- #1
Jonsson
- 79
- 0
Hello there,
I've got a vector field which you can see here: Sketch of the vector field . It is: [itex]\vec{v} = \cos(x)\,\sin(y)\vec{i}-\sin(x)\,\cos(y)\vec{j}[/itex]
Say I want to find the circulation around the square formed by [itex]-\frac{\pi}{2} \, \leq x \leq \, \frac{\pi}{2}[/itex] and [itex]-\frac{\pi}{2} \, \leq y \leq \, \frac{\pi}{2}[/itex]. I think that I should do this by finding the circulation along one of the sides, and multiply by 4 (I can tell from the vector field that the circulation along one of the sides is going to be equal to every other side.
This is where I become unconfident. [itex]\int_{y = -\frac{\pi}{2}}^{y = \frac{\pi}{2}}\vec{v}\,\mathrm{d}\vec{r}[/itex] Please correct me if I am wrong. Any criticism is appreciated:
My [itex]\mathrm{d}\vec{r} = \mathrm{d}y\vec{j}[/itex] along the y-axis, for the first side: [itex](\Delta y = -\frac{\pi}{2}, x = -\frac{\pi}{2})[/itex].
[itex]\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(x)\,\sin(y) \, \mathrm{d}y -\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin(x)\,\cos(y) \, \mathrm{d}y[/itex]
[itex]\cos(-\frac{\pi}{2}) = 0[/itex], so the first integral is equal to zero.
[itex]-\sin(-\frac{\pi}{2})\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(y) \, \mathrm{d}y = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(y) \, \mathrm{d}y = [\sin(y)]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2}) = 2[/itex]
First question:
Is the above correct?
Second question:
Can you think of any additional criticism?
Thank you for your time.
Kind regards,
Marius
I've got a vector field which you can see here: Sketch of the vector field . It is: [itex]\vec{v} = \cos(x)\,\sin(y)\vec{i}-\sin(x)\,\cos(y)\vec{j}[/itex]
Say I want to find the circulation around the square formed by [itex]-\frac{\pi}{2} \, \leq x \leq \, \frac{\pi}{2}[/itex] and [itex]-\frac{\pi}{2} \, \leq y \leq \, \frac{\pi}{2}[/itex]. I think that I should do this by finding the circulation along one of the sides, and multiply by 4 (I can tell from the vector field that the circulation along one of the sides is going to be equal to every other side.
This is where I become unconfident. [itex]\int_{y = -\frac{\pi}{2}}^{y = \frac{\pi}{2}}\vec{v}\,\mathrm{d}\vec{r}[/itex] Please correct me if I am wrong. Any criticism is appreciated:
My [itex]\mathrm{d}\vec{r} = \mathrm{d}y\vec{j}[/itex] along the y-axis, for the first side: [itex](\Delta y = -\frac{\pi}{2}, x = -\frac{\pi}{2})[/itex].
[itex]\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(x)\,\sin(y) \, \mathrm{d}y -\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin(x)\,\cos(y) \, \mathrm{d}y[/itex]
[itex]\cos(-\frac{\pi}{2}) = 0[/itex], so the first integral is equal to zero.
[itex]-\sin(-\frac{\pi}{2})\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(y) \, \mathrm{d}y = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(y) \, \mathrm{d}y = [\sin(y)]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2}) = 2[/itex]
First question:
Is the above correct?
Second question:
Can you think of any additional criticism?
Thank you for your time.
Kind regards,
Marius