F(x,y,z)=xyi+x^2j+z^2k;C is the intersection of the paraboloid z = x^2 + y^2 and the plane z = y with a counter- clockwise orientation looking down the positive z-axis. Use the line integral (parameterizing).
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The intersection is a circle on z = y
z = y = x² + y² ==> x² + y² - y = 0 ==> x² + (y - 1/2)² = 1/4
obtained by completing the square. There are some options as far as parametrization is concerned.
The projection into the plane is r = sinΘ in polar coordinates (solve x² + y² = y ---i.e. r² = r sinΘ). So you can set
x = r cosΘ = sinΘ cosΘ = ½ sin(2Θ),
y = r sinΘ = sin²Θ, and
z = y = sin²Θ.
But be careful! A circle like this is mapped out as 0 ≤ Θ ≤ π. Many common polar curves are mapped out in an interval of length 2π, but this ain't one of 'em. Taking Θ in the natural order maps the circle counterclockwise as required.
This gives
r = x i + y j + z k ==> dr = (cos(2Θ) i + sin(2Θ) j + sin(2Θ) k ) dΘ
I used the ID 2sinΘcosΘ = sin(2Θ). F restricted to the contour is
F = sin^3(Θ)cosΘ i + sin²Θcos²Θ j + sin^4(Θ) k.
So F•dr = (sin^3(Θ)cosΘcos(2Θ) + sin²Θcos²Θsin(2Θ) + sin^4(Θ)sin(2Θ)) dΘ. It's probably best to put that all in terms of Θ instead of 2Θ for the purposes of integrating.
. . . . . . π
∫ F•dr = ∫ (sin^3(Θ)cosΘcos(2Θ) + sin²Θcos²Θsin(2Θ) + sin^4(Θ)sin(2Θ)) dΘ
C . . . .0
= 0.
(This is rather painful compared with using Stokes' Theorem.) You could have opted for an alternative parametrization such as x = cosΘ, y = 1/2 + sinΘ, z = 1/2 + sinΘ with 0 ≤ Θ ≤ 2π. I doubt the integral would be any "nicer".
z = y = x² + y² ==> x² + y² - y = 0 ==> x² + (y - 1/2)² = 1/4
obtained by completing the square. There are some options as far as parametrization is concerned.
The projection into the plane is r = sinΘ in polar coordinates (solve x² + y² = y ---i.e. r² = r sinΘ). So you can set
x = r cosΘ = sinΘ cosΘ = ½ sin(2Θ),
y = r sinΘ = sin²Θ, and
z = y = sin²Θ.
But be careful! A circle like this is mapped out as 0 ≤ Θ ≤ π. Many common polar curves are mapped out in an interval of length 2π, but this ain't one of 'em. Taking Θ in the natural order maps the circle counterclockwise as required.
This gives
r = x i + y j + z k ==> dr = (cos(2Θ) i + sin(2Θ) j + sin(2Θ) k ) dΘ
I used the ID 2sinΘcosΘ = sin(2Θ). F restricted to the contour is
F = sin^3(Θ)cosΘ i + sin²Θcos²Θ j + sin^4(Θ) k.
So F•dr = (sin^3(Θ)cosΘcos(2Θ) + sin²Θcos²Θsin(2Θ) + sin^4(Θ)sin(2Θ)) dΘ. It's probably best to put that all in terms of Θ instead of 2Θ for the purposes of integrating.
. . . . . . π
∫ F•dr = ∫ (sin^3(Θ)cosΘcos(2Θ) + sin²Θcos²Θsin(2Θ) + sin^4(Θ)sin(2Θ)) dΘ
C . . . .0
= 0.
(This is rather painful compared with using Stokes' Theorem.) You could have opted for an alternative parametrization such as x = cosΘ, y = 1/2 + sinΘ, z = 1/2 + sinΘ with 0 ≤ Θ ≤ 2π. I doubt the integral would be any "nicer".