Surface integral and parametrization
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I'm struggling with surface integrals, and I still do not have much confidence with the parameterization of functions. This is the exercise I would like to solve:
Calculate the surface integral of the function
$$f = (x-1)^2 + (y-2)^2$$
extended to the surface $P equiv (1+rho cos vartheta, 2 + rho sin vartheta, 4 - rho^2)$ with $vartheta in [0, 2pi], rho in [sqrt{2}, sqrt{3}]$
without going into detail in the steps and in the final result, how should I proceed? The first thing I should do is find the parameterization of the surface, right? And can I retrieve it?
Thanks in advances
surface-integrals
add a comment |
up vote
0
down vote
favorite
I'm struggling with surface integrals, and I still do not have much confidence with the parameterization of functions. This is the exercise I would like to solve:
Calculate the surface integral of the function
$$f = (x-1)^2 + (y-2)^2$$
extended to the surface $P equiv (1+rho cos vartheta, 2 + rho sin vartheta, 4 - rho^2)$ with $vartheta in [0, 2pi], rho in [sqrt{2}, sqrt{3}]$
without going into detail in the steps and in the final result, how should I proceed? The first thing I should do is find the parameterization of the surface, right? And can I retrieve it?
Thanks in advances
surface-integrals
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm struggling with surface integrals, and I still do not have much confidence with the parameterization of functions. This is the exercise I would like to solve:
Calculate the surface integral of the function
$$f = (x-1)^2 + (y-2)^2$$
extended to the surface $P equiv (1+rho cos vartheta, 2 + rho sin vartheta, 4 - rho^2)$ with $vartheta in [0, 2pi], rho in [sqrt{2}, sqrt{3}]$
without going into detail in the steps and in the final result, how should I proceed? The first thing I should do is find the parameterization of the surface, right? And can I retrieve it?
Thanks in advances
surface-integrals
I'm struggling with surface integrals, and I still do not have much confidence with the parameterization of functions. This is the exercise I would like to solve:
Calculate the surface integral of the function
$$f = (x-1)^2 + (y-2)^2$$
extended to the surface $P equiv (1+rho cos vartheta, 2 + rho sin vartheta, 4 - rho^2)$ with $vartheta in [0, 2pi], rho in [sqrt{2}, sqrt{3}]$
without going into detail in the steps and in the final result, how should I proceed? The first thing I should do is find the parameterization of the surface, right? And can I retrieve it?
Thanks in advances
surface-integrals
surface-integrals
asked 2 days ago
user3204810
1806
1806
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1 Answer
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1
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Your surface $S$ is already given in a parametrized form, i.e. your parametrization is
$$ P(u,v) = (1+ucos v, 2+usin v, 4-v^2), quad uin [sqrt 2, sqrt 3], quad v in [0,2pi]. $$
Now you should compute your surface area element $mathrm dA$, which is defined as
$$ mathrm dA = |partial_u P times partial_v P| , mathrm du , mathrm dv, $$
and the sought integral then becomes
$$ int_S f , mathrm dA = int_{0}^{2pi} int_{sqrt 2}^{sqrt 3} f(P(u,v)) |partial_u P times partial_v P| , mathrm du ,mathrm dv. $$
Can you proceed?
1
Tomorrow I try calmly. In any case, thanks for the tip!
– user3204810
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your surface $S$ is already given in a parametrized form, i.e. your parametrization is
$$ P(u,v) = (1+ucos v, 2+usin v, 4-v^2), quad uin [sqrt 2, sqrt 3], quad v in [0,2pi]. $$
Now you should compute your surface area element $mathrm dA$, which is defined as
$$ mathrm dA = |partial_u P times partial_v P| , mathrm du , mathrm dv, $$
and the sought integral then becomes
$$ int_S f , mathrm dA = int_{0}^{2pi} int_{sqrt 2}^{sqrt 3} f(P(u,v)) |partial_u P times partial_v P| , mathrm du ,mathrm dv. $$
Can you proceed?
1
Tomorrow I try calmly. In any case, thanks for the tip!
– user3204810
2 days ago
add a comment |
up vote
1
down vote
accepted
Your surface $S$ is already given in a parametrized form, i.e. your parametrization is
$$ P(u,v) = (1+ucos v, 2+usin v, 4-v^2), quad uin [sqrt 2, sqrt 3], quad v in [0,2pi]. $$
Now you should compute your surface area element $mathrm dA$, which is defined as
$$ mathrm dA = |partial_u P times partial_v P| , mathrm du , mathrm dv, $$
and the sought integral then becomes
$$ int_S f , mathrm dA = int_{0}^{2pi} int_{sqrt 2}^{sqrt 3} f(P(u,v)) |partial_u P times partial_v P| , mathrm du ,mathrm dv. $$
Can you proceed?
1
Tomorrow I try calmly. In any case, thanks for the tip!
– user3204810
2 days ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your surface $S$ is already given in a parametrized form, i.e. your parametrization is
$$ P(u,v) = (1+ucos v, 2+usin v, 4-v^2), quad uin [sqrt 2, sqrt 3], quad v in [0,2pi]. $$
Now you should compute your surface area element $mathrm dA$, which is defined as
$$ mathrm dA = |partial_u P times partial_v P| , mathrm du , mathrm dv, $$
and the sought integral then becomes
$$ int_S f , mathrm dA = int_{0}^{2pi} int_{sqrt 2}^{sqrt 3} f(P(u,v)) |partial_u P times partial_v P| , mathrm du ,mathrm dv. $$
Can you proceed?
Your surface $S$ is already given in a parametrized form, i.e. your parametrization is
$$ P(u,v) = (1+ucos v, 2+usin v, 4-v^2), quad uin [sqrt 2, sqrt 3], quad v in [0,2pi]. $$
Now you should compute your surface area element $mathrm dA$, which is defined as
$$ mathrm dA = |partial_u P times partial_v P| , mathrm du , mathrm dv, $$
and the sought integral then becomes
$$ int_S f , mathrm dA = int_{0}^{2pi} int_{sqrt 2}^{sqrt 3} f(P(u,v)) |partial_u P times partial_v P| , mathrm du ,mathrm dv. $$
Can you proceed?
answered 2 days ago
MisterRiemann
4,4751622
4,4751622
1
Tomorrow I try calmly. In any case, thanks for the tip!
– user3204810
2 days ago
add a comment |
1
Tomorrow I try calmly. In any case, thanks for the tip!
– user3204810
2 days ago
1
1
Tomorrow I try calmly. In any case, thanks for the tip!
– user3204810
2 days ago
Tomorrow I try calmly. In any case, thanks for the tip!
– user3204810
2 days ago
add a comment |
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