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Solve the line integral without Green's formula
I'm starting on double and line integrals, and I'm stuck at this question.
It asks of me to calculate the following integral without using Green's theorem. Usually with Green's theorem I use polar coordinates easily, but I can't seem to make the same substitution here.
$int_{L}{(2x-y)dx+(x-y)dy}$
$L = {(x,y): x^{2}+y^{2}=2y, xgeq0}cup{(x,y):x^{2}+y^{2}=4, xleq0,ygeq0} $
The curve $L$ is oriented from the point $(0,0)$, that is counter-clockwise.
I am not sure which subsitution to use. I always used Greens theorem so far. How would I go about solving this without it? I thought of using some parameter $t$ but couldn't quite specify its range. Thank you.
integration multivariable-calculus line-integrals
asked Nov 20 '18 at 17:32
math101
597
add a comment |
I'm starting on double and line integrals, and I'm stuck at this question.
It asks of me to calculate the following integral without using Green's theorem. Usually with Green's theorem I use polar coordinates easily, but I can't seem to make the same substitution here.
$int_{L}{(2x-y)dx+(x-y)dy}$
$L = {(x,y): x^{2}+y^{2}=2y, xgeq0}cup{(x,y):x^{2}+y^{2}=4, xleq0,ygeq0} $
The curve $L$ is oriented from the point $(0,0)$, that is counter-clockwise.
I am not sure which subsitution to use. I always used Greens theorem so far. How would I go about solving this without it? I thought of using some parameter $t$ but couldn't quite specify its range. Thank you.
integration multivariable-calculus line-integrals
asked Nov 20 '18 at 17:32
math101
597
Tell your teacher that this is a sick way to define a $1$-chain.
– Christian Blatter
Nov 20 '18 at 19:08
add a comment |
I'm starting on double and line integrals, and I'm stuck at this question.
It asks of me to calculate the following integral without using Green's theorem. Usually with Green's theorem I use polar coordinates easily, but I can't seem to make the same substitution here.
$int_{L}{(2x-y)dx+(x-y)dy}$
$L = {(x,y): x^{2}+y^{2}=2y, xgeq0}cup{(x,y):x^{2}+y^{2}=4, xleq0,ygeq0} $
The curve $L$ is oriented from the point $(0,0)$, that is counter-clockwise.
I am not sure which subsitution to use. I always used Greens theorem so far. How would I go about solving this without it? I thought of using some parameter $t$ but couldn't quite specify its range. Thank you.
integration multivariable-calculus line-integrals
asked Nov 20 '18 at 17:32
math101
597
I'm starting on double and line integrals, and I'm stuck at this question.
It asks of me to calculate the following integral without using Green's theorem. Usually with Green's theorem I use polar coordinates easily, but I can't seem to make the same substitution here.
$int_{L}{(2x-y)dx+(x-y)dy}$
$L = {(x,y): x^{2}+y^{2}=2y, xgeq0}cup{(x,y):x^{2}+y^{2}=4, xleq0,ygeq0} $
The curve $L$ is oriented from the point $(0,0)$, that is counter-clockwise.
I am not sure which subsitution to use. I always used Greens theorem so far. How would I go about solving this without it? I thought of using some parameter $t$ but couldn't quite specify its range. Thank you.
integration multivariable-calculus line-integrals
integration multivariable-calculus line-integrals
asked Nov 20 '18 at 17:32
math101
597
asked Nov 20 '18 at 17:32
math101
597
asked Nov 20 '18 at 17:32
math101
597
asked Nov 20 '18 at 17:32
math101
597
asked Nov 20 '18 at 17:32
math101
597
597
Tell your teacher that this is a sick way to define a $1$-chain.
– Christian Blatter
Nov 20 '18 at 19:08
add a comment |
Tell your teacher that this is a sick way to define a $1$-chain.
– Christian Blatter
Nov 20 '18 at 19:08
Tell your teacher that this is a sick way to define a $1$-chain.
– Christian Blatter
Nov 20 '18 at 19:08
Tell your teacher that this is a sick way to define a $1$-chain.
– Christian Blatter
Nov 20 '18 at 19:08
add a comment |
1 Answer
1
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You have two curves.
first part.
$x = 2sin tcos t = sin 2t\
y = 2sin^2 t = 1-cos 2t$
$int_limits{0}^{frac {pi}{2}} (2sin 2t + 1 - cos 2t )(2cos 2t)+(sin 2t +1-cos 2t)(2sin 2t) dt$
and the second part
$x = 2cos t\
y = 2sin t$
$int_limits{frac {pi}{2}}^{pi} (4cos t - 2sin t)(-2sin t)+(2cos t - sin t)(2cos t) dt$
If you wanted to use greens.
Then the you would add the straight line from $(-2,0)$ to $(0,0)$ to close the contour.
Then
$int_{C_1} P(x,y) dx + Q(x,y) dy + int_{C_2} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA\
int_{C_1} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA- int_{C_2} P(x,y) dx + Q(x,y) dy$
answered Nov 20 '18 at 17:51
Doug M
44k31854
Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$.
– math101
Nov 20 '18 at 22:32
You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = rcos t, y = rsin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2sin t$ And put that back in for $r$ in the equations of $x,y$
– Doug M
Nov 20 '18 at 22:42
So that's it! I see now , thanks!
– math101
Nov 20 '18 at 23:13
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
You have two curves.
first part.
$x = 2sin tcos t = sin 2t\
y = 2sin^2 t = 1-cos 2t$
$int_limits{0}^{frac {pi}{2}} (2sin 2t + 1 - cos 2t )(2cos 2t)+(sin 2t +1-cos 2t)(2sin 2t) dt$
and the second part
$x = 2cos t\
y = 2sin t$
$int_limits{frac {pi}{2}}^{pi} (4cos t - 2sin t)(-2sin t)+(2cos t - sin t)(2cos t) dt$
If you wanted to use greens.
Then the you would add the straight line from $(-2,0)$ to $(0,0)$ to close the contour.
Then
$int_{C_1} P(x,y) dx + Q(x,y) dy + int_{C_2} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA\
int_{C_1} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA- int_{C_2} P(x,y) dx + Q(x,y) dy$
answered Nov 20 '18 at 17:51
Doug M
44k31854
Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$.
– math101
Nov 20 '18 at 22:32
You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = rcos t, y = rsin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2sin t$ And put that back in for $r$ in the equations of $x,y$
– Doug M
Nov 20 '18 at 22:42
So that's it! I see now , thanks!
– math101
Nov 20 '18 at 23:13
add a comment |
You have two curves.
first part.
$x = 2sin tcos t = sin 2t\
y = 2sin^2 t = 1-cos 2t$
$int_limits{0}^{frac {pi}{2}} (2sin 2t + 1 - cos 2t )(2cos 2t)+(sin 2t +1-cos 2t)(2sin 2t) dt$
and the second part
$x = 2cos t\
y = 2sin t$
$int_limits{frac {pi}{2}}^{pi} (4cos t - 2sin t)(-2sin t)+(2cos t - sin t)(2cos t) dt$
If you wanted to use greens.
Then the you would add the straight line from $(-2,0)$ to $(0,0)$ to close the contour.
Then
$int_{C_1} P(x,y) dx + Q(x,y) dy + int_{C_2} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA\
int_{C_1} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA- int_{C_2} P(x,y) dx + Q(x,y) dy$
answered Nov 20 '18 at 17:51
Doug M
44k31854
Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$.
– math101
Nov 20 '18 at 22:32
You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = rcos t, y = rsin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2sin t$ And put that back in for $r$ in the equations of $x,y$
– Doug M
Nov 20 '18 at 22:42
So that's it! I see now , thanks!
– math101
Nov 20 '18 at 23:13
add a comment |
You have two curves.
first part.
$x = 2sin tcos t = sin 2t\
y = 2sin^2 t = 1-cos 2t$
$int_limits{0}^{frac {pi}{2}} (2sin 2t + 1 - cos 2t )(2cos 2t)+(sin 2t +1-cos 2t)(2sin 2t) dt$
and the second part
$x = 2cos t\
y = 2sin t$
$int_limits{frac {pi}{2}}^{pi} (4cos t - 2sin t)(-2sin t)+(2cos t - sin t)(2cos t) dt$
If you wanted to use greens.
Then the you would add the straight line from $(-2,0)$ to $(0,0)$ to close the contour.
Then
$int_{C_1} P(x,y) dx + Q(x,y) dy + int_{C_2} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA\
int_{C_1} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA- int_{C_2} P(x,y) dx + Q(x,y) dy$
answered Nov 20 '18 at 17:51
Doug M
44k31854
You have two curves.
first part.
$x = 2sin tcos t = sin 2t\
y = 2sin^2 t = 1-cos 2t$
$int_limits{0}^{frac {pi}{2}} (2sin 2t + 1 - cos 2t )(2cos 2t)+(sin 2t +1-cos 2t)(2sin 2t) dt$
and the second part
$x = 2cos t\
y = 2sin t$
$int_limits{frac {pi}{2}}^{pi} (4cos t - 2sin t)(-2sin t)+(2cos t - sin t)(2cos t) dt$
If you wanted to use greens.
Then the you would add the straight line from $(-2,0)$ to $(0,0)$ to close the contour.
Then
$int_{C_1} P(x,y) dx + Q(x,y) dy + int_{C_2} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA\
int_{C_1} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA- int_{C_2} P(x,y) dx + Q(x,y) dy$
answered Nov 20 '18 at 17:51
Doug M
44k31854
edited Nov 20 '18 at 17:57
answered Nov 20 '18 at 17:51
Doug M
44k31854
answered Nov 20 '18 at 17:51
Doug M
44k31854
answered Nov 20 '18 at 17:51
Doug M
44k31854
44k31854
Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$.
– math101
Nov 20 '18 at 22:32
You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = rcos t, y = rsin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2sin t$ And put that back in for $r$ in the equations of $x,y$
– Doug M
Nov 20 '18 at 22:42
So that's it! I see now , thanks!
– math101
Nov 20 '18 at 23:13
add a comment |
Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$.
– math101
Nov 20 '18 at 22:32
You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = rcos t, y = rsin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2sin t$ And put that back in for $r$ in the equations of $x,y$
– Doug M
Nov 20 '18 at 22:42
So that's it! I see now , thanks!
– math101
Nov 20 '18 at 23:13
Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$.
– math101
Nov 20 '18 at 22:32
Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$.
– math101
Nov 20 '18 at 22:32
You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = rcos t, y = rsin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2sin t$ And put that back in for $r$ in the equations of $x,y$
– Doug M
Nov 20 '18 at 22:42
You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = rcos t, y = rsin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2sin t$ And put that back in for $r$ in the equations of $x,y$
– Doug M
Nov 20 '18 at 22:42
So that's it! I see now , thanks!
– math101
Nov 20 '18 at 23:13
So that's it! I see now , thanks!
– math101
Nov 20 '18 at 23:13
add a comment |
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Tell your teacher that this is a sick way to define a $1$-chain.
– Christian Blatter
Nov 20 '18 at 19:08