The Electrician
- Joined Oct 9, 2007
- 2,986
Here's what I get using both formulations and both values of R:
Attachments
-
14.4 KB Views: 56
Last edited:
Check your units.I thought I'd try to solve the problem using the Cartesian System ...
Unfortunately I can't produce the same answer given earlier.
I have the "solution" as the integration shown below ...
\(\text{V=\int^{+1}_{-1}\{ \frac{(R^2-x^2)}{2}\[\pi - arcsin\(\sqr{\frac{12-x^2}{R^2-x^2}}\)\]+\sqr{12-x^2} \} \ dx \ = 39.7385 \ cu. \ units}\)
Did anyone else try the Cartesian approach?
Thanks for the "heads up" on the units. Although it looks dubious, I believe the "discrepancy" lies in my having partially substituted some constant (linear) dimensions specific to the problem during the formulation of the result.Check your units.