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Calculate the work and the flux!!

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hello!!! :)
I am given the following exercise:
"Calculate the work and the flux for the path $C: a \cos{\theta}\hat{i}+a \sin{\theta}\hat{j}, 0 \leq \theta \leq \frac{\pi}{2}$ , knowing that $\overrightarrow{F}=x\hat{i}+y\hat{j}$. "

To solve this I thought that I could use the Green's Theorem,like that:

To calculate the work,I did the following:
$$ \oint_C{\overrightarrow{F}}dR=\int \int_R {\nabla \times \overrightarrow{F} \cdot \hat{n}}dA=\iint_R {\nabla \times \overrightarrow{F} \cdot \hat{k}}dA=\iint_R {0}dA=0$$

Is the result for the work correct? :confused:

And,for the flux,I tried this:
$$ \oint_C{\overrightarrow{F} \cdot \hat{n}}ds=\iint_R{\nabla \cdot \overrightarrow{F}}dA=\iint_R{2}dA$$

How can I continue to find the flux? (Thinking)
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hello!!! :)
I am given the following exercise:
"Calculate the work and the flux for the path $C: a \cos{\theta}\hat{i}+a \sin{\theta}\hat{j}, 0 \leq \theta \leq \frac{\pi}{2}$ , knowing that $\overrightarrow{F}=x\hat{i}+y\hat{j}$. "

To solve this I thought that I could use the Green's Theorem,like that:

To calculate the work,I did the following:
$$ \oint_C{\overrightarrow{F}}dR=\int \int_R {\nabla \times \overrightarrow{F} \cdot \hat{n}}dA=\iint_R {\nabla \times \overrightarrow{F} \cdot \hat{k}}dA=\iint_R {0}dA=0$$

Is the result for the work correct? :confused:...
The result for thye work is correct but the procedure probably isn't... The Green's theorem can bu used only for closed path and Your path, that connect the point [a,0] to the point [0,a] with a quarter of circle isn't closed. What You can use is the fact that the field $\displaystyle \overrightarrow{F} = x\ \overrightarrow {i} + y\ \overrightarrow {j}$ obeys to the condition $\displaystyle \frac{\partial {f_{x}}}{\partial {y}} = \frac{\partial {f_{y}}}{\partial {x}}$ so that it is conservative and the work doesn't depend from the path but only from the starting and final points. In this case is...

$\displaystyle W = \int_{a}^{0} x\ dx + \int_{0}^{a} y\ dy = 0\ (1)$


Kind regards


$\chi$ $\sigma$
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
The result for thye work is correct but the procedure probably isn't... The Green's theorem can bu used only for closed path and Your path, that connect the point [a,0] to the point [0,a] with a quarter of circle isn't closed. What You can use is the fact that the field $\displaystyle \overrightarrow{F} = x\ \overrightarrow {i} + y\ \overrightarrow {j}$ obeys to the condition $\displaystyle \frac{\partial {f_{x}}}{\partial {y}} = \frac{\partial {f_{y}}}{\partial {x}}$ so that it is conservative and the work doesn't depend from the path but only from the starting and final points. In this case is...

$\displaystyle W = \int_{a}^{0} x\ dx + \int_{0}^{a} y\ dy = 0\ (1)$


Kind regards


$\chi$ $\sigma$
Why is the formula of $W$ like that?? :confused: And also,why did you take the intervals $[a,0]$ and $[0,a]$ ?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Why is the formula of $W$ like that?? :confused: And also,why did you take the intervals $[a,0]$ and $[0,a]$ ?
Please observe the following picture...



Once You have found that the field of $\overrightarrow{F}$ is conservative, then the work along a path depends only from ne initial and ending point of the path. The original path is from the point [a,0] to the point [0,a] along a quarter of circle, but the path [a,] -> [0,0] and [0,0] -> [0,a] is equivalent to it, so that...


Kind regards


$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hello!!! :)
I am given the following exercise:
"Calculate the work and the flux for the path $C: a \cos{\theta}\hat{i}+a \sin{\theta}\hat{j}, 0 \leq \theta \leq \frac{\pi}{2}$ , knowing that $\overrightarrow{F}=x\hat{i}+y\hat{j}$. "
In order to answer the question about 'flux' it is necessary that somebody explain to me a little detail. I have no problem to undestand the concept of flux of a vector through a surface in $\mathbb{R}^{3}$ but it is difficult for me to undestand the concept of flux of a vector along a path in $\mathbb{R}^{2}$... who helps me?...

Kind regards

$\chi$ $\sigma$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hey!! :eek:

Hello!!! :)
I am given the following exercise:
"Calculate the work and the flux for the path $C: a \cos{\theta}\hat{i}+a \sin{\theta}\hat{j}, 0 \leq \theta \leq \frac{\pi}{2}$ , knowing that $\overrightarrow{F}=x\hat{i}+y\hat{j}$. "

To solve this I thought that I could use the Green's Theorem,like that:

To calculate the work,I did the following:
$$ \oint_C{\overrightarrow{F}}dR=\int \int_R {\nabla \times \overrightarrow{F} \cdot \hat{n}}dA=\iint_R {\nabla \times \overrightarrow{F} \cdot \hat{k}}dA=\iint_R {0}dA=0$$

Is the result for the work correct? :confused:
As chisigma already remarked, you can't apply Green's Theorem here because the path is not closed.

For the work you need to calculate the path integral:
\begin{aligned}W&=\int_C \mathbf F \cdot d\mathbf s \\
&= \int_0^{\pi/2} (a\cos\theta, a\sin\theta) \cdot (-a\sin\theta, a\cos\theta) d\theta \\
&= \int_0^{\pi/2} (-a\cos\theta\cdot a\sin\theta + a\sin\theta \cdot a\cos\theta) d\theta \\
&= \int_0^{\pi/2} 0 d\theta \\
&= 0\end{aligned}

You can make it a bit easier by observing that the force $\mathbf F$ is perpendicular to the path C everywhere.
Therefore $\mathbf F$ does not do any work. (Whew)
So:
$$W = 0$$


And,for the flux,I tried this:
$$ \oint_C{\overrightarrow{F} \cdot \hat{n}}ds=\iint_R{\nabla \cdot \overrightarrow{F}}dA=\iint_R{2}dA$$

How can I continue to find the flux? (Thinking)
Since the curve is not closed, we cannot apply that theorem.

Assuming that with flux you mean the integration of the perpendicular vector over the path, we would have:

\begin{aligned}\text{Flux}&=\int_C \mathbf F \cdot \mathbf {\hat n} ds \\
&= \int_0^{\pi/2} (a\cos\theta, a\sin\theta) \cdot (\cos\theta, \sin\theta) a d\theta \\
&= \int_0^{\pi/2} (a\cos^2\theta + a\sin^2\theta) a d\theta \\
&= \int_0^{\pi/2} a^2 d\theta \\
&= \frac{\pi a^2}{2}\end{aligned}

Alternatively, we can see that the curve has radius $a$ and length $\frac {\pi a} 2$.
The force is perpendicular and has size $a$.
So this flux is $\frac {\pi a} 2 \cdot a = \frac {\pi a^2} 2$.
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
In order to answer the question about 'flux' it is necessary that somebody explain to me a little detail. I have no problem to undestand the concept of flux of a vector through a surface in $\mathbb{R}^{3}$ but it is difficult for me to undestand the concept of flux of a vector along a path in $\mathbb{R}^{2}$... who helps me?...

Kind regards

$\chi$ $\sigma$
Nobody helped me so that I try to work by 'intuition'. Considering to be in $\mathbb{R}^{3}$ the force$\displaystyle \overrightarrow{F} = x\ \overrightarrow {i} + y\ \overrightarrow{j}$ is independent from the z variable and the 'surface' can be considered having the path as 'base' and an unity segment as 'height'. In that case the flux is...

$\displaystyle F = \int_{S} \overrightarrow{F}\ \cdot \overrightarrow {n}\ dS = \int_{0}^{1} \int_{0}^{\frac{\pi}{2}} a\ dz\ d \theta = \frac{\pi}{2}\ a$

Kind regards

$\chi$ $\sigma$
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hey!! :eek:



As chisigma already remarked, you can't apply Green's Theorem here because the path is not closed.

For the work you need to calculate the path integral:
\begin{aligned}W&=\int_C \mathbf F \cdot d\mathbf s \\
&= \int_0^{\pi/2} (a\cos\theta, a\sin\theta) \cdot (-a\sin\theta, a\cos\theta) d\theta \\
&= \int_0^{\pi/2} (-a\cos\theta\cdot a\sin\theta + a\sin\theta \cdot a\cos\theta) d\theta \\
&= \int_0^{\pi/2} 0 d\theta \\
&= 0\end{aligned}

You can make it a bit easier by observing that the force $\mathbf F$ is perpendicular to the path C everywhere.
Therefore $\mathbf F$ does not do any work. (Whew)
So:
$$W = 0$$




Since the curve is not closed, we cannot apply that theorem.

Assuming that with flux you mean the integration of the perpendicular vector over the path, we would have:

\begin{aligned}\text{Flux}&=\int_C \mathbf F \cdot \mathbf {\hat n} ds \\
&= \int_0^{\pi/2} (a\cos\theta, a\sin\theta) \cdot (\cos\theta, \sin\theta) a d\theta \\
&= \int_0^{\pi/2} (a\cos^2\theta + a\sin^2\theta) a d\theta \\
&= \int_0^{\pi/2} a^2 d\theta \\
&= \frac{\pi a^2}{2}\end{aligned}

Alternatively, we can see that the curve has radius $a$ and length $\frac {\pi a} 2$.
The force is perpendicular and has size $a$.
So this flux is $\frac {\pi a} 2 \cdot a = \frac {\pi a^2} 2$.
I calculated it as followed:
\begin{aligned}\text{Flux}&=\int_C \mathbf F \cdot \mathbf {\hat n} ds \\
&= \int_0^{\pi/2} (a\cos\theta, a\sin\theta) \cdot (\cos\theta, \sin\theta) d\theta \\
&= \int_0^{\pi/2} (a\cos^2\theta + a\sin^2\theta) d\theta \\
&= \int_0^{\pi/2} a d\theta \\
&= \frac{\pi a}{2}\end{aligned}
But you have $\int_0^{\pi/2} (a\cos\theta, a\sin\theta) \cdot (\cos\theta, \sin\theta)$ $a$ $d\theta$...Could you explain me why there is also an $a$?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
But you have $\int_0^{\pi/2} (a\cos\theta, a\sin\theta) \cdot (\cos\theta, \sin\theta)$ $a$ $d\theta$...Could you explain me why there is also an $a$?
The line segment $ds = a d\theta$.
It's what happens when you work in polar coordinates.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
The line segment $ds = a d\theta$.
It's what happens when you work in polar coordinates.
When calculating the work,why don't we also replace $ds$ with $a d\theta$ ?? :confused: (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
When calculating the work,why don't we also replace $ds$ with $a d\theta$ ?? :confused: (Thinking)
We could, but note the difference between $d\mathbf s$ and $ds$.
The first is a vector with length ds in the direction of the curve ($\textbf{bold upright}$), whereas the second is a scalar ($italic$).
Perhaps I should have written it as $d\overrightarrow{\mathbf s}$. :eek:
Their relation is:
$$d\vec{\mathbf s} = \hat{\mathbf c} ds$$
where $\hat{\mathbf c}$ is a vector of unit length in the direction of the curve.


Consider the function $\vec{\mathbf s}(\theta)=(a\cos\theta, a\sin\theta)$ that describes your curve.
Then:
$$d\vec{\mathbf s} = \vec{\mathbf{s}}\,'(\theta) d\theta = (-a\sin\theta, a\cos\theta) d\theta$$
whereas:
$$\hat{\mathbf c} ds = (-\sin\theta, \cos\theta) a d\theta$$

See that they are the same?
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hey!! :eek:



As chisigma already remarked, you can't apply Green's Theorem here because the path is not closed.

For the work you need to calculate the path integral:
\begin{aligned}W&=\int_C \mathbf F \cdot d\mathbf s \\
&= \int_0^{\pi/2} (a\cos\theta, a\sin\theta) \cdot (-a\sin\theta, a\cos\theta) d\theta \\
&= \int_0^{\pi/2} (-a\cos\theta\cdot a\sin\theta + a\sin\theta \cdot a\cos\theta) d\theta \\
&= \int_0^{\pi/2} 0 d\theta \\
&= 0\end{aligned}
Is there a difference using this formula: $\oint_C{\overrightarrow{F}}dR$ or this: $\int_C \mathbf F \cdot d\mathbf s $,to calculate the work?? :confused: (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Is there a difference using this formula: $\oint_C{\overrightarrow{F}}dR$ or this: $\int_C \mathbf F \cdot d\mathbf s $,to calculate the work?? :confused: (Thinking)
Yes.

The first version sums the entire force vector along the curve.
The result is a vector, which is not the work.

The second version sums only the component of the force vector that is along the curve, which is how work is calculated.
The result is a scalar.
It does so, by first taking the dot product of the force with a vector in the direction of the curve, and only summing afterward.



Consider for instance $\overrightarrow F = \hat \imath$ and the circle $\overrightarrow s(\theta) = a\cos\theta \hat \imath + a\sin\theta \hat \jmath$.
Then:
$$\oint_C{\overrightarrow{F}}dR = 2\pi a \hat \imath$$
while:
$$\oint_C{\overrightarrow{F}} \cdot d\overrightarrow s = 0$$
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Yes.

The first version sums the entire force vector along the curve.
The result is a vector, which is not the work.

The second version sums only the component of the force vector that is along the curve, which is how work is calculated.
The result is a scalar.
It does so, by first taking the dot product of the force with a vector in the direction of the curve, and only summing afterward.



Consider for instance $\overrightarrow F = \hat \imath$ and the circle $\overrightarrow s(\theta) = a\cos\theta \hat \imath + a\sin\theta \hat \jmath$.
Then:
$$\oint_C{\overrightarrow{F}}dR = 2\pi a \hat \imath$$
while:
$$\oint_C{\overrightarrow{F}} \cdot d\overrightarrow s = 0$$
Aha!! And what's with $\oint_C{\overrightarrow{F}} \cdot d \overrightarrow{R}$ ? (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Aha!! And what's with $\oint_C{\overrightarrow{F}} \cdot d \overrightarrow{R}$ ? (Thinking)
It's the same thing as with $d\overrightarrow s$, just with a different symbol to represent the curve.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720