Integrate and Evaluate Vector Functions | Circle and Straight Line Paths

[thenewbosco]

hello, i just wish to check that i have done the following correctly:

1. Evaluate \int d\overarrow{r} (r is a vector, and its a closed integral) around the circle C represented by x^2 + y^2 = a^2

what i did here was switch to polars and called d\overarrow{r} ->rd\theta then i noted that r=a and integrated from 0 to 2pi to get the answer as 2pi * a.

and

2. If \overarrow{f} = x\hat{x} + y\hat{y} + z\hat{z}
evaluate \int \overarrow{f} \cdot d\overarrow{r} from (0,0,0) to (1,1,1) along
a) a straight line connecting these points
b) a path from (0,0,0) to (1,0,0) to (1,1,0) to (1,1,1)

for both of these i ended up getting 3/2
for a i just replaced y and z by x and dy and dz by dx and integrated 3x dx...
and for b i got 3/2 by adding up three integrals so i think this should be correct? thanks

 

[quasar987]

You should have gotten a hint that your answer is incorect for 1. since what this integral represents is the adding of many many tiny vectors one at the end of another such that the last one of these tiny vector ends where the first one started. Such a sum of vector gives the null vector. (Not to mention that you're summing vectors but your answer is a scalar )

What \oint d\vec{r} represents is actually

\int_{t_1}^{t_2}\frac{d\vec{r}}{dt}dt

for some parametetrisation \vec{r}(t), t\in (t_1,t_2)[/tex] of the circle. So, what this comes down to is finding a parametrisation of the circle and integrate its derivative.<br /> <br /> For 2. there is a good chance that is correct since \nabla \times \vec{f}=0, i.e. the field is conservative, so the line integral should be path-independant.

 

[quasar987]

However, if you had to integrate

\oint_C ||d\vec{r}||

then this would have been 2\pi a since now you're adding the length of all the tiny vectors, which in the limit dr-->0, should give you the circumference of the circle.