Integrating a solid using cartesian, cylindrical and spherical coordinates
$begingroup$
The region $W$ is the cone shown below (see image).
The angle at the vertex is $π/3$, and the top is flat and at a height of $7sqrt{3}$.
Write the limits of integration for $int_W dV$ in the following coordinates (do not reduce the domain of integration by taking advantage of symmetry):
Progress
As seen in the image, the ones highlighted in red are those I cannot figure out. I'm not sure if I have some fundamental misconceptions that allows me to derive some of the boundaries but not others, but if someone can kindly explain how to find those values in red, I would be grateful. Thanks in advance.
integration multivariable-calculus volume
asked Nov 28 '14 at 23:28
eloggingelogging
3829
$endgroup$
add a comment |
$begingroup$
The region $W$ is the cone shown below (see image).
The angle at the vertex is $π/3$, and the top is flat and at a height of $7sqrt{3}$.
Write the limits of integration for $int_W dV$ in the following coordinates (do not reduce the domain of integration by taking advantage of symmetry):
Progress
As seen in the image, the ones highlighted in red are those I cannot figure out. I'm not sure if I have some fundamental misconceptions that allows me to derive some of the boundaries but not others, but if someone can kindly explain how to find those values in red, I would be grateful. Thanks in advance.
integration multivariable-calculus volume
asked Nov 28 '14 at 23:28
eloggingelogging
3829
$endgroup$
1
$begingroup$
Is this your homework or something? There is ""(1 pt)"" attached to the question.
$endgroup$
– Aaron Maroja
Nov 29 '14 at 0:12
add a comment |
$begingroup$
The region $W$ is the cone shown below (see image).
The angle at the vertex is $π/3$, and the top is flat and at a height of $7sqrt{3}$.
Write the limits of integration for $int_W dV$ in the following coordinates (do not reduce the domain of integration by taking advantage of symmetry):
Progress
As seen in the image, the ones highlighted in red are those I cannot figure out. I'm not sure if I have some fundamental misconceptions that allows me to derive some of the boundaries but not others, but if someone can kindly explain how to find those values in red, I would be grateful. Thanks in advance.
integration multivariable-calculus volume
asked Nov 28 '14 at 23:28
eloggingelogging
3829
$endgroup$
The region $W$ is the cone shown below (see image).
The angle at the vertex is $π/3$, and the top is flat and at a height of $7sqrt{3}$.
Write the limits of integration for $int_W dV$ in the following coordinates (do not reduce the domain of integration by taking advantage of symmetry):
Progress
As seen in the image, the ones highlighted in red are those I cannot figure out. I'm not sure if I have some fundamental misconceptions that allows me to derive some of the boundaries but not others, but if someone can kindly explain how to find those values in red, I would be grateful. Thanks in advance.
integration multivariable-calculus volume
integration multivariable-calculus volume
asked Nov 28 '14 at 23:28
eloggingelogging
3829
asked Nov 28 '14 at 23:28
eloggingelogging
3829
edited Dec 12 '14 at 19:32
user147263
asked Nov 28 '14 at 23:28
eloggingelogging
3829
asked Nov 28 '14 at 23:28
eloggingelogging
3829
asked Nov 28 '14 at 23:28
eloggingelogging
3829
3829
1
$begingroup$
Is this your homework or something? There is ""(1 pt)"" attached to the question.
$endgroup$
– Aaron Maroja
Nov 29 '14 at 0:12
add a comment |
1
$begingroup$
Is this your homework or something? There is ""(1 pt)"" attached to the question.
$endgroup$
– Aaron Maroja
Nov 29 '14 at 0:12
1
1
$begingroup$
Is this your homework or something? There is ""(1 pt)"" attached to the question.
$endgroup$
– Aaron Maroja
Nov 29 '14 at 0:12
$begingroup$
Is this your homework or something? There is ""(1 pt)"" attached to the question.
$endgroup$
– Aaron Maroja
Nov 29 '14 at 0:12
add a comment |
1 Answer
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The cross-sectional radius of the cone increases linearly with $z$, so the lower limit of $z$ also increases linearly. So $1/7$th of the way up, the lower $z$ limit needs to be $sqrt{3}$. From your bounds on $x$ and $y$, $sqrt{x^2 + y^2} = sqrt{2}$ at $z=1$, so the limits of $z$ are $left(sqrt{3/2}sqrt{x^2+y^2}right)$ to $7sqrt{3}.$
For the cylindrical coordinates, it's much the same. Replace $sqrt{x^2+y^2}$ with $r$. (You should have $r$ as the integrand instead of $rho$.)
For the spherical coordinates, the $rho$ dependence is with $phi$ only. At $phi=0$, $rho = 7sqrt{3}$, while at $phi=pi/6$, $rho=7sqrt{3}/cos(pi/6).$ So your upper limit is $7sqrt{3}/cosphi$.
The integrand is determined from the spherical volume element, which is
$$dV = rho^2 sin theta; drho; dtheta; dphi.$$
answered Nov 29 '14 at 0:32
JohnJohn
22.6k32450
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$begingroup$
The cross-sectional radius of the cone increases linearly with $z$, so the lower limit of $z$ also increases linearly. So $1/7$th of the way up, the lower $z$ limit needs to be $sqrt{3}$. From your bounds on $x$ and $y$, $sqrt{x^2 + y^2} = sqrt{2}$ at $z=1$, so the limits of $z$ are $left(sqrt{3/2}sqrt{x^2+y^2}right)$ to $7sqrt{3}.$
For the cylindrical coordinates, it's much the same. Replace $sqrt{x^2+y^2}$ with $r$. (You should have $r$ as the integrand instead of $rho$.)
For the spherical coordinates, the $rho$ dependence is with $phi$ only. At $phi=0$, $rho = 7sqrt{3}$, while at $phi=pi/6$, $rho=7sqrt{3}/cos(pi/6).$ So your upper limit is $7sqrt{3}/cosphi$.
The integrand is determined from the spherical volume element, which is
$$dV = rho^2 sin theta; drho; dtheta; dphi.$$
answered Nov 29 '14 at 0:32
JohnJohn
22.6k32450
$endgroup$
add a comment |
$begingroup$
The cross-sectional radius of the cone increases linearly with $z$, so the lower limit of $z$ also increases linearly. So $1/7$th of the way up, the lower $z$ limit needs to be $sqrt{3}$. From your bounds on $x$ and $y$, $sqrt{x^2 + y^2} = sqrt{2}$ at $z=1$, so the limits of $z$ are $left(sqrt{3/2}sqrt{x^2+y^2}right)$ to $7sqrt{3}.$
For the cylindrical coordinates, it's much the same. Replace $sqrt{x^2+y^2}$ with $r$. (You should have $r$ as the integrand instead of $rho$.)
For the spherical coordinates, the $rho$ dependence is with $phi$ only. At $phi=0$, $rho = 7sqrt{3}$, while at $phi=pi/6$, $rho=7sqrt{3}/cos(pi/6).$ So your upper limit is $7sqrt{3}/cosphi$.
The integrand is determined from the spherical volume element, which is
$$dV = rho^2 sin theta; drho; dtheta; dphi.$$
answered Nov 29 '14 at 0:32
JohnJohn
22.6k32450
$endgroup$
add a comment |
$begingroup$
The cross-sectional radius of the cone increases linearly with $z$, so the lower limit of $z$ also increases linearly. So $1/7$th of the way up, the lower $z$ limit needs to be $sqrt{3}$. From your bounds on $x$ and $y$, $sqrt{x^2 + y^2} = sqrt{2}$ at $z=1$, so the limits of $z$ are $left(sqrt{3/2}sqrt{x^2+y^2}right)$ to $7sqrt{3}.$
For the cylindrical coordinates, it's much the same. Replace $sqrt{x^2+y^2}$ with $r$. (You should have $r$ as the integrand instead of $rho$.)
For the spherical coordinates, the $rho$ dependence is with $phi$ only. At $phi=0$, $rho = 7sqrt{3}$, while at $phi=pi/6$, $rho=7sqrt{3}/cos(pi/6).$ So your upper limit is $7sqrt{3}/cosphi$.
The integrand is determined from the spherical volume element, which is
$$dV = rho^2 sin theta; drho; dtheta; dphi.$$
answered Nov 29 '14 at 0:32
JohnJohn
22.6k32450
$endgroup$
The cross-sectional radius of the cone increases linearly with $z$, so the lower limit of $z$ also increases linearly. So $1/7$th of the way up, the lower $z$ limit needs to be $sqrt{3}$. From your bounds on $x$ and $y$, $sqrt{x^2 + y^2} = sqrt{2}$ at $z=1$, so the limits of $z$ are $left(sqrt{3/2}sqrt{x^2+y^2}right)$ to $7sqrt{3}.$
For the cylindrical coordinates, it's much the same. Replace $sqrt{x^2+y^2}$ with $r$. (You should have $r$ as the integrand instead of $rho$.)
For the spherical coordinates, the $rho$ dependence is with $phi$ only. At $phi=0$, $rho = 7sqrt{3}$, while at $phi=pi/6$, $rho=7sqrt{3}/cos(pi/6).$ So your upper limit is $7sqrt{3}/cosphi$.
The integrand is determined from the spherical volume element, which is
$$dV = rho^2 sin theta; drho; dtheta; dphi.$$
answered Nov 29 '14 at 0:32
JohnJohn
22.6k32450
edited Nov 29 '14 at 0:38
answered Nov 29 '14 at 0:32
JohnJohn
22.6k32450
answered Nov 29 '14 at 0:32
JohnJohn
22.6k32450
answered Nov 29 '14 at 0:32
JohnJohn
22.6k32450
22.6k32450
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$begingroup$
Is this your homework or something? There is ""(1 pt)"" attached to the question.
$endgroup$
– Aaron Maroja
Nov 29 '14 at 0:12