Calculate the area of the helicoid defined by the image of $phi:Dsubset mathbb{R}to mathbb{R}^3$; $phi (u,...
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Calculate the area of the helicoid defined by the image of $phi:Dsubset mathbb{R}to mathbb{R}^3$; $phi (u, v) = (u (cos v), u (sin v), v)$, with $D={(u, v) ∈ mathbb{R}^2 : 0 ≤ u ≤ 1; 0 ≤ v ≤ 2π}$.
I have thought about doing the following but I am not sure:
I have to calculate P where $intint_{phi(D)}dA=intint_D|det(D_{phi})|dudv$, but $D_{phi}$ is not square, am I making an error? How can I do this? Thank you.
calculus integration multivariable-calculus change-of-variable
asked Nov 17 at 20:26
user424241
20819
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Calculate the area of the helicoid defined by the image of $phi:Dsubset mathbb{R}to mathbb{R}^3$; $phi (u, v) = (u (cos v), u (sin v), v)$, with $D={(u, v) ∈ mathbb{R}^2 : 0 ≤ u ≤ 1; 0 ≤ v ≤ 2π}$.
I have thought about doing the following but I am not sure:
I have to calculate P where $intint_{phi(D)}dA=intint_D|det(D_{phi})|dudv$, but $D_{phi}$ is not square, am I making an error? How can I do this? Thank you.
calculus integration multivariable-calculus change-of-variable
asked Nov 17 at 20:26
user424241
20819
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Calculate the area of the helicoid defined by the image of $phi:Dsubset mathbb{R}to mathbb{R}^3$; $phi (u, v) = (u (cos v), u (sin v), v)$, with $D={(u, v) ∈ mathbb{R}^2 : 0 ≤ u ≤ 1; 0 ≤ v ≤ 2π}$.
I have thought about doing the following but I am not sure:
I have to calculate P where $intint_{phi(D)}dA=intint_D|det(D_{phi})|dudv$, but $D_{phi}$ is not square, am I making an error? How can I do this? Thank you.
calculus integration multivariable-calculus change-of-variable
asked Nov 17 at 20:26
user424241
20819
Calculate the area of the helicoid defined by the image of $phi:Dsubset mathbb{R}to mathbb{R}^3$; $phi (u, v) = (u (cos v), u (sin v), v)$, with $D={(u, v) ∈ mathbb{R}^2 : 0 ≤ u ≤ 1; 0 ≤ v ≤ 2π}$.
I have thought about doing the following but I am not sure:
I have to calculate P where $intint_{phi(D)}dA=intint_D|det(D_{phi})|dudv$, but $D_{phi}$ is not square, am I making an error? How can I do this? Thank you.
calculus integration multivariable-calculus change-of-variable
calculus integration multivariable-calculus change-of-variable
asked Nov 17 at 20:26
user424241
20819
asked Nov 17 at 20:26
user424241
20819
asked Nov 17 at 20:26
user424241
20819
asked Nov 17 at 20:26
user424241
20819
asked Nov 17 at 20:26
user424241
20819
20819
add a comment |
add a comment |
1 Answer
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Your surface area element $mathrm d A$ is wrong.
The area of a paramerized surface surface $phi:Dto mathbb{R}^3$ is defined as
$$ A(phi) = int_{phi(D)} , mathrm dA = iint_D |phi_u times phi_v| , mathrm du, mathrm dv. $$
We have
$$ mathrm dphi = begin{pmatrix}phi_u & phi_vend{pmatrix} = begin{pmatrix}
cos v & -usin v\
sin v & ucos v\
0 & 1
end{pmatrix},$$
so that
begin{align}
|phi_utimes phi_v| = |(sin v, -cos v, u)| = sqrt{1+u^2}.
end{align}
Thus
$$ A(phi) = iint_D sqrt{1+u^2}, mathrm du,mathrm dv = 2pi int_0^1 sqrt{1+u^2}, mathrm du. $$
Can you compute this integral on your own?
answered Nov 17 at 20:56
MisterRiemann
5,1981623
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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 area element $mathrm d A$ is wrong.
The area of a paramerized surface surface $phi:Dto mathbb{R}^3$ is defined as
$$ A(phi) = int_{phi(D)} , mathrm dA = iint_D |phi_u times phi_v| , mathrm du, mathrm dv. $$
We have
$$ mathrm dphi = begin{pmatrix}phi_u & phi_vend{pmatrix} = begin{pmatrix}
cos v & -usin v\
sin v & ucos v\
0 & 1
end{pmatrix},$$
so that
begin{align}
|phi_utimes phi_v| = |(sin v, -cos v, u)| = sqrt{1+u^2}.
end{align}
Thus
$$ A(phi) = iint_D sqrt{1+u^2}, mathrm du,mathrm dv = 2pi int_0^1 sqrt{1+u^2}, mathrm du. $$
Can you compute this integral on your own?
answered Nov 17 at 20:56
MisterRiemann
5,1981623
add a comment |
up vote
1
down vote
accepted
Your surface area element $mathrm d A$ is wrong.
The area of a paramerized surface surface $phi:Dto mathbb{R}^3$ is defined as
$$ A(phi) = int_{phi(D)} , mathrm dA = iint_D |phi_u times phi_v| , mathrm du, mathrm dv. $$
We have
$$ mathrm dphi = begin{pmatrix}phi_u & phi_vend{pmatrix} = begin{pmatrix}
cos v & -usin v\
sin v & ucos v\
0 & 1
end{pmatrix},$$
so that
begin{align}
|phi_utimes phi_v| = |(sin v, -cos v, u)| = sqrt{1+u^2}.
end{align}
Thus
$$ A(phi) = iint_D sqrt{1+u^2}, mathrm du,mathrm dv = 2pi int_0^1 sqrt{1+u^2}, mathrm du. $$
Can you compute this integral on your own?
answered Nov 17 at 20:56
MisterRiemann
5,1981623
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your surface area element $mathrm d A$ is wrong.
The area of a paramerized surface surface $phi:Dto mathbb{R}^3$ is defined as
$$ A(phi) = int_{phi(D)} , mathrm dA = iint_D |phi_u times phi_v| , mathrm du, mathrm dv. $$
We have
$$ mathrm dphi = begin{pmatrix}phi_u & phi_vend{pmatrix} = begin{pmatrix}
cos v & -usin v\
sin v & ucos v\
0 & 1
end{pmatrix},$$
so that
begin{align}
|phi_utimes phi_v| = |(sin v, -cos v, u)| = sqrt{1+u^2}.
end{align}
Thus
$$ A(phi) = iint_D sqrt{1+u^2}, mathrm du,mathrm dv = 2pi int_0^1 sqrt{1+u^2}, mathrm du. $$
Can you compute this integral on your own?
answered Nov 17 at 20:56
MisterRiemann
5,1981623
Your surface area element $mathrm d A$ is wrong.
The area of a paramerized surface surface $phi:Dto mathbb{R}^3$ is defined as
$$ A(phi) = int_{phi(D)} , mathrm dA = iint_D |phi_u times phi_v| , mathrm du, mathrm dv. $$
We have
$$ mathrm dphi = begin{pmatrix}phi_u & phi_vend{pmatrix} = begin{pmatrix}
cos v & -usin v\
sin v & ucos v\
0 & 1
end{pmatrix},$$
so that
begin{align}
|phi_utimes phi_v| = |(sin v, -cos v, u)| = sqrt{1+u^2}.
end{align}
Thus
$$ A(phi) = iint_D sqrt{1+u^2}, mathrm du,mathrm dv = 2pi int_0^1 sqrt{1+u^2}, mathrm du. $$
Can you compute this integral on your own?
answered Nov 17 at 20:56
MisterRiemann
5,1981623
edited Nov 17 at 21:05
answered Nov 17 at 20:56
MisterRiemann
5,1981623
answered Nov 17 at 20:56
MisterRiemann
5,1981623
answered Nov 17 at 20:56
MisterRiemann
5,1981623
5,1981623
add a comment |
add a comment |
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