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Line integral path independence proof check

Finding work via Line Integralsfinding the parametric path for line integralparameterise line segment or functionCompute line integral for vector function along the curveFinding work done by force with line integralsBasic Line Integral for Workderiving the formula of the torsion of a curveCalculation of the line integral of a vector field along a pathTried a line integral and then did it using Stokes theorem, but I'm getting different results…Computing the line integral of the oriented curve

1

$begingroup$

Find the work done by the force $F(x, y, z) = (x^4y^5, x^3)$ along the curve
C given by the part of the graph of $y$ = $(x^3)$ from $(0, 0)$ to $(-1, -1)$.

I first checked for independence, which did not work.

Next I parameterized the curve by $r(t) = [x(t),y(t)]$,
beginalign*
x(t) &= t, \
y(t)&=t^3
endalign*
which has
$$dr= (1, 3t^2). $$

Computing the work is then
$$int_0^-1 (t^19,t^3)cdot (1,3t^2),dt = 11/20.$$
Is this correct?

calculus proof-verification line-integrals

share|cite|improve this question

edited Mar 16 at 10:45

Benjamin

625419

asked Mar 16 at 5:12

MasterYoshiMasterYoshi

807

$endgroup$

add a comment | 

1

$begingroup$

Find the work done by the force $F(x, y, z) = (x^4y^5, x^3)$ along the curve
C given by the part of the graph of $y$ = $(x^3)$ from $(0, 0)$ to $(-1, -1)$.

I first checked for independence, which did not work.

Next I parameterized the curve by $r(t) = [x(t),y(t)]$,
beginalign*
x(t) &= t, \
y(t)&=t^3
endalign*
which has
$$dr= (1, 3t^2). $$

Computing the work is then
$$int_0^-1 (t^19,t^3)cdot (1,3t^2),dt = 11/20.$$
Is this correct?

calculus proof-verification line-integrals

share|cite|improve this question

edited Mar 16 at 10:45

Benjamin

625419

asked Mar 16 at 5:12

MasterYoshiMasterYoshi

807

$endgroup$

add a comment | 

$begingroup$

Find the work done by the force $F(x, y, z) = (x^4y^5, x^3)$ along the curve
C given by the part of the graph of $y$ = $(x^3)$ from $(0, 0)$ to $(-1, -1)$.

I first checked for independence, which did not work.

Next I parameterized the curve by $r(t) = [x(t),y(t)]$,
beginalign*
x(t) &= t, \
y(t)&=t^3
endalign*
which has
$$dr= (1, 3t^2). $$

Computing the work is then
$$int_0^-1 (t^19,t^3)cdot (1,3t^2),dt = 11/20.$$
Is this correct?

calculus proof-verification line-integrals

share|cite|improve this question

edited Mar 16 at 10:45

Benjamin

625419

asked Mar 16 at 5:12

MasterYoshiMasterYoshi

807

$endgroup$

share|cite|improve this question

edited Mar 16 at 10:45

Benjamin

625419

asked Mar 16 at 5:12

MasterYoshiMasterYoshi

807

share|cite|improve this question

edited Mar 16 at 10:45

Benjamin

625419

asked Mar 16 at 5:12

MasterYoshiMasterYoshi

807

edited Mar 16 at 10:45

Benjamin

625419

edited Mar 16 at 10:45

Benjamin

625419

asked Mar 16 at 5:12

MasterYoshiMasterYoshi

807

asked Mar 16 at 5:12

MasterYoshiMasterYoshi

807

1 Answer
1

active

oldest

votes

1

$begingroup$

Looks fine to me. By "check independence" I assume you mean that you computed that $nabla times F neq 0$ so that $F$ has no chance of being integrable.

As a sanity check, when $x<0,y<0$ then $F$ points to the bottom-left, so that a positive work along the given curve segment is expected.

share|cite|improve this answer

answered Mar 16 at 7:34

user7530user7530

35.1k761114

$endgroup$

add a comment | 

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1

$begingroup$

Looks fine to me. By "check independence" I assume you mean that you computed that $nabla times F neq 0$ so that $F$ has no chance of being integrable.

As a sanity check, when $x<0,y<0$ then $F$ points to the bottom-left, so that a positive work along the given curve segment is expected.

share|cite|improve this answer

answered Mar 16 at 7:34

user7530user7530

35.1k761114

$endgroup$

add a comment | 

1

$begingroup$

Looks fine to me. By "check independence" I assume you mean that you computed that $nabla times F neq 0$ so that $F$ has no chance of being integrable.

As a sanity check, when $x<0,y<0$ then $F$ points to the bottom-left, so that a positive work along the given curve segment is expected.

share|cite|improve this answer

answered Mar 16 at 7:34

user7530user7530

35.1k761114

$endgroup$

add a comment | 

$begingroup$

Looks fine to me. By "check independence" I assume you mean that you computed that $nabla times F neq 0$ so that $F$ has no chance of being integrable.

As a sanity check, when $x<0,y<0$ then $F$ points to the bottom-left, so that a positive work along the given curve segment is expected.

share|cite|improve this answer

answered Mar 16 at 7:34

user7530user7530

35.1k761114

$endgroup$

share|cite|improve this answer

answered Mar 16 at 7:34

user7530user7530

35.1k761114

answered Mar 16 at 7:34

user7530user7530

35.1k761114

answered Mar 16 at 7:34

user7530user7530

35.1k761114