How to calculate the line integral with respect to the circle in counterclockwise direction
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Consider the vector field $F=<y,-x>$.
Compute the line integral $int_cFcdot dr$ where $C$ is the circle of radius $3$ centered at the origin counterclockwise.
My Try:
The circle is $x^2+y^2=9$
$x=3cos t$
$y=3sin t$ for $0le tle2pi$
Now how do I calculate $int_cFcdot dr$?
Can anyone explain how to solve this.
calculus integration multivariable-calculus line-integrals
asked Nov 26 at 2:02
user982787
897
add a comment |
up vote
1
down vote
favorite
Consider the vector field $F=<y,-x>$.
Compute the line integral $int_cFcdot dr$ where $C$ is the circle of radius $3$ centered at the origin counterclockwise.
My Try:
The circle is $x^2+y^2=9$
$x=3cos t$
$y=3sin t$ for $0le tle2pi$
Now how do I calculate $int_cFcdot dr$?
Can anyone explain how to solve this.
calculus integration multivariable-calculus line-integrals
asked Nov 26 at 2:02
user982787
897
Just follow the definition. Can you look up the definition?
– Zachary Selk
Nov 26 at 2:09
@ZacharySelk Here $F=<y,-x>$. Then what about $r(t)$?
– user982787
Nov 26 at 2:16
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the vector field $F=<y,-x>$.
Compute the line integral $int_cFcdot dr$ where $C$ is the circle of radius $3$ centered at the origin counterclockwise.
My Try:
The circle is $x^2+y^2=9$
$x=3cos t$
$y=3sin t$ for $0le tle2pi$
Now how do I calculate $int_cFcdot dr$?
Can anyone explain how to solve this.
calculus integration multivariable-calculus line-integrals
asked Nov 26 at 2:02
user982787
897
Consider the vector field $F=<y,-x>$.
Compute the line integral $int_cFcdot dr$ where $C$ is the circle of radius $3$ centered at the origin counterclockwise.
My Try:
The circle is $x^2+y^2=9$
$x=3cos t$
$y=3sin t$ for $0le tle2pi$
Now how do I calculate $int_cFcdot dr$?
Can anyone explain how to solve this.
calculus integration multivariable-calculus line-integrals
calculus integration multivariable-calculus line-integrals
asked Nov 26 at 2:02
user982787
897
asked Nov 26 at 2:02
user982787
897
asked Nov 26 at 2:02
user982787
897
asked Nov 26 at 2:02
user982787
897
asked Nov 26 at 2:02
user982787
897
897
Just follow the definition. Can you look up the definition?
– Zachary Selk
Nov 26 at 2:09
@ZacharySelk Here $F=<y,-x>$. Then what about $r(t)$?
– user982787
Nov 26 at 2:16
add a comment |
Just follow the definition. Can you look up the definition?
– Zachary Selk
Nov 26 at 2:09
@ZacharySelk Here $F=<y,-x>$. Then what about $r(t)$?
– user982787
Nov 26 at 2:16
Just follow the definition. Can you look up the definition?
– Zachary Selk
Nov 26 at 2:09
Just follow the definition. Can you look up the definition?
– Zachary Selk
Nov 26 at 2:09
@ZacharySelk Here $F=<y,-x>$. Then what about $r(t)$?
– user982787
Nov 26 at 2:16
@ZacharySelk Here $F=<y,-x>$. Then what about $r(t)$?
– user982787
Nov 26 at 2:16
add a comment |
1 Answer
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1
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The radial vector is
$vec r = begin{pmatrix} x \ y end{pmatrix}; tag 1$
also,
$F = begin{pmatrix} y \ - x end{pmatrix}; tag 2$
with
$x = 3 cos t, tag 3$
$y = 3 sin t, tag 4$
we have
$vec r = begin{pmatrix} 3 cos t \ 3sin t end{pmatrix}, tag 5$
$d vec r = dfrac{d vec r}{dt} dt = begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt, tag 5$
and
$F = begin{pmatrix} 3sin t \ -3cos t end{pmatrix}; tag 6$
then
$F cdot d vec r = begin{pmatrix} 3sin t \ -3cos t end{pmatrix} cdot begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt$
$= (-9 sin^2 t - 9 cos^2 t)dt = -9(sin^2 t + cos^2 t)dt = -9(1)dt = -9dt; tag 7$
finally,
$displaystyle int_C F cdot d vec r = int_C -9 dt = -9 int_C dt = -9(2pi) = -18pi. tag 8$
answered Nov 26 at 2:26
Robert Lewis
42.3k22761
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
The radial vector is
$vec r = begin{pmatrix} x \ y end{pmatrix}; tag 1$
also,
$F = begin{pmatrix} y \ - x end{pmatrix}; tag 2$
with
$x = 3 cos t, tag 3$
$y = 3 sin t, tag 4$
we have
$vec r = begin{pmatrix} 3 cos t \ 3sin t end{pmatrix}, tag 5$
$d vec r = dfrac{d vec r}{dt} dt = begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt, tag 5$
and
$F = begin{pmatrix} 3sin t \ -3cos t end{pmatrix}; tag 6$
then
$F cdot d vec r = begin{pmatrix} 3sin t \ -3cos t end{pmatrix} cdot begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt$
$= (-9 sin^2 t - 9 cos^2 t)dt = -9(sin^2 t + cos^2 t)dt = -9(1)dt = -9dt; tag 7$
finally,
$displaystyle int_C F cdot d vec r = int_C -9 dt = -9 int_C dt = -9(2pi) = -18pi. tag 8$
answered Nov 26 at 2:26
Robert Lewis
42.3k22761
add a comment |
up vote
1
down vote
accepted
The radial vector is
$vec r = begin{pmatrix} x \ y end{pmatrix}; tag 1$
also,
$F = begin{pmatrix} y \ - x end{pmatrix}; tag 2$
with
$x = 3 cos t, tag 3$
$y = 3 sin t, tag 4$
we have
$vec r = begin{pmatrix} 3 cos t \ 3sin t end{pmatrix}, tag 5$
$d vec r = dfrac{d vec r}{dt} dt = begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt, tag 5$
and
$F = begin{pmatrix} 3sin t \ -3cos t end{pmatrix}; tag 6$
then
$F cdot d vec r = begin{pmatrix} 3sin t \ -3cos t end{pmatrix} cdot begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt$
$= (-9 sin^2 t - 9 cos^2 t)dt = -9(sin^2 t + cos^2 t)dt = -9(1)dt = -9dt; tag 7$
finally,
$displaystyle int_C F cdot d vec r = int_C -9 dt = -9 int_C dt = -9(2pi) = -18pi. tag 8$
answered Nov 26 at 2:26
Robert Lewis
42.3k22761
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The radial vector is
$vec r = begin{pmatrix} x \ y end{pmatrix}; tag 1$
also,
$F = begin{pmatrix} y \ - x end{pmatrix}; tag 2$
with
$x = 3 cos t, tag 3$
$y = 3 sin t, tag 4$
we have
$vec r = begin{pmatrix} 3 cos t \ 3sin t end{pmatrix}, tag 5$
$d vec r = dfrac{d vec r}{dt} dt = begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt, tag 5$
and
$F = begin{pmatrix} 3sin t \ -3cos t end{pmatrix}; tag 6$
then
$F cdot d vec r = begin{pmatrix} 3sin t \ -3cos t end{pmatrix} cdot begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt$
$= (-9 sin^2 t - 9 cos^2 t)dt = -9(sin^2 t + cos^2 t)dt = -9(1)dt = -9dt; tag 7$
finally,
$displaystyle int_C F cdot d vec r = int_C -9 dt = -9 int_C dt = -9(2pi) = -18pi. tag 8$
answered Nov 26 at 2:26
Robert Lewis
42.3k22761
The radial vector is
$vec r = begin{pmatrix} x \ y end{pmatrix}; tag 1$
also,
$F = begin{pmatrix} y \ - x end{pmatrix}; tag 2$
with
$x = 3 cos t, tag 3$
$y = 3 sin t, tag 4$
we have
$vec r = begin{pmatrix} 3 cos t \ 3sin t end{pmatrix}, tag 5$
$d vec r = dfrac{d vec r}{dt} dt = begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt, tag 5$
and
$F = begin{pmatrix} 3sin t \ -3cos t end{pmatrix}; tag 6$
then
$F cdot d vec r = begin{pmatrix} 3sin t \ -3cos t end{pmatrix} cdot begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt$
$= (-9 sin^2 t - 9 cos^2 t)dt = -9(sin^2 t + cos^2 t)dt = -9(1)dt = -9dt; tag 7$
finally,
$displaystyle int_C F cdot d vec r = int_C -9 dt = -9 int_C dt = -9(2pi) = -18pi. tag 8$
answered Nov 26 at 2:26
Robert Lewis
42.3k22761
answered Nov 26 at 2:26
Robert Lewis
42.3k22761
answered Nov 26 at 2:26
Robert Lewis
42.3k22761
answered Nov 26 at 2:26
Robert Lewis
42.3k22761
42.3k22761
add a comment |
add a comment |
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Just follow the definition. Can you look up the definition?
– Zachary Selk
Nov 26 at 2:09
@ZacharySelk Here $F=<y,-x>$. Then what about $r(t)$?
– user982787
Nov 26 at 2:16