We know #r = 5h# for the cylinder

The formula for the volume of a cylinder is:

#V = pir^2h#

We can substitute #5h# for #r# giving:

#V = pi(5h)^2h#

#V = pi25h^2h#

#V = pi25h^3#

#V = 25pih^3#

The formula for the surface area of a cylinder is:

#A = 2pirh + 2pir^2#

Again, we can substitute #5h# for #r# giving:

#A = (2pi5h xx h) + 2pi(5h)^2#

#A = 10pih^2 + 2pi25h^2#

#A = 10pih^2 + 50pih^2#

#A = (10 + 50)pih^2#

#A = 60pih^2#

Because the Area is equal to the volume we can equate the two and solve for #h#:

#60pih^2 = 25pih^3#

#(60pih^2)/(25pih^2) = (25pih^3)/(25pih^2)#

#(60color(red)(cancel(color(black)(pih^2))))/(25color(red)(cancel(color(black)(pih^2)))) = (color(red)(cancel(color(black)(25pi)))h^3)/(color(red)(cancel(color(black)(25pi)))h^2)#

#60/25 = h^3/h^2#

#(5 xx 12)/(5 xx 5) = (h^2 xx h)/h^2#

#(color(red)(cancel(color(black)(5))) xx 12)/(color(red)(cancel(color(black)(5))) xx 5) = (color(red)(cancel(color(black)(h^2))) xx h)/color(red)(cancel(color(black)(h^2)))#

#12/5 = h#

#h = 12/5#

The we can calculate #r# as #r = 5 xx 12/5 = 12#

We can substitute these back into the formula for the volume of a cylinder and calculate #V#:

#V = pir^2h# becomes:

#V = pi xx 12^2 xx 12/5#

#V = pi xx 144 xx 12/5#

#V = pi xx 1728/5#

#V = pi xx 345.6#

#V = 345.6pi#

Approximating #pi# with 3.14 gives:

#V = 345.6pi ~= 345.6 xx 3.14 ~= 1085.184#