The radius, the cone's vertical height, and the slant height form a right triangle. Use the Pythagorean theorem after finding $r$.

Do not include the circular face where the cone and cylinder are attached. Include the bottom base, both lateral surfaces, and no other circular face.

Algebra is the fastest method. To verify in Desmos, graph $y=\frac{22}{9}x^3$ and $y=1782$. The positive $x$-coordinate of their intersection is the radius $r$.

After finding the radius, enter $r=9$ and $\ell=\sqrt{r^2+\left(\frac{4r}{3}\right)^2}$. Then enter $k=r^2+2r(2r)+r\ell$ to add the bottom-base area and the two lateral areas while excluding the attached circular face.

The cylinder volume is $\pi r^2(2r)=2\pi r^3$. The cone volume is $\frac{1}{3}\pi r^2\left(\frac{4r}{3}\right)=\frac{4}{9}\pi r^3$. Therefore, $2\pi r^3+\frac{4}{9}\pi r^3=1782\pi$. Dividing by $\pi$ gives $\frac{22}{9}r^3=1782$, so $r^3=729$ and $r=9$.

The cone's slant height $\ell$ is the hypotenuse of a right triangle with legs $r=9$ and $\frac{4r}{3}=12$. Thus, $\ell=\sqrt{9^2+12^2}=15$.

The bottom base has area $\pi r^2=81\pi$. The cylinder's lateral area is $2\pi r(2r)=324\pi$, and the cone's lateral area is $\pi r\ell=135\pi$. Therefore, $81\pi+324\pi+135\pi=540\pi$, so the value of $k$ is $540$.