FOR GMS ICSE 2nd Sept

Ch 22 (Continued)

Today, we wish to take a step forward in Ch.22

Problems with a Cylinder considering the thickness of the cylinder

The last day, we worked at problems where we took it for granted that the cylinder had no thickness.

Today we take a real situation where most cylindrical shaped objects have some measurable thickness, for instance a water pipe ! You notice that then there will be

• an external radius / diameter / circumference / surface area
• an internal radius / diameter / circumference / surface area
• you will be able to find the volume of metal used 

We use the same formula

• external surface area = 2πRh (where R is the larger external radius)
• internal surface area = 2πrh (where r is the smaller internal radius)
• total surface area = [2πh(R + r) + 2π(R² – r²)]
• 2πh(R + r) is the surface area of the external + internal length of the cylinder
• 2π(R² – r²) is the surface area of the circular top and base of the cylinder, which is the area of the external circle - the area of the internal circle and this is doubled because it has a top and a base.
• volume of the metal in the cylinder will be πh(R² – r²) which is the volume of the external cylinder - the volume of the internal air space.

Now, kindly do the sum on pg.323 example 3 and Ex.22(A) nos 13 and 14 and in all three sums find

1. the total surface area of the open cylinder
2. the volume of the open cylinder.

So, these aren't too many sums to be done today are there ? All the best.