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How to Find the area of the portion of the sphere $ x^2 + y^2 + z^2 = 1$ between the two parallel planes .
$begingroup$
Find the area of the portion of the sphere $ x^2 + y^2 + z^2 = 1$ between the two parallel planes $ z = a$ and $z = b$ where $-1 < a < b < 1$ are parameters.
How to solve this question using surface integral ? i got $0$ when parametrizing the sphere but it can't be right
multivariable-calculus surfaces surface-integrals
asked Jan 8 at 10:16
Mather Mather
3047
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add a comment |
$begingroup$
Find the area of the portion of the sphere $ x^2 + y^2 + z^2 = 1$ between the two parallel planes $ z = a$ and $z = b$ where $-1 < a < b < 1$ are parameters.
How to solve this question using surface integral ? i got $0$ when parametrizing the sphere but it can't be right
multivariable-calculus surfaces surface-integrals
asked Jan 8 at 10:16
Mather Mather
3047
$endgroup$
add a comment |
$begingroup$
Find the area of the portion of the sphere $ x^2 + y^2 + z^2 = 1$ between the two parallel planes $ z = a$ and $z = b$ where $-1 < a < b < 1$ are parameters.
How to solve this question using surface integral ? i got $0$ when parametrizing the sphere but it can't be right
multivariable-calculus surfaces surface-integrals
asked Jan 8 at 10:16
Mather Mather
3047
$endgroup$
Find the area of the portion of the sphere $ x^2 + y^2 + z^2 = 1$ between the two parallel planes $ z = a$ and $z = b$ where $-1 < a < b < 1$ are parameters.
How to solve this question using surface integral ? i got $0$ when parametrizing the sphere but it can't be right
multivariable-calculus surfaces surface-integrals
multivariable-calculus surfaces surface-integrals
asked Jan 8 at 10:16
Mather Mather
3047
asked Jan 8 at 10:16
Mather Mather
3047
asked Jan 8 at 10:16
Mather Mather
3047
asked Jan 8 at 10:16
Mather Mather
3047
asked Jan 8 at 10:16
Mather Mather
3047
3047
add a comment |
add a comment |
1 Answer
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$begingroup$
So between the planes $z=a$ and $z=b$ we have a portion of a sphere. $z=a$ lies below $z=b$ from the fact $-1 < a < b < 1$. Now, we aren't dealing with any sections of the portion of the sphere cut away. Another way of explaining is if you cut through any section of the shape, every cross section would be a circle. So we have $0le varphi le 2pi$
Now, $theta$ measures the angle from the positive $z$-axis down to the surface of the sphere at a particular $z$ value.
$$theta = arccosleft(frac{z}{r}right)$$
For this sphere, $r=1$, so $theta = arccos(z)$
For $z=a$ and $z=b$, we have $theta = arccos(a)$ and $theta = arccos(b)$ respectively
Since the plane $z=b$ lies above the plane $z=a$, the $theta$ angle to the plane $z=b$ will be smaller than that for the plane $z=a$ so we have $$arccos(b)le theta le arccos(a)$$
The Jacobian for spherical coordinates is $r^2 sin(theta) = sin(theta)$ since $r=1$
The surface integral is therefore $$int_{theta = arccos(b)}^{theta = arccos(a)} int_{varphi = 0}^{varphi=2pi} sin(theta) mathrm d varphi mathrm d theta$$
answered Jan 8 at 11:19
Patrick JankowskiPatrick Jankowski
471115
$endgroup$
$begingroup$
thank you i got the same answer but i fliped the names thats why i got an integral of $ 0 $ value
$endgroup$
– Mather
Jan 8 at 12:12
add a comment |
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1 Answer
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1 Answer
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$begingroup$
So between the planes $z=a$ and $z=b$ we have a portion of a sphere. $z=a$ lies below $z=b$ from the fact $-1 < a < b < 1$. Now, we aren't dealing with any sections of the portion of the sphere cut away. Another way of explaining is if you cut through any section of the shape, every cross section would be a circle. So we have $0le varphi le 2pi$
Now, $theta$ measures the angle from the positive $z$-axis down to the surface of the sphere at a particular $z$ value.
$$theta = arccosleft(frac{z}{r}right)$$
For this sphere, $r=1$, so $theta = arccos(z)$
For $z=a$ and $z=b$, we have $theta = arccos(a)$ and $theta = arccos(b)$ respectively
Since the plane $z=b$ lies above the plane $z=a$, the $theta$ angle to the plane $z=b$ will be smaller than that for the plane $z=a$ so we have $$arccos(b)le theta le arccos(a)$$
The Jacobian for spherical coordinates is $r^2 sin(theta) = sin(theta)$ since $r=1$
The surface integral is therefore $$int_{theta = arccos(b)}^{theta = arccos(a)} int_{varphi = 0}^{varphi=2pi} sin(theta) mathrm d varphi mathrm d theta$$
answered Jan 8 at 11:19
Patrick JankowskiPatrick Jankowski
471115
$endgroup$
$begingroup$
thank you i got the same answer but i fliped the names thats why i got an integral of $ 0 $ value
$endgroup$
– Mather
Jan 8 at 12:12
add a comment |
$begingroup$
So between the planes $z=a$ and $z=b$ we have a portion of a sphere. $z=a$ lies below $z=b$ from the fact $-1 < a < b < 1$. Now, we aren't dealing with any sections of the portion of the sphere cut away. Another way of explaining is if you cut through any section of the shape, every cross section would be a circle. So we have $0le varphi le 2pi$
Now, $theta$ measures the angle from the positive $z$-axis down to the surface of the sphere at a particular $z$ value.
$$theta = arccosleft(frac{z}{r}right)$$
For this sphere, $r=1$, so $theta = arccos(z)$
For $z=a$ and $z=b$, we have $theta = arccos(a)$ and $theta = arccos(b)$ respectively
Since the plane $z=b$ lies above the plane $z=a$, the $theta$ angle to the plane $z=b$ will be smaller than that for the plane $z=a$ so we have $$arccos(b)le theta le arccos(a)$$
The Jacobian for spherical coordinates is $r^2 sin(theta) = sin(theta)$ since $r=1$
The surface integral is therefore $$int_{theta = arccos(b)}^{theta = arccos(a)} int_{varphi = 0}^{varphi=2pi} sin(theta) mathrm d varphi mathrm d theta$$
answered Jan 8 at 11:19
Patrick JankowskiPatrick Jankowski
471115
$endgroup$
$begingroup$
thank you i got the same answer but i fliped the names thats why i got an integral of $ 0 $ value
$endgroup$
– Mather
Jan 8 at 12:12
add a comment |
$begingroup$
So between the planes $z=a$ and $z=b$ we have a portion of a sphere. $z=a$ lies below $z=b$ from the fact $-1 < a < b < 1$. Now, we aren't dealing with any sections of the portion of the sphere cut away. Another way of explaining is if you cut through any section of the shape, every cross section would be a circle. So we have $0le varphi le 2pi$
Now, $theta$ measures the angle from the positive $z$-axis down to the surface of the sphere at a particular $z$ value.
$$theta = arccosleft(frac{z}{r}right)$$
For this sphere, $r=1$, so $theta = arccos(z)$
For $z=a$ and $z=b$, we have $theta = arccos(a)$ and $theta = arccos(b)$ respectively
Since the plane $z=b$ lies above the plane $z=a$, the $theta$ angle to the plane $z=b$ will be smaller than that for the plane $z=a$ so we have $$arccos(b)le theta le arccos(a)$$
The Jacobian for spherical coordinates is $r^2 sin(theta) = sin(theta)$ since $r=1$
The surface integral is therefore $$int_{theta = arccos(b)}^{theta = arccos(a)} int_{varphi = 0}^{varphi=2pi} sin(theta) mathrm d varphi mathrm d theta$$
answered Jan 8 at 11:19
Patrick JankowskiPatrick Jankowski
471115
$endgroup$
So between the planes $z=a$ and $z=b$ we have a portion of a sphere. $z=a$ lies below $z=b$ from the fact $-1 < a < b < 1$. Now, we aren't dealing with any sections of the portion of the sphere cut away. Another way of explaining is if you cut through any section of the shape, every cross section would be a circle. So we have $0le varphi le 2pi$
Now, $theta$ measures the angle from the positive $z$-axis down to the surface of the sphere at a particular $z$ value.
$$theta = arccosleft(frac{z}{r}right)$$
For this sphere, $r=1$, so $theta = arccos(z)$
For $z=a$ and $z=b$, we have $theta = arccos(a)$ and $theta = arccos(b)$ respectively
Since the plane $z=b$ lies above the plane $z=a$, the $theta$ angle to the plane $z=b$ will be smaller than that for the plane $z=a$ so we have $$arccos(b)le theta le arccos(a)$$
The Jacobian for spherical coordinates is $r^2 sin(theta) = sin(theta)$ since $r=1$
The surface integral is therefore $$int_{theta = arccos(b)}^{theta = arccos(a)} int_{varphi = 0}^{varphi=2pi} sin(theta) mathrm d varphi mathrm d theta$$
answered Jan 8 at 11:19
Patrick JankowskiPatrick Jankowski
471115
answered Jan 8 at 11:19
Patrick JankowskiPatrick Jankowski
471115
answered Jan 8 at 11:19
Patrick JankowskiPatrick Jankowski
471115
answered Jan 8 at 11:19
Patrick JankowskiPatrick Jankowski
471115
471115
$begingroup$
thank you i got the same answer but i fliped the names thats why i got an integral of $ 0 $ value
$endgroup$
– Mather
Jan 8 at 12:12
add a comment |
$begingroup$
thank you i got the same answer but i fliped the names thats why i got an integral of $ 0 $ value
$endgroup$
– Mather
Jan 8 at 12:12
$begingroup$
thank you i got the same answer but i fliped the names thats why i got an integral of $ 0 $ value
$endgroup$
– Mather
Jan 8 at 12:12
$begingroup$
thank you i got the same answer but i fliped the names thats why i got an integral of $ 0 $ value
$endgroup$
– Mather
Jan 8 at 12:12
add a comment |
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