Mixing
Volume of a segment of a sphere
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Let a sphere of radius r be cut by a plane,therefore forming a segment of height h.fine the volume of the segment
I wonder how to start. shell or disk method,i thought shell was appropriate.but IDK what is the volume it asked?from 0 to h or h to the top of the sphere? THANK YOU
calculus
asked Jan 3 at 3:39
KizaruKizaru
274
$endgroup$
add a comment |
$begingroup$
Let a sphere of radius r be cut by a plane,therefore forming a segment of height h.fine the volume of the segment
I wonder how to start. shell or disk method,i thought shell was appropriate.but IDK what is the volume it asked?from 0 to h or h to the top of the sphere? THANK YOU
calculus
asked Jan 3 at 3:39
KizaruKizaru
274
$endgroup$
$begingroup$
How are you with spherical coordinates? Are you comfortable with Jacobians?
$endgroup$
– Theo Bendit
Jan 3 at 3:41
$begingroup$
You know the sum of the volumes of the two proposed segments is equal to $frac{4 times pi times r^3}{3}$ so finding one is essentially equivalent to finding the other...
$endgroup$
– Neeyanth Kopparapu
Jan 3 at 3:54
add a comment |
$begingroup$
Let a sphere of radius r be cut by a plane,therefore forming a segment of height h.fine the volume of the segment
I wonder how to start. shell or disk method,i thought shell was appropriate.but IDK what is the volume it asked?from 0 to h or h to the top of the sphere? THANK YOU
calculus
asked Jan 3 at 3:39
KizaruKizaru
274
$endgroup$
Let a sphere of radius r be cut by a plane,therefore forming a segment of height h.fine the volume of the segment
I wonder how to start. shell or disk method,i thought shell was appropriate.but IDK what is the volume it asked?from 0 to h or h to the top of the sphere? THANK YOU
calculus
calculus
asked Jan 3 at 3:39
KizaruKizaru
274
asked Jan 3 at 3:39
KizaruKizaru
274
asked Jan 3 at 3:39
KizaruKizaru
274
asked Jan 3 at 3:39
KizaruKizaru
274
asked Jan 3 at 3:39
KizaruKizaru
274
274
$begingroup$
How are you with spherical coordinates? Are you comfortable with Jacobians?
$endgroup$
– Theo Bendit
Jan 3 at 3:41
$begingroup$
You know the sum of the volumes of the two proposed segments is equal to $frac{4 times pi times r^3}{3}$ so finding one is essentially equivalent to finding the other...
$endgroup$
– Neeyanth Kopparapu
Jan 3 at 3:54
add a comment |
$begingroup$
How are you with spherical coordinates? Are you comfortable with Jacobians?
$endgroup$
– Theo Bendit
Jan 3 at 3:41
$begingroup$
You know the sum of the volumes of the two proposed segments is equal to $frac{4 times pi times r^3}{3}$ so finding one is essentially equivalent to finding the other...
$endgroup$
– Neeyanth Kopparapu
Jan 3 at 3:54
$begingroup$
How are you with spherical coordinates? Are you comfortable with Jacobians?
$endgroup$
– Theo Bendit
Jan 3 at 3:41
$begingroup$
How are you with spherical coordinates? Are you comfortable with Jacobians?
$endgroup$
– Theo Bendit
Jan 3 at 3:41
$begingroup$
You know the sum of the volumes of the two proposed segments is equal to $frac{4 times pi times r^3}{3}$ so finding one is essentially equivalent to finding the other...
$endgroup$
– Neeyanth Kopparapu
Jan 3 at 3:54
$begingroup$
You know the sum of the volumes of the two proposed segments is equal to $frac{4 times pi times r^3}{3}$ so finding one is essentially equivalent to finding the other...
$endgroup$
– Neeyanth Kopparapu
Jan 3 at 3:54
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
I think the easiest way to find the volume is to chop the segments into circles with an infinitesimally small width. Lets just assume for the sake of argument that this segment is contained entirely in the upper hemisphere. For any cross-sectional circle whose center is $x$ distance away from the center of the sphere, we can use the Pythagorean theorem to determine that $$r`^2 + x^2 = r^2,$$ where $r`$ is the radius of the cross-sectional circle.
That means the area of these circles is equivalent to $pi times r`^2 = pi cdot (r^2-x^2)$. We are now tasked to find the integral of $$int_{d_1}^{d_2} pi cdot (r^2-x^2) dx$$ To find the range of integration, we think about the problem statement. If the segment is of height $h$, then the top of the segment is $r$ above the center of the sphere, and thus the bottom is $r-h$ above the center of the sphere (note that if it is below, $r-h < 0$ but that's fine because we are squaring the quantity so we are only concerned with its magnitude). We can solve this integral then to determine $$int_{r-h}^r pi cdot (r^2-x^2)dx = pi timesBig[(r^2(r-(r-h)) + frac{1}{3}((r-h)^3-r^3))Big] = pibig[rh^2-frac{h^3}{3}big].$$ You can confirm this by plugging in $h=r$ (implying the entire hemisphere) to get $frac{2pi r^3}{3}$, which is half the volume of the sphere.
EDIT: More information here: https://en.wikipedia.org/wiki/Spherical_cap
edited Jan 3 at 4:56
Shubham Johri
4,734717
answered Jan 3 at 4:05
Neeyanth KopparapuNeeyanth Kopparapu
3016
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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active
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votes
$begingroup$
I think the easiest way to find the volume is to chop the segments into circles with an infinitesimally small width. Lets just assume for the sake of argument that this segment is contained entirely in the upper hemisphere. For any cross-sectional circle whose center is $x$ distance away from the center of the sphere, we can use the Pythagorean theorem to determine that $$r`^2 + x^2 = r^2,$$ where $r`$ is the radius of the cross-sectional circle.
That means the area of these circles is equivalent to $pi times r`^2 = pi cdot (r^2-x^2)$. We are now tasked to find the integral of $$int_{d_1}^{d_2} pi cdot (r^2-x^2) dx$$ To find the range of integration, we think about the problem statement. If the segment is of height $h$, then the top of the segment is $r$ above the center of the sphere, and thus the bottom is $r-h$ above the center of the sphere (note that if it is below, $r-h < 0$ but that's fine because we are squaring the quantity so we are only concerned with its magnitude). We can solve this integral then to determine $$int_{r-h}^r pi cdot (r^2-x^2)dx = pi timesBig[(r^2(r-(r-h)) + frac{1}{3}((r-h)^3-r^3))Big] = pibig[rh^2-frac{h^3}{3}big].$$ You can confirm this by plugging in $h=r$ (implying the entire hemisphere) to get $frac{2pi r^3}{3}$, which is half the volume of the sphere.
EDIT: More information here: https://en.wikipedia.org/wiki/Spherical_cap
edited Jan 3 at 4:56
Shubham Johri
4,734717
answered Jan 3 at 4:05
Neeyanth KopparapuNeeyanth Kopparapu
3016
$endgroup$
add a comment |
$begingroup$
I think the easiest way to find the volume is to chop the segments into circles with an infinitesimally small width. Lets just assume for the sake of argument that this segment is contained entirely in the upper hemisphere. For any cross-sectional circle whose center is $x$ distance away from the center of the sphere, we can use the Pythagorean theorem to determine that $$r`^2 + x^2 = r^2,$$ where $r`$ is the radius of the cross-sectional circle.
That means the area of these circles is equivalent to $pi times r`^2 = pi cdot (r^2-x^2)$. We are now tasked to find the integral of $$int_{d_1}^{d_2} pi cdot (r^2-x^2) dx$$ To find the range of integration, we think about the problem statement. If the segment is of height $h$, then the top of the segment is $r$ above the center of the sphere, and thus the bottom is $r-h$ above the center of the sphere (note that if it is below, $r-h < 0$ but that's fine because we are squaring the quantity so we are only concerned with its magnitude). We can solve this integral then to determine $$int_{r-h}^r pi cdot (r^2-x^2)dx = pi timesBig[(r^2(r-(r-h)) + frac{1}{3}((r-h)^3-r^3))Big] = pibig[rh^2-frac{h^3}{3}big].$$ You can confirm this by plugging in $h=r$ (implying the entire hemisphere) to get $frac{2pi r^3}{3}$, which is half the volume of the sphere.
EDIT: More information here: https://en.wikipedia.org/wiki/Spherical_cap
edited Jan 3 at 4:56
Shubham Johri
4,734717
answered Jan 3 at 4:05
Neeyanth KopparapuNeeyanth Kopparapu
3016
$endgroup$
add a comment |
$begingroup$
I think the easiest way to find the volume is to chop the segments into circles with an infinitesimally small width. Lets just assume for the sake of argument that this segment is contained entirely in the upper hemisphere. For any cross-sectional circle whose center is $x$ distance away from the center of the sphere, we can use the Pythagorean theorem to determine that $$r`^2 + x^2 = r^2,$$ where $r`$ is the radius of the cross-sectional circle.
That means the area of these circles is equivalent to $pi times r`^2 = pi cdot (r^2-x^2)$. We are now tasked to find the integral of $$int_{d_1}^{d_2} pi cdot (r^2-x^2) dx$$ To find the range of integration, we think about the problem statement. If the segment is of height $h$, then the top of the segment is $r$ above the center of the sphere, and thus the bottom is $r-h$ above the center of the sphere (note that if it is below, $r-h < 0$ but that's fine because we are squaring the quantity so we are only concerned with its magnitude). We can solve this integral then to determine $$int_{r-h}^r pi cdot (r^2-x^2)dx = pi timesBig[(r^2(r-(r-h)) + frac{1}{3}((r-h)^3-r^3))Big] = pibig[rh^2-frac{h^3}{3}big].$$ You can confirm this by plugging in $h=r$ (implying the entire hemisphere) to get $frac{2pi r^3}{3}$, which is half the volume of the sphere.
EDIT: More information here: https://en.wikipedia.org/wiki/Spherical_cap
edited Jan 3 at 4:56
Shubham Johri
4,734717
answered Jan 3 at 4:05
Neeyanth KopparapuNeeyanth Kopparapu
3016
$endgroup$
I think the easiest way to find the volume is to chop the segments into circles with an infinitesimally small width. Lets just assume for the sake of argument that this segment is contained entirely in the upper hemisphere. For any cross-sectional circle whose center is $x$ distance away from the center of the sphere, we can use the Pythagorean theorem to determine that $$r`^2 + x^2 = r^2,$$ where $r`$ is the radius of the cross-sectional circle.
That means the area of these circles is equivalent to $pi times r`^2 = pi cdot (r^2-x^2)$. We are now tasked to find the integral of $$int_{d_1}^{d_2} pi cdot (r^2-x^2) dx$$ To find the range of integration, we think about the problem statement. If the segment is of height $h$, then the top of the segment is $r$ above the center of the sphere, and thus the bottom is $r-h$ above the center of the sphere (note that if it is below, $r-h < 0$ but that's fine because we are squaring the quantity so we are only concerned with its magnitude). We can solve this integral then to determine $$int_{r-h}^r pi cdot (r^2-x^2)dx = pi timesBig[(r^2(r-(r-h)) + frac{1}{3}((r-h)^3-r^3))Big] = pibig[rh^2-frac{h^3}{3}big].$$ You can confirm this by plugging in $h=r$ (implying the entire hemisphere) to get $frac{2pi r^3}{3}$, which is half the volume of the sphere.
EDIT: More information here: https://en.wikipedia.org/wiki/Spherical_cap
edited Jan 3 at 4:56
Shubham Johri
4,734717
answered Jan 3 at 4:05
Neeyanth KopparapuNeeyanth Kopparapu
3016
edited Jan 3 at 4:56
Shubham Johri
4,734717
edited Jan 3 at 4:56
Shubham Johri
4,734717
edited Jan 3 at 4:56
Shubham Johri
4,734717
4,734717
answered Jan 3 at 4:05
Neeyanth KopparapuNeeyanth Kopparapu
3016
answered Jan 3 at 4:05
Neeyanth KopparapuNeeyanth Kopparapu
3016
answered Jan 3 at 4:05
Neeyanth KopparapuNeeyanth Kopparapu
3016
3016
add a comment |
add a comment |
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$begingroup$
How are you with spherical coordinates? Are you comfortable with Jacobians?
$endgroup$
– Theo Bendit
Jan 3 at 3:41
$begingroup$
You know the sum of the volumes of the two proposed segments is equal to $frac{4 times pi times r^3}{3}$ so finding one is essentially equivalent to finding the other...
$endgroup$
– Neeyanth Kopparapu
Jan 3 at 3:54