Question

The diameter of internal and external surface of hollow hemispherical bowl are 8cm and 10cm find the cost of painting the bowl completely @ rs 1.50 per square cm 0

Answer 1

Answer:

The cost of the painting a hemispherical bowl is $428.61 .

Step-by-step explanation:

Formula

$Area\ of\ the\ hemisphere = 2\pi a^{2}$

$Area\ of\ the\ top\ of\ the\ ring = \pi a^{2}$

Where a is the radius of the hemisphere .

Let us assume that the external radius of the hemisphere is denoted by u .

Let us assume thet the internal radius of the hemisphere is denoted by v .

As given

The diameter of internal and external surface of hollow hemispherical bowl are 8cm and 10cm .

$Radius(u)= \frac{10}{2}$

                     = 5 cm

$Radius(v)= \frac{8}{2}$

                     = 4 cm

Thus

$Total\ area\ painted = 2\pi u^{2} + 2\pi v^{2} + \pi u^{2} - \pi v^{2}$

Simplify the above

$Total\ area\ painted = 3\pi u^{2} + \pi v^{2}$

$\pi = 3.14$

Put all the values in the above

$Total\ area\ painted = 3\times 3.14\times 5\times 5+3.14\times 4\times 4$

$Total\ area\ painted =235.5+50.24$

                                        = 285.74 cm

As given

The cost of painting the bowl completely @ rs 1.50 per square cm .

Cost of the painting the bowl = 285.74 × Cost per  square cm

                                                 = 285.74 × 1.50

                                                 = $ 428.61

Therefore the cost of the painting a hemispherical bowl is $428.61 .