Calculate the line integral on the negatively oriented unit circumference (like the hands of the clock) under...
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Calculate the line integral on the negatively oriented unit circumference (like the hands of the clock) under the $F(x,y)=(e^x+x^2y,e^y-xy^2)$ field
I have thought to do the following to solve this:
$int_{C'}Fcdot dr=int_{-C}Fcdot dr=-int_CFcdot dr$, where $C$ is a positively oriented curve.
But $int_CFcdot dr=intint_D(frac{partial Q}{partial x}-frac{partial P}{partial y})dA=intint_D(-y^2-x^2)dA=-intint_D(x^2+y^2)dA=-int_{0}^{2pi}int_{0}^{1}r^3drdtheta=-int_{0}^{2pi}1/4dtheta=-2pi/4=-pi/2$.
Then $int_{C'}Fcdot dr=pi/2$
Is this fine? Thank you.
calculus integration multivariable-calculus greens-theorem
asked Nov 16 at 19:10
user402543
375212
add a comment |
up vote
0
down vote
favorite
Calculate the line integral on the negatively oriented unit circumference (like the hands of the clock) under the $F(x,y)=(e^x+x^2y,e^y-xy^2)$ field
I have thought to do the following to solve this:
$int_{C'}Fcdot dr=int_{-C}Fcdot dr=-int_CFcdot dr$, where $C$ is a positively oriented curve.
But $int_CFcdot dr=intint_D(frac{partial Q}{partial x}-frac{partial P}{partial y})dA=intint_D(-y^2-x^2)dA=-intint_D(x^2+y^2)dA=-int_{0}^{2pi}int_{0}^{1}r^3drdtheta=-int_{0}^{2pi}1/4dtheta=-2pi/4=-pi/2$.
Then $int_{C'}Fcdot dr=pi/2$
Is this fine? Thank you.
calculus integration multivariable-calculus greens-theorem
asked Nov 16 at 19:10
user402543
375212
How do you include the positively oriented condition?
– Rafa Budría
Nov 17 at 8:46
Pues, cambio la curva por la misma pero orientada positivamente, esto sí se puede hacer así?
– user402543
Nov 17 at 19:56
Ah, yes, I didn't see that. Sorry.
– Rafa Budría
Nov 17 at 20:15
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Calculate the line integral on the negatively oriented unit circumference (like the hands of the clock) under the $F(x,y)=(e^x+x^2y,e^y-xy^2)$ field
I have thought to do the following to solve this:
$int_{C'}Fcdot dr=int_{-C}Fcdot dr=-int_CFcdot dr$, where $C$ is a positively oriented curve.
But $int_CFcdot dr=intint_D(frac{partial Q}{partial x}-frac{partial P}{partial y})dA=intint_D(-y^2-x^2)dA=-intint_D(x^2+y^2)dA=-int_{0}^{2pi}int_{0}^{1}r^3drdtheta=-int_{0}^{2pi}1/4dtheta=-2pi/4=-pi/2$.
Then $int_{C'}Fcdot dr=pi/2$
Is this fine? Thank you.
calculus integration multivariable-calculus greens-theorem
asked Nov 16 at 19:10
user402543
375212
Calculate the line integral on the negatively oriented unit circumference (like the hands of the clock) under the $F(x,y)=(e^x+x^2y,e^y-xy^2)$ field
I have thought to do the following to solve this:
$int_{C'}Fcdot dr=int_{-C}Fcdot dr=-int_CFcdot dr$, where $C$ is a positively oriented curve.
But $int_CFcdot dr=intint_D(frac{partial Q}{partial x}-frac{partial P}{partial y})dA=intint_D(-y^2-x^2)dA=-intint_D(x^2+y^2)dA=-int_{0}^{2pi}int_{0}^{1}r^3drdtheta=-int_{0}^{2pi}1/4dtheta=-2pi/4=-pi/2$.
Then $int_{C'}Fcdot dr=pi/2$
Is this fine? Thank you.
calculus integration multivariable-calculus greens-theorem
calculus integration multivariable-calculus greens-theorem
asked Nov 16 at 19:10
user402543
375212
asked Nov 16 at 19:10
user402543
375212
edited Nov 16 at 22:38
asked Nov 16 at 19:10
user402543
375212
asked Nov 16 at 19:10
user402543
375212
asked Nov 16 at 19:10
user402543
375212
375212
How do you include the positively oriented condition?
– Rafa Budría
Nov 17 at 8:46
Pues, cambio la curva por la misma pero orientada positivamente, esto sí se puede hacer así?
– user402543
Nov 17 at 19:56
Ah, yes, I didn't see that. Sorry.
– Rafa Budría
Nov 17 at 20:15
add a comment |
How do you include the positively oriented condition?
– Rafa Budría
Nov 17 at 8:46
Pues, cambio la curva por la misma pero orientada positivamente, esto sí se puede hacer así?
– user402543
Nov 17 at 19:56
Ah, yes, I didn't see that. Sorry.
– Rafa Budría
Nov 17 at 20:15
How do you include the positively oriented condition?
– Rafa Budría
Nov 17 at 8:46
How do you include the positively oriented condition?
– Rafa Budría
Nov 17 at 8:46
Pues, cambio la curva por la misma pero orientada positivamente, esto sí se puede hacer así?
– user402543
Nov 17 at 19:56
Pues, cambio la curva por la misma pero orientada positivamente, esto sí se puede hacer así?
– user402543
Nov 17 at 19:56
Ah, yes, I didn't see that. Sorry.
– Rafa Budría
Nov 17 at 20:15
Ah, yes, I didn't see that. Sorry.
– Rafa Budría
Nov 17 at 20:15
add a comment |
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How do you include the positively oriented condition?
– Rafa Budría
Nov 17 at 8:46
Pues, cambio la curva por la misma pero orientada positivamente, esto sí se puede hacer así?
– user402543
Nov 17 at 19:56
Ah, yes, I didn't see that. Sorry.
– Rafa Budría
Nov 17 at 20:15