Find the work done by the force field $F(x,y)=(x^2+xy)bar{i}+(xy^2)bar{j}$ along a path
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Find the work done by the force field $F(x,y)=(x^2+xy)bar{i}+(xy^2)bar{j}$ when a particle moves from the origin, along the $x$ axis to the $(1,0)$, then on the line segment that joins the $(1,0)$ with the $(0,1)$ and finally returns along the $y$ axis to origin.
I have thought to do the following:
By Green's theorem we have to
$int_CFcdot rdr=intint_D(y^2-x)dA=int_{0}^{1}int_{0}^{1}(y^2-x)dxdy=int_{0}^{1}[y^2x-x^2/2]_{0}^{1}dy=int_{0}^{1}(y^2-1/2)dy=y^3/3-y/2]_{0}^{1}=1/3-1/2=-1/6$
Is this fine? Thank you.
calculus integration multivariable-calculus greens-theorem
edited Nov 18 at 18:52
David G. Stork
9,07321232
asked Nov 16 at 18:30
Nash
47049
add a comment |
up vote
0
down vote
favorite
Find the work done by the force field $F(x,y)=(x^2+xy)bar{i}+(xy^2)bar{j}$ when a particle moves from the origin, along the $x$ axis to the $(1,0)$, then on the line segment that joins the $(1,0)$ with the $(0,1)$ and finally returns along the $y$ axis to origin.
I have thought to do the following:
By Green's theorem we have to
$int_CFcdot rdr=intint_D(y^2-x)dA=int_{0}^{1}int_{0}^{1}(y^2-x)dxdy=int_{0}^{1}[y^2x-x^2/2]_{0}^{1}dy=int_{0}^{1}(y^2-1/2)dy=y^3/3-y/2]_{0}^{1}=1/3-1/2=-1/6$
Is this fine? Thank you.
calculus integration multivariable-calculus greens-theorem
edited Nov 18 at 18:52
David G. Stork
9,07321232
asked Nov 16 at 18:30
Nash
47049
You're integrating over a square; the integral over the triangle enclosed by the path is $int_0^1 int_0^{1 - x} (y^2 - x) dy dx$ (also, $F cdot r dr$ should be $mathbf F cdot dmathbf r$).
– Maxim
Nov 18 at 19:59
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find the work done by the force field $F(x,y)=(x^2+xy)bar{i}+(xy^2)bar{j}$ when a particle moves from the origin, along the $x$ axis to the $(1,0)$, then on the line segment that joins the $(1,0)$ with the $(0,1)$ and finally returns along the $y$ axis to origin.
I have thought to do the following:
By Green's theorem we have to
$int_CFcdot rdr=intint_D(y^2-x)dA=int_{0}^{1}int_{0}^{1}(y^2-x)dxdy=int_{0}^{1}[y^2x-x^2/2]_{0}^{1}dy=int_{0}^{1}(y^2-1/2)dy=y^3/3-y/2]_{0}^{1}=1/3-1/2=-1/6$
Is this fine? Thank you.
calculus integration multivariable-calculus greens-theorem
edited Nov 18 at 18:52
David G. Stork
9,07321232
asked Nov 16 at 18:30
Nash
47049
Find the work done by the force field $F(x,y)=(x^2+xy)bar{i}+(xy^2)bar{j}$ when a particle moves from the origin, along the $x$ axis to the $(1,0)$, then on the line segment that joins the $(1,0)$ with the $(0,1)$ and finally returns along the $y$ axis to origin.
I have thought to do the following:
By Green's theorem we have to
$int_CFcdot rdr=intint_D(y^2-x)dA=int_{0}^{1}int_{0}^{1}(y^2-x)dxdy=int_{0}^{1}[y^2x-x^2/2]_{0}^{1}dy=int_{0}^{1}(y^2-1/2)dy=y^3/3-y/2]_{0}^{1}=1/3-1/2=-1/6$
Is this fine? Thank you.
calculus integration multivariable-calculus greens-theorem
calculus integration multivariable-calculus greens-theorem
edited Nov 18 at 18:52
David G. Stork
9,07321232
asked Nov 16 at 18:30
Nash
47049
edited Nov 18 at 18:52
David G. Stork
9,07321232
asked Nov 16 at 18:30
Nash
47049
edited Nov 18 at 18:52
David G. Stork
9,07321232
edited Nov 18 at 18:52
David G. Stork
9,07321232
edited Nov 18 at 18:52
David G. Stork
9,07321232
9,07321232
asked Nov 16 at 18:30
Nash
47049
asked Nov 16 at 18:30
Nash
47049
asked Nov 16 at 18:30
Nash
47049
47049
You're integrating over a square; the integral over the triangle enclosed by the path is $int_0^1 int_0^{1 - x} (y^2 - x) dy dx$ (also, $F cdot r dr$ should be $mathbf F cdot dmathbf r$).
– Maxim
Nov 18 at 19:59
add a comment |
You're integrating over a square; the integral over the triangle enclosed by the path is $int_0^1 int_0^{1 - x} (y^2 - x) dy dx$ (also, $F cdot r dr$ should be $mathbf F cdot dmathbf r$).
– Maxim
Nov 18 at 19:59
You're integrating over a square; the integral over the triangle enclosed by the path is $int_0^1 int_0^{1 - x} (y^2 - x) dy dx$ (also, $F cdot r dr$ should be $mathbf F cdot dmathbf r$).
– Maxim
Nov 18 at 19:59
You're integrating over a square; the integral over the triangle enclosed by the path is $int_0^1 int_0^{1 - x} (y^2 - x) dy dx$ (also, $F cdot r dr$ should be $mathbf F cdot dmathbf r$).
– Maxim
Nov 18 at 19:59
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Just perform the three integrals:
I: $intlimits_{x=0}^1 (x^2 + x y) dx = intlimits_{x=0}^1 x^2 dx = {1 over 3}$
where along the first path $y=0$.
II: $intlimits_{x=1}^0 x^2 + x y dx + intlimits_{y=0}^1 x y^2 dy$
where along the second path $y = 1-x$ so
$intlimits_{x=1}^0 x^2 + x (1 - x) dx + intlimits_{y=0}^1 (1-y) y^2 dy = -{5 over 12}$
III: $intlimits_{y=1}^0 x y^2 dy = 0$
where along the third path $x=0$:
Total = $-{1 over 12}$
answered Nov 16 at 19:11
David G. Stork
9,07321232
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Just perform the three integrals:
I: $intlimits_{x=0}^1 (x^2 + x y) dx = intlimits_{x=0}^1 x^2 dx = {1 over 3}$
where along the first path $y=0$.
II: $intlimits_{x=1}^0 x^2 + x y dx + intlimits_{y=0}^1 x y^2 dy$
where along the second path $y = 1-x$ so
$intlimits_{x=1}^0 x^2 + x (1 - x) dx + intlimits_{y=0}^1 (1-y) y^2 dy = -{5 over 12}$
III: $intlimits_{y=1}^0 x y^2 dy = 0$
where along the third path $x=0$:
Total = $-{1 over 12}$
answered Nov 16 at 19:11
David G. Stork
9,07321232
add a comment |
up vote
2
down vote
accepted
Just perform the three integrals:
I: $intlimits_{x=0}^1 (x^2 + x y) dx = intlimits_{x=0}^1 x^2 dx = {1 over 3}$
where along the first path $y=0$.
II: $intlimits_{x=1}^0 x^2 + x y dx + intlimits_{y=0}^1 x y^2 dy$
where along the second path $y = 1-x$ so
$intlimits_{x=1}^0 x^2 + x (1 - x) dx + intlimits_{y=0}^1 (1-y) y^2 dy = -{5 over 12}$
III: $intlimits_{y=1}^0 x y^2 dy = 0$
where along the third path $x=0$:
Total = $-{1 over 12}$
answered Nov 16 at 19:11
David G. Stork
9,07321232
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Just perform the three integrals:
I: $intlimits_{x=0}^1 (x^2 + x y) dx = intlimits_{x=0}^1 x^2 dx = {1 over 3}$
where along the first path $y=0$.
II: $intlimits_{x=1}^0 x^2 + x y dx + intlimits_{y=0}^1 x y^2 dy$
where along the second path $y = 1-x$ so
$intlimits_{x=1}^0 x^2 + x (1 - x) dx + intlimits_{y=0}^1 (1-y) y^2 dy = -{5 over 12}$
III: $intlimits_{y=1}^0 x y^2 dy = 0$
where along the third path $x=0$:
Total = $-{1 over 12}$
answered Nov 16 at 19:11
David G. Stork
9,07321232
Just perform the three integrals:
I: $intlimits_{x=0}^1 (x^2 + x y) dx = intlimits_{x=0}^1 x^2 dx = {1 over 3}$
where along the first path $y=0$.
II: $intlimits_{x=1}^0 x^2 + x y dx + intlimits_{y=0}^1 x y^2 dy$
where along the second path $y = 1-x$ so
$intlimits_{x=1}^0 x^2 + x (1 - x) dx + intlimits_{y=0}^1 (1-y) y^2 dy = -{5 over 12}$
III: $intlimits_{y=1}^0 x y^2 dy = 0$
where along the third path $x=0$:
Total = $-{1 over 12}$
answered Nov 16 at 19:11
David G. Stork
9,07321232
answered Nov 16 at 19:11
David G. Stork
9,07321232
answered Nov 16 at 19:11
David G. Stork
9,07321232
answered Nov 16 at 19:11
David G. Stork
9,07321232
9,07321232
add a comment |
add a comment |
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You're integrating over a square; the integral over the triangle enclosed by the path is $int_0^1 int_0^{1 - x} (y^2 - x) dy dx$ (also, $F cdot r dr$ should be $mathbf F cdot dmathbf r$).
– Maxim
Nov 18 at 19:59