Resistance of a wire with changing radius

[brainslush]

Homework Statement

The radius r of a wire of length L increases according to r = a * exp(bx^2), x is the distance from one end to the other end of the wire. What is the resistance of the wire?

Homework Equations

$R =\frac{L * \rho}{A}$

The Attempt at a Solution

$dR =\frac{dx * \rho}{A}$

$A(r) = \pi * r^2$

$r(x) = a* e^{bx^2}$

$A(x) =\pi a^2 * e^{2bx^2}$

$R =\int^{L}_{0}\frac{\rho}{\pi a^2 * e^{2bx^2}}*dx$

Two questions
First of all. Is this approach correct? And second, how does one integrate an errorfunction? (Of course one can use WolframAlpha but how does one get this solution?)
Thanks

 

[L-x]

Remember that A is a function of r when you try to work out dR

 

[brainslush]

Of course, A is a function of r but r is also a function of x. Therefore the substitution yields what I have written above. Or am I wrong?

 

[L-x]

$$R =\frac{x \rho}{A}=\frac{x \rho}{\pi r^2}=\frac{x \rho}{\pi a^2 e^{2bx^2}}$$

Are you sure your expression for dR is correct?

 

[brainslush]

Our professor just told us that r = a * exp(b*x) and i found the mistake... A is dependent on r and therefore I need to consider this in my calculations...

$dR = \rho d(\frac{L}{A})$

$d(\frac{L}{A}) = \frac{dL}{A} - \frac{L*dA}{A^2}$

$dA = 2*\pi*r*dr$ , $dL = dx$

$dr = a*b*e^{bx} dx$

$dR =\rho \int^{L}_{0} (\frac{e^{-2bx}}{\pi*a} - \frac{2*x*b*e^{-2*b*x}}{\pi*a^2})dx$