Simple Curve Integral using Parametrization
$begingroup$
I know the answer but cannot for the love of everything figure out how the book got the answer and the solution is only the answer nothing step-by step at all.
$$
F(D) = F*dr
$$calculate curve integral using the parametrisation
where F is
$$
F(x,y,z) = yz, xz, xy
$$
over domain D, where D is
$$
(x,y,z) = cos(t), sin(t), t
$$
where t is between (0, pi/4)
So im supposed to use the parametrization that is already there
I get this:
$$
F(x,y,z) =tsin(t), tcos(t), sin(t)cos(t) $$
$$
r'(t) =-sin(t), cos(t), 1 $$
I know im supposed to multiply F with r'(t) but I can't simplify in any way that get the answer pi/8 that should be the final answer for the integral
$$
int_0^{pi/4}-tsin^2(t) + tcos^2(t) + cos(t)sin(t)$$
Any idea how to get from there to pi/8?
integration algebraic-curves parametrization
asked Dec 18 '18 at 15:38
user3514461user3514461
82
$endgroup$
add a comment |
$begingroup$
I know the answer but cannot for the love of everything figure out how the book got the answer and the solution is only the answer nothing step-by step at all.
$$
F(D) = F*dr
$$calculate curve integral using the parametrisation
where F is
$$
F(x,y,z) = yz, xz, xy
$$
over domain D, where D is
$$
(x,y,z) = cos(t), sin(t), t
$$
where t is between (0, pi/4)
So im supposed to use the parametrization that is already there
I get this:
$$
F(x,y,z) =tsin(t), tcos(t), sin(t)cos(t) $$
$$
r'(t) =-sin(t), cos(t), 1 $$
I know im supposed to multiply F with r'(t) but I can't simplify in any way that get the answer pi/8 that should be the final answer for the integral
$$
int_0^{pi/4}-tsin^2(t) + tcos^2(t) + cos(t)sin(t)$$
Any idea how to get from there to pi/8?
integration algebraic-curves parametrization
asked Dec 18 '18 at 15:38
user3514461user3514461
82
$endgroup$
add a comment |
$begingroup$
I know the answer but cannot for the love of everything figure out how the book got the answer and the solution is only the answer nothing step-by step at all.
$$
F(D) = F*dr
$$calculate curve integral using the parametrisation
where F is
$$
F(x,y,z) = yz, xz, xy
$$
over domain D, where D is
$$
(x,y,z) = cos(t), sin(t), t
$$
where t is between (0, pi/4)
So im supposed to use the parametrization that is already there
I get this:
$$
F(x,y,z) =tsin(t), tcos(t), sin(t)cos(t) $$
$$
r'(t) =-sin(t), cos(t), 1 $$
I know im supposed to multiply F with r'(t) but I can't simplify in any way that get the answer pi/8 that should be the final answer for the integral
$$
int_0^{pi/4}-tsin^2(t) + tcos^2(t) + cos(t)sin(t)$$
Any idea how to get from there to pi/8?
integration algebraic-curves parametrization
asked Dec 18 '18 at 15:38
user3514461user3514461
82
$endgroup$
I know the answer but cannot for the love of everything figure out how the book got the answer and the solution is only the answer nothing step-by step at all.
$$
F(D) = F*dr
$$calculate curve integral using the parametrisation
where F is
$$
F(x,y,z) = yz, xz, xy
$$
over domain D, where D is
$$
(x,y,z) = cos(t), sin(t), t
$$
where t is between (0, pi/4)
So im supposed to use the parametrization that is already there
I get this:
$$
F(x,y,z) =tsin(t), tcos(t), sin(t)cos(t) $$
$$
r'(t) =-sin(t), cos(t), 1 $$
I know im supposed to multiply F with r'(t) but I can't simplify in any way that get the answer pi/8 that should be the final answer for the integral
$$
int_0^{pi/4}-tsin^2(t) + tcos^2(t) + cos(t)sin(t)$$
Any idea how to get from there to pi/8?
integration algebraic-curves parametrization
integration algebraic-curves parametrization
asked Dec 18 '18 at 15:38
user3514461user3514461
82
asked Dec 18 '18 at 15:38
user3514461user3514461
82
asked Dec 18 '18 at 15:38
user3514461user3514461
82
asked Dec 18 '18 at 15:38
user3514461user3514461
82
asked Dec 18 '18 at 15:38
user3514461user3514461
82
82
add a comment |
add a comment |
1 Answer
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$begingroup$
You are given the vector field ${bf F}(x,y,z):=(yz,xz,xy)$ and the path $$gamma:quad {bf r}(t):=(cos t,sin t, t)qquadleft(0leq tleq{piover4}right) .$$ The line integral of thie field ${bf F}$ along $gamma$ is given by
$$int_gamma{bf F}cdot d{bf r}=int_0^{piover4}{bf F}bigl({bf r}(t)bigr)cdot{bf r}'(t)>dt=int_0^{piover4}bigl(-tsin^2 t+tcos^2 t+sin tcos t)>dt={piover8} .$$
For doing the integral you can use $t(cos^2 t-sin^2 t)=tcos(2t)$ and proceed by partial integration; furthermore $sin tcos t={1over2}sin(2t)$.
answered Dec 18 '18 at 17:10
Christian BlatterChristian Blatter
175k8115327
$endgroup$
$begingroup$
Still get pi^2/16 + 1/4...
$endgroup$
– user3514461
Dec 18 '18 at 19:35
$begingroup$
Worked, had a lot of problem integrating the tcos2t but in the end only (1/2)tsin2t is left and setting t=pi/4
$endgroup$
– user3514461
Dec 18 '18 at 21:32
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You are given the vector field ${bf F}(x,y,z):=(yz,xz,xy)$ and the path $$gamma:quad {bf r}(t):=(cos t,sin t, t)qquadleft(0leq tleq{piover4}right) .$$ The line integral of thie field ${bf F}$ along $gamma$ is given by
$$int_gamma{bf F}cdot d{bf r}=int_0^{piover4}{bf F}bigl({bf r}(t)bigr)cdot{bf r}'(t)>dt=int_0^{piover4}bigl(-tsin^2 t+tcos^2 t+sin tcos t)>dt={piover8} .$$
For doing the integral you can use $t(cos^2 t-sin^2 t)=tcos(2t)$ and proceed by partial integration; furthermore $sin tcos t={1over2}sin(2t)$.
answered Dec 18 '18 at 17:10
Christian BlatterChristian Blatter
175k8115327
$endgroup$
$begingroup$
Still get pi^2/16 + 1/4...
$endgroup$
– user3514461
Dec 18 '18 at 19:35
$begingroup$
Worked, had a lot of problem integrating the tcos2t but in the end only (1/2)tsin2t is left and setting t=pi/4
$endgroup$
– user3514461
Dec 18 '18 at 21:32
add a comment |
$begingroup$
You are given the vector field ${bf F}(x,y,z):=(yz,xz,xy)$ and the path $$gamma:quad {bf r}(t):=(cos t,sin t, t)qquadleft(0leq tleq{piover4}right) .$$ The line integral of thie field ${bf F}$ along $gamma$ is given by
$$int_gamma{bf F}cdot d{bf r}=int_0^{piover4}{bf F}bigl({bf r}(t)bigr)cdot{bf r}'(t)>dt=int_0^{piover4}bigl(-tsin^2 t+tcos^2 t+sin tcos t)>dt={piover8} .$$
For doing the integral you can use $t(cos^2 t-sin^2 t)=tcos(2t)$ and proceed by partial integration; furthermore $sin tcos t={1over2}sin(2t)$.
answered Dec 18 '18 at 17:10
Christian BlatterChristian Blatter
175k8115327
$endgroup$
$begingroup$
Still get pi^2/16 + 1/4...
$endgroup$
– user3514461
Dec 18 '18 at 19:35
$begingroup$
Worked, had a lot of problem integrating the tcos2t but in the end only (1/2)tsin2t is left and setting t=pi/4
$endgroup$
– user3514461
Dec 18 '18 at 21:32
add a comment |
$begingroup$
You are given the vector field ${bf F}(x,y,z):=(yz,xz,xy)$ and the path $$gamma:quad {bf r}(t):=(cos t,sin t, t)qquadleft(0leq tleq{piover4}right) .$$ The line integral of thie field ${bf F}$ along $gamma$ is given by
$$int_gamma{bf F}cdot d{bf r}=int_0^{piover4}{bf F}bigl({bf r}(t)bigr)cdot{bf r}'(t)>dt=int_0^{piover4}bigl(-tsin^2 t+tcos^2 t+sin tcos t)>dt={piover8} .$$
For doing the integral you can use $t(cos^2 t-sin^2 t)=tcos(2t)$ and proceed by partial integration; furthermore $sin tcos t={1over2}sin(2t)$.
answered Dec 18 '18 at 17:10
Christian BlatterChristian Blatter
175k8115327
$endgroup$
You are given the vector field ${bf F}(x,y,z):=(yz,xz,xy)$ and the path $$gamma:quad {bf r}(t):=(cos t,sin t, t)qquadleft(0leq tleq{piover4}right) .$$ The line integral of thie field ${bf F}$ along $gamma$ is given by
$$int_gamma{bf F}cdot d{bf r}=int_0^{piover4}{bf F}bigl({bf r}(t)bigr)cdot{bf r}'(t)>dt=int_0^{piover4}bigl(-tsin^2 t+tcos^2 t+sin tcos t)>dt={piover8} .$$
For doing the integral you can use $t(cos^2 t-sin^2 t)=tcos(2t)$ and proceed by partial integration; furthermore $sin tcos t={1over2}sin(2t)$.
answered Dec 18 '18 at 17:10
Christian BlatterChristian Blatter
175k8115327
edited Dec 18 '18 at 20:00
answered Dec 18 '18 at 17:10
Christian BlatterChristian Blatter
175k8115327
answered Dec 18 '18 at 17:10
Christian BlatterChristian Blatter
175k8115327
answered Dec 18 '18 at 17:10
Christian BlatterChristian Blatter
175k8115327
175k8115327
$begingroup$
Still get pi^2/16 + 1/4...
$endgroup$
– user3514461
Dec 18 '18 at 19:35
$begingroup$
Worked, had a lot of problem integrating the tcos2t but in the end only (1/2)tsin2t is left and setting t=pi/4
$endgroup$
– user3514461
Dec 18 '18 at 21:32
add a comment |
$begingroup$
Still get pi^2/16 + 1/4...
$endgroup$
– user3514461
Dec 18 '18 at 19:35
$begingroup$
Worked, had a lot of problem integrating the tcos2t but in the end only (1/2)tsin2t is left and setting t=pi/4
$endgroup$
– user3514461
Dec 18 '18 at 21:32
$begingroup$
Still get pi^2/16 + 1/4...
$endgroup$
– user3514461
Dec 18 '18 at 19:35
$begingroup$
Still get pi^2/16 + 1/4...
$endgroup$
– user3514461
Dec 18 '18 at 19:35
$begingroup$
Worked, had a lot of problem integrating the tcos2t but in the end only (1/2)tsin2t is left and setting t=pi/4
$endgroup$
– user3514461
Dec 18 '18 at 21:32
$begingroup$
Worked, had a lot of problem integrating the tcos2t but in the end only (1/2)tsin2t is left and setting t=pi/4
$endgroup$
– user3514461
Dec 18 '18 at 21:32
add a comment |
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