Recent content by justaboy

Oh, I don't mean r1 and r2 as being two separate radii, I just mean they are the limits of integration, and no two limits of integration could give me the correct answer for the potential if that last step is correct. Is it only by evaluating over the surface that the potential diverges?

Anyone?

Cheers, great help! :)

I would normally, but the question explicitly states to solve for V(z) using the first integral. In addition, I'm not given any dimensions.

Homework Statement Calculate potential.

found my mistake... thanks

Thanks very much. I end up with this: \frac{dz}{dx}=\frac{a^2-z^2}{z^2} Which has turned out to be difficult to solve.

Does anyone see a way of solving for y=f(x)? My teacher is quite positive that it can be done. Apparently this is an old geometry problem that Fermat, Descartes, Bernoulli and Leibniz worked on in the late 17th century... the curve y(x) should be such that the length of the segment of the...

You just forgot a factor of 1/2 outside the integral. The integral for polar area is I = 1/2 integral of r^2 dtheta.

The question asks to solve for a curve y=y(x) explicitly.

Okay... after many trig subs, I found a\log(a^3 y)+\sqrt{a^2-y^2}-a\log(2a\sqrt{a^2-y^2}+a) = -x How do I go about finding a y(x)?

If you separate it, you get \frac{\sqrt{a^2-y^2}}{y}dy=-dx but I still have the same problem... how do I integrate the left side?

Homework Statement Solve the ODE: y'(x) = - \frac{y(x)}{\sqrt{a^2-y(x)^2}} The Attempt at a Solution To be honest I'm having trouble even classifying this ODE. My teacher hinted that the substitution z^2=a^2-y^2 could be helpful, but once I make the substitution, I can't seem to take the...