Answer:
[tex]h=15.23 \:\mathrm{in^2}[/tex]
Step-by-step explanation:
The surface area of this cylinder consist of two circles and a 3D outer edge with respect to the circumference. The area of a circle is given as [tex]r^2\pi[/tex]. Using [tex]\pi=3.14[/tex] as requested in the problem, the area of both these circles is:
[tex]2\cdot6^2\cdot 3.14=226.08\:\mathrm{in^2}\\[/tex].
The 3D outer edge is then [tex]800-226.08=573.92\:\mathrm{in^2}[/tex].
The area of this 3D outer edge is represented by [tex]2r\pi\cdot h[/tex]. Therefore, we set up the following equation:
[tex]2r\pi\cdot h=573.92, \\h=\frac{573.92}{2r\pi}, \\h=\fbox{$15.23\:\mathrm{in^2}$}[/tex].
