Question

Find the volume and surface area of a solid in the form of a right circular cylinder with hemispherical ends whose extreme length is 21 dm and diameter is 2.5 dm.

Answer 1

Given:

Extreme (total) length = 21 dm

Diameter = 2.5 dm

Find radius and cylinder height.

Radius r = `"diameter"/2 = 2.5/2` = 1.25 dm.

Two hemispheres together equal one full sphere; their axial length = 2r = 2.5 dm.

Cylinder height h = total length − 2r

= 21 − 2.5

= 18.5 dm

Use the formula for cylinder volume and curved surface area

V_cyl = πr²h

= 1.25² = 1.5625, so V_cyl

= π × 1.5625 × 18.5 = 28.90625π dm³

Two hemispheres = one sphere, volume

`V_(sph) = (4/3)π r^3`   ......[`"hemisphere"/"sphere formula used"`]

= `1.953125, so V_(sph) = (4/3)π × 1.953125`

= 2.6041666667π dm³    

Total volume V = V_cyl + V_sph

= (28.90625 + 2.6041666667)π

= 31.5104166667π dm³

Numerical value

V ≈ 31.51041667 × `22/7`

= 99.0327 dm³

= 99.03 dm³

Curved (lateral) area of the cylinder (ends are covered by hemispheres): A_cyl = 2π r h

= 2π × 1.25 × 18.5

= 46.25π dm²

Surface area of the two hemispheres = surface area of the full sphere 

= 4π r²

= 4π × 1.5625

= 6.25π dm²

Total surface area

A = 46.25π + 6.25π

= 52.5π dm²

A = 52.5 × `22/7` 

= 165.0 dm²

= 165 dm²