The magnitude of the current density J flowing normally through a circular area of radius R varies with the radial distance r < R is 

                                J = 3ar + 2b 

where a and b are constants. The current through the circular area is 

Consider a small circular strip of radius r and thickness dr as shown in Fig.  

The current through the small circular strip is

                      $dI={\int }_{ }J.dA={\int }_{ }JdA \cos \theta$

where dA = $\hat{n}$ dA;   $\hat{n}$ is a unit vector perpendicular to dA. Hence $\theta$= 0°. Also dA = $\pi$(r+ dr)²$-$$\pi r^2$ = 2$\mathrm{\pi }$rdr. Therefore,-

                   $$\begin{gathered} dI={\int }_0^R\left(3\mathrm{ar}+2\mathrm{b}\right)\times 2\mathrm{\pi r} \mathrm{dr}\times \cos 0^0 \\ =2\mathrm{\pi }{\int }_0^{\mathrm{R}}\left(3\mathrm{ar}+2\mathrm{b}\right)\mathrm{r} \mathrm{dr} \\ =6\mathrm{\pi a}{\int }_0^{\mathrm{R}}{\mathrm{r}}^2\mathrm{dr}+4\mathrm{\pi b}{\int }_0^{\mathrm{R}}\mathrm{rdr} \\ =6\mathrm{\pi a}\frac{{\mathrm{R}}^3}{3}+4\mathrm{\pi b}\frac{{\mathrm{R}}^2}{2} \\ =2{\mathrm{\pi aR}}^3+2{\mathrm{\pi bR}}^2 \\ =2{\mathrm{\pi R}}^2\left(\mathrm{aR}+\mathrm{b}\right) \end{gathered}$$