
$begingroup$
I think this will depend on the position of the light origin as well. Your diagrams suggest that you're thinking of it as located far away (infinitely far), so that, when the circle lies parallel to the $x, z$ plane, the area of the shadow is equal to that of the circle. Am I right?
$endgroup$
– Patricio
Dec 9 '18 at 12:47
$begingroup$
Yes, you are right.
$endgroup$
– Larry
Dec 9 '18 at 12:49
$begingroup$
Yes, you are right.
$endgroup$
– Larry
Dec 9 '18 at 12:49
1
1
$begingroup$
As a 3-d object, your circle is convex and has total surface area $2pi r^2$, so the average area of its projection along different orientations is $frac14$ of that, i.e $frac{pi}{2}{r^2}$. In general, when you randomly rotate your circle and its normal vector is making an angle with the $z$-axis, the area of its projection onto $xy$-plane equals to $pi r^2|costheta|$. So the average area of projection is $$pi r^2 times frac{int_0^{2pi}int_0^{pi} |costheta| sintheta dtheta dphi}{int_0^{2pi}int_0^pi sintheta dtheta dphi} = frac{pi}{2}r^2$$
$endgroup$
– achille hui
Dec 9 '18 at 13:21
$begingroup$
As a 3-d object, your circle is convex and has total surface area $2pi r^2$, so the average area of its projection along different orientations is $frac14$ of that, i.e $frac{pi}{2}{r^2}$. In general, when you randomly rotate your circle and its normal vector is making an angle with the $z$-axis, the area of its projection onto $xy$-plane equals to $pi r^2|costheta|$. So the average area of projection is $$pi r^2 times frac{int_0^{2pi}int_0^{pi} |costheta| sintheta dtheta dphi}{int_0^{2pi}int_0^pi sintheta dtheta dphi} = frac{pi}{2}r^2$$
$endgroup$
– achille hui
Dec 9 '18 at 13:21
$begingroup$
@achillehui: Thank you, this is really helpful. Could you explain which angle is $theta$ and which angle is $phi$? I am not very familiar with 3D vector.
$endgroup$
– Larry
Dec 9 '18 at 13:32
$begingroup$
@achillehui: Thank you, this is really helpful. Could you explain which angle is $theta$ and which angle is $phi$? I am not very familiar with 3D vector.
$endgroup$
– Larry
Dec 9 '18 at 13:32
1
1
$begingroup$
I am using physicist's version of spherical coordinate system $$(r,theta,phi) mapsto (x,y,z) = (r sintheta cosphi,r sintheta sinphi, rcostheta)$$ where $r ge 0, theta in [0,pi]$ and $phi in [0,2pi)$. $theta$ is the angle between the point and +ve $z$-axis.
$endgroup$
– achille hui
Dec 9 '18 at 13:43
$begingroup$
I am using physicist's version of spherical coordinate system $$(r,theta,phi) mapsto (x,y,z) = (r sintheta cosphi,r sintheta sinphi, rcostheta)$$ where $r ge 0, theta in [0,pi]$ and $phi in [0,2pi)$. $theta$ is the angle between the point and +ve $z$-axis.
$endgroup$
– achille hui
Dec 9 '18 at 13:43
add a comment |
1 Answer
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The problem can be reformulated probabilistically:
Uniformly choose a random unit vector and take the unit circle perpendicular to that vector. What is the expected area of the circle's projection onto the $xy$-plane?
By symmetry, we can consider only random vectors with positive $z$-coordinate. Let the angle the chosen unit vector makes with the $xy$-plane be $varphi$, then its pdf is $cosvarphi$ for $0levarphilepi/2$. At a fixed $varphi$ the circle's projection is an ellipse with semi-axes 1 and $sinvarphi$, thus area $pisinvarphi$. Therefore the expected area of the projection is
$$int_0^{pi/2}pisinvarphicosvarphi,dvarphi=fracpi2$$
answered Dec 9 '18 at 14:10
Parcly TaxelParcly Taxel
41.7k137299
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add a comment |
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$begingroup$
The problem can be reformulated probabilistically:
Uniformly choose a random unit vector and take the unit circle perpendicular to that vector. What is the expected area of the circle's projection onto the $xy$-plane?
By symmetry, we can consider only random vectors with positive $z$-coordinate. Let the angle the chosen unit vector makes with the $xy$-plane be $varphi$, then its pdf is $cosvarphi$ for $0levarphilepi/2$. At a fixed $varphi$ the circle's projection is an ellipse with semi-axes 1 and $sinvarphi$, thus area $pisinvarphi$. Therefore the expected area of the projection is
$$int_0^{pi/2}pisinvarphicosvarphi,dvarphi=fracpi2$$
answered Dec 9 '18 at 14:10
Parcly TaxelParcly Taxel
41.7k137299
$endgroup$
add a comment |
$begingroup$
The problem can be reformulated probabilistically:
Uniformly choose a random unit vector and take the unit circle perpendicular to that vector. What is the expected area of the circle's projection onto the $xy$-plane?
By symmetry, we can consider only random vectors with positive $z$-coordinate. Let the angle the chosen unit vector makes with the $xy$-plane be $varphi$, then its pdf is $cosvarphi$ for $0levarphilepi/2$. At a fixed $varphi$ the circle's projection is an ellipse with semi-axes 1 and $sinvarphi$, thus area $pisinvarphi$. Therefore the expected area of the projection is
$$int_0^{pi/2}pisinvarphicosvarphi,dvarphi=fracpi2$$
answered Dec 9 '18 at 14:10
Parcly TaxelParcly Taxel
41.7k137299
$endgroup$
add a comment |
$begingroup$
The problem can be reformulated probabilistically:
Uniformly choose a random unit vector and take the unit circle perpendicular to that vector. What is the expected area of the circle's projection onto the $xy$-plane?
By symmetry, we can consider only random vectors with positive $z$-coordinate. Let the angle the chosen unit vector makes with the $xy$-plane be $varphi$, then its pdf is $cosvarphi$ for $0levarphilepi/2$. At a fixed $varphi$ the circle's projection is an ellipse with semi-axes 1 and $sinvarphi$, thus area $pisinvarphi$. Therefore the expected area of the projection is
$$int_0^{pi/2}pisinvarphicosvarphi,dvarphi=fracpi2$$
answered Dec 9 '18 at 14:10
Parcly TaxelParcly Taxel
41.7k137299
$endgroup$
The problem can be reformulated probabilistically:
Uniformly choose a random unit vector and take the unit circle perpendicular to that vector. What is the expected area of the circle's projection onto the $xy$-plane?
By symmetry, we can consider only random vectors with positive $z$-coordinate. Let the angle the chosen unit vector makes with the $xy$-plane be $varphi$, then its pdf is $cosvarphi$ for $0levarphilepi/2$. At a fixed $varphi$ the circle's projection is an ellipse with semi-axes 1 and $sinvarphi$, thus area $pisinvarphi$. Therefore the expected area of the projection is
$$int_0^{pi/2}pisinvarphicosvarphi,dvarphi=fracpi2$$
answered Dec 9 '18 at 14:10
Parcly TaxelParcly Taxel
41.7k137299
answered Dec 9 '18 at 14:10
Parcly TaxelParcly Taxel
41.7k137299
answered Dec 9 '18 at 14:10
Parcly TaxelParcly Taxel
41.7k137299
answered Dec 9 '18 at 14:10
Parcly TaxelParcly Taxel
41.7k137299
41.7k137299
add a comment |
add a comment |
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$begingroup$
I think this will depend on the position of the light origin as well. Your diagrams suggest that you're thinking of it as located far away (infinitely far), so that, when the circle lies parallel to the $x, z$ plane, the area of the shadow is equal to that of the circle. Am I right?
$endgroup$
– Patricio
Dec 9 '18 at 12:47
$begingroup$
Yes, you are right.
$endgroup$
– Larry
Dec 9 '18 at 12:49
1
$begingroup$
As a 3-d object, your circle is convex and has total surface area $2pi r^2$, so the average area of its projection along different orientations is $frac14$ of that, i.e $frac{pi}{2}{r^2}$. In general, when you randomly rotate your circle and its normal vector is making an angle with the $z$-axis, the area of its projection onto $xy$-plane equals to $pi r^2|costheta|$. So the average area of projection is $$pi r^2 times frac{int_0^{2pi}int_0^{pi} |costheta| sintheta dtheta dphi}{int_0^{2pi}int_0^pi sintheta dtheta dphi} = frac{pi}{2}r^2$$
$endgroup$
– achille hui
Dec 9 '18 at 13:21
$begingroup$
@achillehui: Thank you, this is really helpful. Could you explain which angle is $theta$ and which angle is $phi$? I am not very familiar with 3D vector.
$endgroup$
– Larry
Dec 9 '18 at 13:32
1
$begingroup$
I am using physicist's version of spherical coordinate system $$(r,theta,phi) mapsto (x,y,z) = (r sintheta cosphi,r sintheta sinphi, rcostheta)$$ where $r ge 0, theta in [0,pi]$ and $phi in [0,2pi)$. $theta$ is the angle between the point and +ve $z$-axis.
$endgroup$
– achille hui
Dec 9 '18 at 13:43