1. ## Integration ln(x)

I am new to this forum. Can anyone help me to integrate ln(x) over a circle having radius r. X is the distance from the center to the point, like r. I appreciate any help!

2. Originally Posted by fanta
I am new to this forum. Can anyone help me to integrate ln(x) over a circle having radius r. X is the distance from the center to the point, like r. I appreciate any help!
Use polars $\displaystyle (r, \theta)$, then you seem to be asking for:

$\displaystyle \int_{\theta=0}^{2 \pi} \ln(r) d\theta = 2 \pi \ln(r)$

RonL

3. Thanks CaptainBlack! I am asked to find the average of ln(x) over a circle having radius r. x is the radial distance starting from center to any point. Can the answer 2pi*ln(r) be the answer? thanks for your help.

4. Originally Posted by fanta
Thanks CaptainBlack! I am asked to find the average of ln(x) over a circle having radius r. x is the radial distance starting from center to any point. Can the answer 2pi*ln(r) be the answer? thanks for your help.
No its not the average, the average is just $\displaystyle \ln(r)$ as this is
a constant on the circle.

RonL

5. Thanks again CaptainBlack! In the case of ln(x) --- x is a variable depending upon the distance from the center that is 0<= x <= r. Where r is the radius of the circle. I want to find the average ln(x) over the whole area of the circle for example at the curcumferance ln(x) = ln(r) at the center ln(x) = ln(0)-undefined? How can I get the average ln(x) for the entire circle -- sorry for the misunderstanding and thanks for your help.

6. Originally Posted by fanta
Thanks again CaptainBlack! In the case of ln(x) --- x is a variable depending upon the distance from the center that is 0<= x <= r. Where r is the radius of the circle. I want to find the average ln(x) over the whole area of the circle for example at the curcumferance ln(x) = ln(r) at the center ln(x) = ln(0)-undefined? How can I get the average ln(x) for the entire circle -- sorry for the misunderstanding and thanks for your help.
Ahh.. slightly different problem.

Now we need:

$\displaystyle \mu = (1/(\pi r^2) \int_{R=0}^r \int_{\theta=0}^{2 \pi} R\ \ln(R)\ d \theta dR$

which is the integral of $\displaystyle \ln(R)$ over the circle of radius $\displaystyle r$ in polars (where the area element is $\displaystyle R\ dR d\theta$) divited by the area of the circle.

RonL

7. is there a solution for this integral then?

8. Originally Posted by fanta
is there a solution for this integral then?
O-yes, but rather than slog through it by hand, I will ues machine assistance
for the final step. First change the order of integration and integrate out $\displaystyle \theta$

..... $\displaystyle =\frac{2}{r^2} \int_{R=0}^r R\ \ln(R) \ dR$

Now using QuickMath this becomes:

..... $\displaystyle = \ln(r)-1/2$

9. CaptainBlack . Thanks so much that's what I want!

10. CaptainBlack! I am quite gratful for your help and satisfied with your answer; however, I have got the same question but the circle diplaced from the origin by a distance x. The center of the circle is displaced by a distance x from the origion where the this distance, x is larger than the radius of the circle.

11. Originally Posted by fanta
CaptainBlack! I am quite gratful for your help and satisfied with your answer; however, I have got the same question but the circle diplaced from the origin by a distance x. The center of the circle is displaced by a distance x from the origion where the this distance, x is larger than the radius of the circle.
It is not clear to me what you want integrated over which region, as you
seem to be using x for two different things in this diagram.

RonL

12. Thanks for your reply! I am interested to find the average lan(x) in the circle. At any point inside the circle the function varies as lan(x) where x is the distance from the origin to any point in the circle. I want to find the average lan(x) inside the circle.[In the previous case we find the average lan(r) where the center of the circle is at the origin.]

13. I described it here graphically.