Check integral answers

  • #1
thenewbosco
187
0
hello, i just wish to check that i have done the following correctly:

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

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

and

2. If [tex]\overarrow{f} = x\hat{x} + y\hat{y} + z\hat{z}[/tex]
evaluate [tex]\int \overarrow{f} \cdot d\overarrow{r} [/tex] 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
 
Last edited:

Answers and Replies

  • #2
quasar987
Science Advisor
Homework Helper
Gold Member
4,790
20
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 :grumpy:)

What [itex]\oint d\vec{r}[/itex] represents is actually

[tex]\int_{t_1}^{t_2}\frac{d\vec{r}}{dt}dt[/tex]

for some parametetrisation [itex]\vec{r}(t)[/itex], [itex]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.

For 2. there is a good chance that is correct since [itex]\nabla \times \vec{f}=0[/itex], i.e. the field is conservative, so the line integral should be path-independant.
 
Last edited:
  • #3
quasar987
Science Advisor
Homework Helper
Gold Member
4,790
20
However, if you had to integrate

[tex]\oint_C ||d\vec{r}||[/tex]

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

Suggested for: Check integral answers

Replies
3
Views
441
Replies
5
Views
550
Replies
5
Views
145
Replies
6
Views
502
Replies
2
Views
301
Replies
47
Views
1K
Replies
18
Views
583
Replies
7
Views
143
Replies
2
Views
413
Replies
6
Views
409
Top