Resistance of a circular segment!

The small change in resistance $\Delta R$ at $r$ is given by $\Delta R = \dfrac{\rho \Delta r} {A}$ , where $A$ is the curved cross-sectional area $A = \dfrac{45^\circ}{360^\circ} \times 2\pi rt = \dfrac{\pi rt}{4}$ . Therefore,

$\displaystyle \Delta R = \dfrac{4\rho \Delta r} {\pi rt} \quad \Rightarrow \int_0^a \dfrac{4\rho} {\pi rt} dr = \int_0^a \dfrac{4\rho_0 r} {\pi rt} dr = \int_0^a \dfrac {4 \rho_0} {\pi t} dr = \boxed {\dfrac {4 \rho_0 a} {\pi t}}$

Let's divide the plate into infinite parallel differential conic wires of length $a$ , thickness $t$ and width of $rd\theta$ at a point with distance $r$ from the center $O$ . The resistance of a differential section of length $dr$ of the wire at a distance $r$ from the center $O$ is $\rho \frac{l}{A}$ , where $l$ is the length of the section and $A$ is its area, therefore $\rho \frac{l}{A} = \rho _o r\frac{{dr}}{{rtd\theta }} = \rho _o \frac{{dr}}{{td\theta }}$ . Integrating over the differential wire we obtain its resistence: $dR = \int\limits_0^a {\frac{{\rho _o }}{{td\theta }}dr} = \frac{{\rho _o a}}{{td\theta }}$ The conductance of the differential wire is: $dG = \frac{1}{{dR}} = \frac{{td\theta }}{{\rho _o a}}$ The total conductance of the plate is: $G = \int\limits_0^{\frac{\pi }{4}} {\frac{t}{{\rho _o a}}} d\theta = \frac{{\pi t}}{{4\rho _o a}}$ Finally, the total resistence of the plate is: $R = \frac{1}{G} = \frac{{4\rho _o a}}{{\pi t}}$