The area of the curved surface of an iron -sphere is equal to the area of the total surface of a cylinder having length and radius?

The surface area of the cylinder is:

A=2pr^2+2prh

A=2pr(r+h) and r=2.25 and h=22.25 so

A=4.5p(24.5)

A=110.25p

Now that area is equal to the area of the sphere which is:

A=4pr^2 so we can find the radius of the sphere...

4pr^2=110.25p

r^2=27.5625

r=√27.5625

Now the volume of that sphere is:

V=(4pr^3)/3 or (4p/3)r^3 which with r found above is:

V=(4p/3)(27.5625)^(3/2)

Since the density is 7.8g/cm^3 we can find the weight (actually this is mass since it is in grams :P)

W=Vd

W=7.8(4p/3)(27.5625)^(3/2)

W=10.4p(27.5625)^(3/2)g

W~4727.822g

So the answer you had was obviously in kg not g...

W~4.73kg

A(cyl) = 2pirh + 2pir^2 = 2pir(h + r)

h = 22.25
r = 2,25
A(cyl) = 2(2.25)pi(22.25 + 2.25)
A(cyl) = 2(2.25)(24.5)pi = 110.25 pi
A(sph) = A(cyl)
Hence
A(sph) = 110.25 pi
The general eq'n for A(sph ) = 4pi r^2
Hence
4pi r^2 = 110.25 pi
'pi' cancels down
4r^2 = 110..25
r^2 = 110.25/4 = 27.562...
r = sqrt(27.2562 = 5.25 (is the radius if the sphere.).

Volume(sph) = (4/3)pi r^3
V(sph) = (4/3)pi (5.25)^3
V(sph) = 606.131 ... cm^3

Density = mass/ volume
mass = density x volume
mass = 7.8 g cm^-3 X 606.131 cm^3
mass = 4,727.8 g = 4.7278 kg.