Determine $int_C Fcdot dx$ where $F(x,y,z)=(z,-y,x)$ and $C$ is the line segment from $(5,0,2)$ to $(5,3,4)$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Use Stokes' Theorem to evaluate integralEvaluate the line integral $int_C x^2 dx+(x+y)dy $Line integrals and path independenceFind a plane that passes through a given point and contains a given lineParametric line segment in 3-spaceEvaluate the line integral $int_C (x^2-y+3z);ds$Maximizing the line integral over a line segmentWork on a line segment - Final stepFinding the distance between a point and line (3-D)Formula of curvature not defined in arc length
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Determine $int_C Fcdot dx$ where $F(x,y,z)=(z,-y,x)$ and $C$ is the line segment from $(5,0,2)$ to $(5,3,4)$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Use Stokes' Theorem to evaluate integralEvaluate the line integral $int_C x^2 dx+(x+y)dy $Line integrals and path independenceFind a plane that passes through a given point and contains a given lineParametric line segment in 3-spaceEvaluate the line integral $int_C (x^2-y+3z);ds$Maximizing the line integral over a line segmentWork on a line segment - Final stepFinding the distance between a point and line (3-D)Formula of curvature not defined in arc length
$begingroup$
So first thing I did was parametrize the line. I used $gamma(t)=(5,t,2+frac23t)$ For $tin [0,3]$
I then have $F(gamma(t))=(2+frac23t,-t,5)$
and $dotgamma(t)=(0,1,frac23)$
Then $int_C Fdx=int_0^3 langle F,Trangle=int_0^3 -t+frac103dt=-frac23$
I'm pretty sure my parametrization is incorrect as I dont think I should get this negative.
multivariable-calculus
asked Mar 27 at 17:06
AColoredReptileAColoredReptile
401210
$endgroup$
add a comment |
$begingroup$
So first thing I did was parametrize the line. I used $gamma(t)=(5,t,2+frac23t)$ For $tin [0,3]$
I then have $F(gamma(t))=(2+frac23t,-t,5)$
and $dotgamma(t)=(0,1,frac23)$
Then $int_C Fdx=int_0^3 langle F,Trangle=int_0^3 -t+frac103dt=-frac23$
I'm pretty sure my parametrization is incorrect as I dont think I should get this negative.
multivariable-calculus
asked Mar 27 at 17:06
AColoredReptileAColoredReptile
401210
$endgroup$
add a comment |
$begingroup$
So first thing I did was parametrize the line. I used $gamma(t)=(5,t,2+frac23t)$ For $tin [0,3]$
I then have $F(gamma(t))=(2+frac23t,-t,5)$
and $dotgamma(t)=(0,1,frac23)$
Then $int_C Fdx=int_0^3 langle F,Trangle=int_0^3 -t+frac103dt=-frac23$
I'm pretty sure my parametrization is incorrect as I dont think I should get this negative.
multivariable-calculus
asked Mar 27 at 17:06
AColoredReptileAColoredReptile
401210
$endgroup$
So first thing I did was parametrize the line. I used $gamma(t)=(5,t,2+frac23t)$ For $tin [0,3]$
I then have $F(gamma(t))=(2+frac23t,-t,5)$
and $dotgamma(t)=(0,1,frac23)$
Then $int_C Fdx=int_0^3 langle F,Trangle=int_0^3 -t+frac103dt=-frac23$
I'm pretty sure my parametrization is incorrect as I dont think I should get this negative.
multivariable-calculus
multivariable-calculus
asked Mar 27 at 17:06
AColoredReptileAColoredReptile
401210
asked Mar 27 at 17:06
AColoredReptileAColoredReptile
401210
edited Mar 27 at 17:12
AColoredReptile
asked Mar 27 at 17:06
AColoredReptileAColoredReptile
401210
asked Mar 27 at 17:06
AColoredReptileAColoredReptile
401210
asked Mar 27 at 17:06
AColoredReptileAColoredReptile
401210
401210
add a comment |
add a comment |
1 Answer
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Your calculations are correct; except the last step of the integration, which is $$int_0^3left(frac103-tright)mathrm dt=frac112.$$
answered Mar 27 at 17:11
st.mathst.math
1,268115
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$begingroup$
Your calculations are correct; except the last step of the integration, which is $$int_0^3left(frac103-tright)mathrm dt=frac112.$$
answered Mar 27 at 17:11
st.mathst.math
1,268115
$endgroup$
add a comment |
$begingroup$
Your calculations are correct; except the last step of the integration, which is $$int_0^3left(frac103-tright)mathrm dt=frac112.$$
answered Mar 27 at 17:11
st.mathst.math
1,268115
$endgroup$
add a comment |
$begingroup$
Your calculations are correct; except the last step of the integration, which is $$int_0^3left(frac103-tright)mathrm dt=frac112.$$
answered Mar 27 at 17:11
st.mathst.math
1,268115
$endgroup$
Your calculations are correct; except the last step of the integration, which is $$int_0^3left(frac103-tright)mathrm dt=frac112.$$
answered Mar 27 at 17:11
st.mathst.math
1,268115
answered Mar 27 at 17:11
st.mathst.math
1,268115
answered Mar 27 at 17:11
st.mathst.math
1,268115
answered Mar 27 at 17:11
st.mathst.math
1,268115
1,268115
add a comment |
add a comment |
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