Why does the integral of the surface area of a sphere equal its volume? [duplicate]
$begingroup$
This question already has an answer here:
Why is the derivative of a circle's area its perimeter (and similarly for spheres)?
8 answers
Why does the integral of the surface area of a sphere equal it's volume?
$$int{4pi r^2 mathrm{d}r =frac{4}{3}pi r^3}$$
I don't quite understand why the relationship between surface area and volume is this way. Is this true for any other shapes?
volume area
edited May 2 '14 at 5:34
Bennett Gardiner
3,05911540
asked May 2 '14 at 5:26
codedudecodedude
3472720
$endgroup$
marked as duplicate by Najib Idrissi, Grigory M, user63181, mau, vonbrand May 2 '14 at 9:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Why is the derivative of a circle's area its perimeter (and similarly for spheres)?
8 answers
Why does the integral of the surface area of a sphere equal it's volume?
$$int{4pi r^2 mathrm{d}r =frac{4}{3}pi r^3}$$
I don't quite understand why the relationship between surface area and volume is this way. Is this true for any other shapes?
volume area
edited May 2 '14 at 5:34
Bennett Gardiner
3,05911540
asked May 2 '14 at 5:26
codedudecodedude
3472720
$endgroup$
marked as duplicate by Najib Idrissi, Grigory M, user63181, mau, vonbrand May 2 '14 at 9:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
$begingroup$
True for circles as well. Area = $pi r^2$. Circumference = $2pi r$. So here's the question: does it hold for all $n$-spheres? Other kinds of surfaces?
$endgroup$
– Kaj Hansen
May 2 '14 at 5:27
$begingroup$
Integrating the surface equation gives you the volume that the surface encloses like when you integrate the function of a graph you obtain the area under the curve.
$endgroup$
– Test123
May 2 '14 at 5:30
$begingroup$
So would this apply to say a square-because integrating 6x^2 doesn't give me x^3
$endgroup$
– codedude
May 2 '14 at 5:31
$begingroup$
I'm sorry... I meant a cube. Clearly it's too late for math. :)
$endgroup$
– codedude
May 2 '14 at 5:40
1
$begingroup$
If you consider the radius of a cube to be the distance $r$ from center to middle of a face, then its volume is $(2r)^3$ and its surface area is $6(2r)^2$. Integrate from $r=0$ to $r=r$.
$endgroup$
– DanielV
May 2 '14 at 5:42
add a comment |
$begingroup$
This question already has an answer here:
Why is the derivative of a circle's area its perimeter (and similarly for spheres)?
8 answers
Why does the integral of the surface area of a sphere equal it's volume?
$$int{4pi r^2 mathrm{d}r =frac{4}{3}pi r^3}$$
I don't quite understand why the relationship between surface area and volume is this way. Is this true for any other shapes?
volume area
edited May 2 '14 at 5:34
Bennett Gardiner
3,05911540
asked May 2 '14 at 5:26
codedudecodedude
3472720
$endgroup$
This question already has an answer here:
Why is the derivative of a circle's area its perimeter (and similarly for spheres)?
8 answers
Why does the integral of the surface area of a sphere equal it's volume?
$$int{4pi r^2 mathrm{d}r =frac{4}{3}pi r^3}$$
I don't quite understand why the relationship between surface area and volume is this way. Is this true for any other shapes?
This question already has an answer here:
Why is the derivative of a circle's area its perimeter (and similarly for spheres)?
8 answers
volume area
volume area
edited May 2 '14 at 5:34
Bennett Gardiner
3,05911540
asked May 2 '14 at 5:26
codedudecodedude
3472720
edited May 2 '14 at 5:34
Bennett Gardiner
3,05911540
asked May 2 '14 at 5:26
codedudecodedude
3472720
edited May 2 '14 at 5:34
Bennett Gardiner
3,05911540
edited May 2 '14 at 5:34
Bennett Gardiner
3,05911540
edited May 2 '14 at 5:34
Bennett Gardiner
3,05911540
3,05911540
asked May 2 '14 at 5:26
codedudecodedude
3472720
asked May 2 '14 at 5:26
codedudecodedude
3472720
asked May 2 '14 at 5:26
codedudecodedude
3472720
3472720
marked as duplicate by Najib Idrissi, Grigory M, user63181, mau, vonbrand May 2 '14 at 9:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Najib Idrissi, Grigory M, user63181, mau, vonbrand May 2 '14 at 9:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
$begingroup$
True for circles as well. Area = $pi r^2$. Circumference = $2pi r$. So here's the question: does it hold for all $n$-spheres? Other kinds of surfaces?
$endgroup$
– Kaj Hansen
May 2 '14 at 5:27
$begingroup$
Integrating the surface equation gives you the volume that the surface encloses like when you integrate the function of a graph you obtain the area under the curve.
$endgroup$
– Test123
May 2 '14 at 5:30
$begingroup$
So would this apply to say a square-because integrating 6x^2 doesn't give me x^3
$endgroup$
– codedude
May 2 '14 at 5:31
$begingroup$
I'm sorry... I meant a cube. Clearly it's too late for math. :)
$endgroup$
– codedude
May 2 '14 at 5:40
1
$begingroup$
If you consider the radius of a cube to be the distance $r$ from center to middle of a face, then its volume is $(2r)^3$ and its surface area is $6(2r)^2$. Integrate from $r=0$ to $r=r$.
$endgroup$
– DanielV
May 2 '14 at 5:42
add a comment |
3
$begingroup$
True for circles as well. Area = $pi r^2$. Circumference = $2pi r$. So here's the question: does it hold for all $n$-spheres? Other kinds of surfaces?
$endgroup$
– Kaj Hansen
May 2 '14 at 5:27
$begingroup$
Integrating the surface equation gives you the volume that the surface encloses like when you integrate the function of a graph you obtain the area under the curve.
$endgroup$
– Test123
May 2 '14 at 5:30
$begingroup$
So would this apply to say a square-because integrating 6x^2 doesn't give me x^3
$endgroup$
– codedude
May 2 '14 at 5:31
$begingroup$
I'm sorry... I meant a cube. Clearly it's too late for math. :)
$endgroup$
– codedude
May 2 '14 at 5:40
1
$begingroup$
If you consider the radius of a cube to be the distance $r$ from center to middle of a face, then its volume is $(2r)^3$ and its surface area is $6(2r)^2$. Integrate from $r=0$ to $r=r$.
$endgroup$
– DanielV
May 2 '14 at 5:42
3
3
$begingroup$
True for circles as well. Area = $pi r^2$. Circumference = $2pi r$. So here's the question: does it hold for all $n$-spheres? Other kinds of surfaces?
$endgroup$
– Kaj Hansen
May 2 '14 at 5:27
$begingroup$
True for circles as well. Area = $pi r^2$. Circumference = $2pi r$. So here's the question: does it hold for all $n$-spheres? Other kinds of surfaces?
$endgroup$
– Kaj Hansen
May 2 '14 at 5:27
$begingroup$
Integrating the surface equation gives you the volume that the surface encloses like when you integrate the function of a graph you obtain the area under the curve.
$endgroup$
– Test123
May 2 '14 at 5:30
$begingroup$
Integrating the surface equation gives you the volume that the surface encloses like when you integrate the function of a graph you obtain the area under the curve.
$endgroup$
– Test123
May 2 '14 at 5:30
$begingroup$
So would this apply to say a square-because integrating 6x^2 doesn't give me x^3
$endgroup$
– codedude
May 2 '14 at 5:31
$begingroup$
So would this apply to say a square-because integrating 6x^2 doesn't give me x^3
$endgroup$
– codedude
May 2 '14 at 5:31
$begingroup$
I'm sorry... I meant a cube. Clearly it's too late for math. :)
$endgroup$
– codedude
May 2 '14 at 5:40
$begingroup$
I'm sorry... I meant a cube. Clearly it's too late for math. :)
$endgroup$
– codedude
May 2 '14 at 5:40
1
1
$begingroup$
If you consider the radius of a cube to be the distance $r$ from center to middle of a face, then its volume is $(2r)^3$ and its surface area is $6(2r)^2$. Integrate from $r=0$ to $r=r$.
$endgroup$
– DanielV
May 2 '14 at 5:42
$begingroup$
If you consider the radius of a cube to be the distance $r$ from center to middle of a face, then its volume is $(2r)^3$ and its surface area is $6(2r)^2$. Integrate from $r=0$ to $r=r$.
$endgroup$
– DanielV
May 2 '14 at 5:42
add a comment |
2 Answers
2
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I'm sure some will have an issue with the technical aspects of the following explanation, but I believe that it should intuitively help.
One way to think about the area under a curve (and the definite integral), is to picture a vertical line from the horizontal axis to the graph of your function. Starting at say $x=0$, picture this line moving to the right. As it does, and its height changes, it will touch along the way a set of points in the plane (shaded below). Perhaps think about this line as being covered in paint, and as it moves along it paints the area under the curve.
You can view the definite integral $$int_0^xf(t)dt$$ as suggesting the motion of this line of changing height on the horizontal axis from 0 to $x$ and giving you the area of the region that it "paints".
The same thought process works for other shapes. Consider a square centered at the origin.
The perimeter of this square is $8x$. Think of $x$ starting at zero and increasing. As the square increases in size, its perimeter will also "paint" a collection of points in the plane, and we can think about the area of the square as being the painted points. In the same vein as above, the integral $$int_0^x8tdt=4x^2$$ suggests the painting of these points and gives us the area of the resulting collection of painted points (you can easily see that $4x^2$ is the area of the shaded square).
Likewise for a circle when integrating the circumference gives the area: $$int_0^r2pi tdt=pi r^2$$
Three dimensional shapes are a little harder to picture (and definitely harder to draw), but you can think about them in the same way. Extending the square example to a cube, we can now think about the faces of the cube (its surface area) as "painting" a volume of points. Consider a cube centered at the origin, with the horizontal axis going through the center of opposing faces. The surface area of such a cube is $24t^2$ (check) when it has "radius" $t$. Integrating $$int_0^x24t^2dt=8x^3$$ gives the volume of points touched by the faces of the cube as it expands from radius 0 to radius $x$.
Hopefully, by using the same sort of thought process on a sphere, you'll find that it makes a little more intuitive sense that the integral of its surface area gives us its volume.
answered May 2 '14 at 7:11
Scott H.Scott H.
446717
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add a comment |
$begingroup$
The reason is essentially the same as the reason why the volume of an onion is the same as the volume of all its layers. Imagine a spherical ball as being composed of many thin spherical layers, like an onion. The volume of the layer between radius $r$ and radius $r+delta r$ is (to a close approximation if $delta r$ is small enough) the area $4pi r^2$ times its thickness: $4pi r^2cdotdelta r$. Summing all these small volumes from the centre of the ball ($r=0$) to its outer radius ($r=a$, say) and taking the limit as the $delta r$ tend to $0$ gives us the limit
$$int_0^a 4pi r^2mathrm dr.$$As you know, this evaluates to $frac43!pi a^3$, the volume of the sphere.
answered May 2 '14 at 8:03
John BentinJohn Bentin
11.3k22554
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'm sure some will have an issue with the technical aspects of the following explanation, but I believe that it should intuitively help.
One way to think about the area under a curve (and the definite integral), is to picture a vertical line from the horizontal axis to the graph of your function. Starting at say $x=0$, picture this line moving to the right. As it does, and its height changes, it will touch along the way a set of points in the plane (shaded below). Perhaps think about this line as being covered in paint, and as it moves along it paints the area under the curve.
You can view the definite integral $$int_0^xf(t)dt$$ as suggesting the motion of this line of changing height on the horizontal axis from 0 to $x$ and giving you the area of the region that it "paints".
The same thought process works for other shapes. Consider a square centered at the origin.
The perimeter of this square is $8x$. Think of $x$ starting at zero and increasing. As the square increases in size, its perimeter will also "paint" a collection of points in the plane, and we can think about the area of the square as being the painted points. In the same vein as above, the integral $$int_0^x8tdt=4x^2$$ suggests the painting of these points and gives us the area of the resulting collection of painted points (you can easily see that $4x^2$ is the area of the shaded square).
Likewise for a circle when integrating the circumference gives the area: $$int_0^r2pi tdt=pi r^2$$
Three dimensional shapes are a little harder to picture (and definitely harder to draw), but you can think about them in the same way. Extending the square example to a cube, we can now think about the faces of the cube (its surface area) as "painting" a volume of points. Consider a cube centered at the origin, with the horizontal axis going through the center of opposing faces. The surface area of such a cube is $24t^2$ (check) when it has "radius" $t$. Integrating $$int_0^x24t^2dt=8x^3$$ gives the volume of points touched by the faces of the cube as it expands from radius 0 to radius $x$.
Hopefully, by using the same sort of thought process on a sphere, you'll find that it makes a little more intuitive sense that the integral of its surface area gives us its volume.
answered May 2 '14 at 7:11
Scott H.Scott H.
446717
$endgroup$
add a comment |
$begingroup$
I'm sure some will have an issue with the technical aspects of the following explanation, but I believe that it should intuitively help.
One way to think about the area under a curve (and the definite integral), is to picture a vertical line from the horizontal axis to the graph of your function. Starting at say $x=0$, picture this line moving to the right. As it does, and its height changes, it will touch along the way a set of points in the plane (shaded below). Perhaps think about this line as being covered in paint, and as it moves along it paints the area under the curve.
You can view the definite integral $$int_0^xf(t)dt$$ as suggesting the motion of this line of changing height on the horizontal axis from 0 to $x$ and giving you the area of the region that it "paints".
The same thought process works for other shapes. Consider a square centered at the origin.
The perimeter of this square is $8x$. Think of $x$ starting at zero and increasing. As the square increases in size, its perimeter will also "paint" a collection of points in the plane, and we can think about the area of the square as being the painted points. In the same vein as above, the integral $$int_0^x8tdt=4x^2$$ suggests the painting of these points and gives us the area of the resulting collection of painted points (you can easily see that $4x^2$ is the area of the shaded square).
Likewise for a circle when integrating the circumference gives the area: $$int_0^r2pi tdt=pi r^2$$
Three dimensional shapes are a little harder to picture (and definitely harder to draw), but you can think about them in the same way. Extending the square example to a cube, we can now think about the faces of the cube (its surface area) as "painting" a volume of points. Consider a cube centered at the origin, with the horizontal axis going through the center of opposing faces. The surface area of such a cube is $24t^2$ (check) when it has "radius" $t$. Integrating $$int_0^x24t^2dt=8x^3$$ gives the volume of points touched by the faces of the cube as it expands from radius 0 to radius $x$.
Hopefully, by using the same sort of thought process on a sphere, you'll find that it makes a little more intuitive sense that the integral of its surface area gives us its volume.
answered May 2 '14 at 7:11
Scott H.Scott H.
446717
$endgroup$
add a comment |
$begingroup$
I'm sure some will have an issue with the technical aspects of the following explanation, but I believe that it should intuitively help.
One way to think about the area under a curve (and the definite integral), is to picture a vertical line from the horizontal axis to the graph of your function. Starting at say $x=0$, picture this line moving to the right. As it does, and its height changes, it will touch along the way a set of points in the plane (shaded below). Perhaps think about this line as being covered in paint, and as it moves along it paints the area under the curve.
You can view the definite integral $$int_0^xf(t)dt$$ as suggesting the motion of this line of changing height on the horizontal axis from 0 to $x$ and giving you the area of the region that it "paints".
The same thought process works for other shapes. Consider a square centered at the origin.
The perimeter of this square is $8x$. Think of $x$ starting at zero and increasing. As the square increases in size, its perimeter will also "paint" a collection of points in the plane, and we can think about the area of the square as being the painted points. In the same vein as above, the integral $$int_0^x8tdt=4x^2$$ suggests the painting of these points and gives us the area of the resulting collection of painted points (you can easily see that $4x^2$ is the area of the shaded square).
Likewise for a circle when integrating the circumference gives the area: $$int_0^r2pi tdt=pi r^2$$
Three dimensional shapes are a little harder to picture (and definitely harder to draw), but you can think about them in the same way. Extending the square example to a cube, we can now think about the faces of the cube (its surface area) as "painting" a volume of points. Consider a cube centered at the origin, with the horizontal axis going through the center of opposing faces. The surface area of such a cube is $24t^2$ (check) when it has "radius" $t$. Integrating $$int_0^x24t^2dt=8x^3$$ gives the volume of points touched by the faces of the cube as it expands from radius 0 to radius $x$.
Hopefully, by using the same sort of thought process on a sphere, you'll find that it makes a little more intuitive sense that the integral of its surface area gives us its volume.
answered May 2 '14 at 7:11
Scott H.Scott H.
446717
$endgroup$
I'm sure some will have an issue with the technical aspects of the following explanation, but I believe that it should intuitively help.
One way to think about the area under a curve (and the definite integral), is to picture a vertical line from the horizontal axis to the graph of your function. Starting at say $x=0$, picture this line moving to the right. As it does, and its height changes, it will touch along the way a set of points in the plane (shaded below). Perhaps think about this line as being covered in paint, and as it moves along it paints the area under the curve.
You can view the definite integral $$int_0^xf(t)dt$$ as suggesting the motion of this line of changing height on the horizontal axis from 0 to $x$ and giving you the area of the region that it "paints".
The same thought process works for other shapes. Consider a square centered at the origin.
The perimeter of this square is $8x$. Think of $x$ starting at zero and increasing. As the square increases in size, its perimeter will also "paint" a collection of points in the plane, and we can think about the area of the square as being the painted points. In the same vein as above, the integral $$int_0^x8tdt=4x^2$$ suggests the painting of these points and gives us the area of the resulting collection of painted points (you can easily see that $4x^2$ is the area of the shaded square).
Likewise for a circle when integrating the circumference gives the area: $$int_0^r2pi tdt=pi r^2$$
Three dimensional shapes are a little harder to picture (and definitely harder to draw), but you can think about them in the same way. Extending the square example to a cube, we can now think about the faces of the cube (its surface area) as "painting" a volume of points. Consider a cube centered at the origin, with the horizontal axis going through the center of opposing faces. The surface area of such a cube is $24t^2$ (check) when it has "radius" $t$. Integrating $$int_0^x24t^2dt=8x^3$$ gives the volume of points touched by the faces of the cube as it expands from radius 0 to radius $x$.
Hopefully, by using the same sort of thought process on a sphere, you'll find that it makes a little more intuitive sense that the integral of its surface area gives us its volume.
answered May 2 '14 at 7:11
Scott H.Scott H.
446717
answered May 2 '14 at 7:11
Scott H.Scott H.
446717
answered May 2 '14 at 7:11
Scott H.Scott H.
446717
answered May 2 '14 at 7:11
Scott H.Scott H.
446717
446717
add a comment |
add a comment |
$begingroup$
The reason is essentially the same as the reason why the volume of an onion is the same as the volume of all its layers. Imagine a spherical ball as being composed of many thin spherical layers, like an onion. The volume of the layer between radius $r$ and radius $r+delta r$ is (to a close approximation if $delta r$ is small enough) the area $4pi r^2$ times its thickness: $4pi r^2cdotdelta r$. Summing all these small volumes from the centre of the ball ($r=0$) to its outer radius ($r=a$, say) and taking the limit as the $delta r$ tend to $0$ gives us the limit
$$int_0^a 4pi r^2mathrm dr.$$As you know, this evaluates to $frac43!pi a^3$, the volume of the sphere.
answered May 2 '14 at 8:03
John BentinJohn Bentin
11.3k22554
$endgroup$
add a comment |
$begingroup$
The reason is essentially the same as the reason why the volume of an onion is the same as the volume of all its layers. Imagine a spherical ball as being composed of many thin spherical layers, like an onion. The volume of the layer between radius $r$ and radius $r+delta r$ is (to a close approximation if $delta r$ is small enough) the area $4pi r^2$ times its thickness: $4pi r^2cdotdelta r$. Summing all these small volumes from the centre of the ball ($r=0$) to its outer radius ($r=a$, say) and taking the limit as the $delta r$ tend to $0$ gives us the limit
$$int_0^a 4pi r^2mathrm dr.$$As you know, this evaluates to $frac43!pi a^3$, the volume of the sphere.
answered May 2 '14 at 8:03
John BentinJohn Bentin
11.3k22554
$endgroup$
add a comment |
$begingroup$
The reason is essentially the same as the reason why the volume of an onion is the same as the volume of all its layers. Imagine a spherical ball as being composed of many thin spherical layers, like an onion. The volume of the layer between radius $r$ and radius $r+delta r$ is (to a close approximation if $delta r$ is small enough) the area $4pi r^2$ times its thickness: $4pi r^2cdotdelta r$. Summing all these small volumes from the centre of the ball ($r=0$) to its outer radius ($r=a$, say) and taking the limit as the $delta r$ tend to $0$ gives us the limit
$$int_0^a 4pi r^2mathrm dr.$$As you know, this evaluates to $frac43!pi a^3$, the volume of the sphere.
answered May 2 '14 at 8:03
John BentinJohn Bentin
11.3k22554
$endgroup$
The reason is essentially the same as the reason why the volume of an onion is the same as the volume of all its layers. Imagine a spherical ball as being composed of many thin spherical layers, like an onion. The volume of the layer between radius $r$ and radius $r+delta r$ is (to a close approximation if $delta r$ is small enough) the area $4pi r^2$ times its thickness: $4pi r^2cdotdelta r$. Summing all these small volumes from the centre of the ball ($r=0$) to its outer radius ($r=a$, say) and taking the limit as the $delta r$ tend to $0$ gives us the limit
$$int_0^a 4pi r^2mathrm dr.$$As you know, this evaluates to $frac43!pi a^3$, the volume of the sphere.
answered May 2 '14 at 8:03
John BentinJohn Bentin
11.3k22554
answered May 2 '14 at 8:03
John BentinJohn Bentin
11.3k22554
answered May 2 '14 at 8:03
John BentinJohn Bentin
11.3k22554
answered May 2 '14 at 8:03
John BentinJohn Bentin
11.3k22554
11.3k22554
add a comment |
add a comment |
3
$begingroup$
True for circles as well. Area = $pi r^2$. Circumference = $2pi r$. So here's the question: does it hold for all $n$-spheres? Other kinds of surfaces?
$endgroup$
– Kaj Hansen
May 2 '14 at 5:27
$begingroup$
Integrating the surface equation gives you the volume that the surface encloses like when you integrate the function of a graph you obtain the area under the curve.
$endgroup$
– Test123
May 2 '14 at 5:30
$begingroup$
So would this apply to say a square-because integrating 6x^2 doesn't give me x^3
$endgroup$
– codedude
May 2 '14 at 5:31
$begingroup$
I'm sorry... I meant a cube. Clearly it's too late for math. :)
$endgroup$
– codedude
May 2 '14 at 5:40
1
$begingroup$
If you consider the radius of a cube to be the distance $r$ from center to middle of a face, then its volume is $(2r)^3$ and its surface area is $6(2r)^2$. Integrate from $r=0$ to $r=r$.
$endgroup$
– DanielV
May 2 '14 at 5:42