Chapter 12.6, Problem 8PSB

(a)

To determine

To calculate: The total volume enclosed by a hemispherical dome.

Answer to Problem 8PSB

The total volume enclosed by a hemispherical dome is $\text{56548}{\text{.668 m}}^{\text{3}}.$

Explanation of Solution

Given: A hemispherical dome is having radius $\left(r\right)=30\text{ m}$ as height is just same as the radius of its base.

Formula Used:

Volume of hemisphere $\left(V\right)=\left(\frac{2}{3}\right)\pi r^3$ …………………. (i)

where r = radius of the hemisphere

Calculation:

Volume of hemispherical dome can be calculated by using the given value of radius $\left(r\right)=30\text{ m}$ in equation (i)

  $\begin{array}{l}V=\left(\frac{2}{3}\right)\times \pi \times {\text{30}}^{\text{3}}{\text{ m}}^{\text{3}}\\ V=56548.668{\text{ m}}^{\text{3}}\end{array}$

 Hence, the total volume enclosed by a hemispherical dome is $\text{56548}{\text{.668 m}}^{\text{3}}.$

(b)

To determine

To calculate: The area of ground covered by the dome.

Answer to Problem 8PSB

The area of ground covered by the dome is $\text{2827}{\text{.433 m}}^{\text{2}}.$

Explanation of Solution

Given: A hemispherical dome is having radius $\left(r\right)=30\text{ m}$ as height is just same as the radius of its base.

Formula Used:

Area of Circular base $\left(A\right)=\pi r^2$ ………………. (ii)

where r = radius of the circular base

Calculation:

Here the area of ground covered by the hemispherical dome is the area of circular base of hemisphere.

Substituting the given value of radius $\left(r\right)=30\text{ m}$ in equation (ii) to get,

  $\begin{array}{l}A=\pi \times {30}^2{\text{ m}}^{\text{2}}\\ A=2827.433{\text{ m}}^{\text{2}}\end{array}$

 Hence, the area of ground covered by the dome is $\text{2827}{\text{.433 m}}^{\text{2}}.$

(c)

To determine

To calculate: The extra amount of paint needed to paint the dome than to paint the floor.

Answer to Problem 8PSB

The extra amount of paint needed to paint the dome than to paint the floor is $\text{2827}{\text{.433 m}}^{\text{2}}.$

Explanation of Solution

Given: A hemispherical dome is having radius $\left(r\right)=30\text{ m}$ as height is just same as the radius of its base.

Formula Used:

Curved Surface Area of Hemisphere $\left(A\right)=2\pi r^2$ ………... (iii)

where r = radius of Hemisphere

Area of Circular base $\left(A\right)=\pi r^2$ …………………………. (iv)

where r = radius of the circular base

Calculation:

The extra amount of paint required (S)= Curved Surface Area of Hemisphere - Area of Circular base

Now using the equations (iii) and (iv) in above equation to get,

  $S=2\pi r^2-\pi r^2$

  $S=\pi r^2$ …………………………….. (v)

Using the given value of radius $\left(r\right)=30\text{ m}$ in equation (v)

  $\begin{array}{l}S=\pi \times {30}^2{\text{ m}}^{\text{2}}\\ S=2827.433{\text{ m}}^{\text{2}}\end{array}$

 Hence, the extra amount of paint needed to paint the dome than to paint the floor is $\text{2827}{\text{.433 m}}^{\text{2}}.$

(d)

To determine

To calculate: The radius of a dome that covers double the area of ground covered by the given one.

Answer to Problem 8PSB

The radius of a dome that covers double the area of ground covered by the given one is $\text{42}\text{.426 m}.$

Explanation of Solution

Given: A hemispherical dome is having radius $\left(r\right)=30\text{ m}$ as height is just same as the radius of its base. Let us assume the radius of required dome be R.

Formula Used:

Area of Circular base $\left(A\right)=\pi r^2$

where r = radius of the circular base

Calculation:

According to question,

Area of the circular base of required dome (A)= $2\times$ Area of the circular base of given dome

  $\pi R^2=2\times \pi r^2$

  $R^2=2r^2$

Taking the square root of the terms on both sides of the above equation

  $R=\sqrt{2}r$ ……………………. (vi)

Using the given value of radius $\left(r\right)=30\text{ m}$ in equation (vi)

  $\begin{array}{l}R=\sqrt{2}\times 30\text{ m}\\ R=42.426\text{ m}\end{array}$

 Hence, the radius of a dome that covers double the area of ground covered by the given one is $\text{42}\text{.426 m}.$