Chapter 12.6, Problem 7PSB

(a)

To determine

To calculate: The total volume of the given solid.

Answer to Problem 7PSB

The total volume of the given solid is $753.982{\text{ unit}}^{\text{3}}.$

Explanation of Solution

Given: In the given solid, cone is having height $\left(h\right)=8\text{ units}$ and radius $\left(r\right)=6\text{ units}\text{.}$

Formula Used:

Volume of Cone $\left(V\right)=\left(\frac{1}{3}\right)\pi r^{2}h$ ……………………(i)

where r = radius of base of cone

h = height of cone

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

where r = radius of base of hemisphere

  $$

Calculation:

As we can see that the solid is formed by placing a cone over the base of hemisphere which means that the radii of the cone and hemisphere would be same. Then,

Total volume of given solid $\left(V\right)=$ Volume of Cone + Volume of Hemisphere ……………(iii)

By using equations (i) and (ii) in equation (iii) we get,

  $V=\left(\frac{1}{3}\right)\pi r^{2}h+\left(\frac{2}{3}\right)\pi r^3$ ……………………………(iv)

Substituting the given values $h=8\text{ units}$ and $r=6\text{ units }$ in equation (iv)

  $\begin{array}{l}V=\left\{\left(\frac{1}{3}\right)\times \pi \times 6^2\times 8\right\}{\text{ unit}}^{\text{3}}+\left\{\left(\frac{2}{3}\right)\times \pi \times 6^3\right\}{\text{ unit}}^{\text{3}}\\ V=301.593{\text{ unit}}^{\text{3}}+452.389{\text{ unit}}^{\text{3}}\\ V=753.982{\text{ unit}}^{\text{3}}\end{array}$

Hence, the total volume of the given solid is $753.982{\text{ unit}}^{\text{3}}.$

(b)

To determine

To calculate: The total surface area of the given solid.

Answer to Problem 7PSB

The total surface area of the given solid is $\text{414}{\text{.690 unit}}^{\text{2}}.$

Explanation of Solution

Given: In the given solid, cone is having height $h=8\text{ units}$ and radius $r=6\text{ units}\text{.}$

Formula Used:

Curved Surface Area of Cone $\left(A\right)=\pi rl$ …………………. (i)

where r = radius of base of the cone

l = slant height of the cone

Slant height of the Cone $\left(l\right)=\sqrt{h^2+r^2}$ …………………. (ii)

where h = height of the cone

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

where r = radius of Hemisphere

Calculation: As we can see that the given solid is formed by placing a cone over the base of hemisphere which means that the radii of the cone and hemisphere would be same. So,

Total Surface Area of the given solid $\left(A\right)=$ Curved Surface Area of Cone + Curved Surface Area of Hemisphere …………………. (iv)

By using equation (i) and (iii) in equation (iv) we get,

  $A=\pi rl+2\pi r^2$ …………………………… (v)

Now replacing l in equation (v) by equation (ii),

  $A=\pi \times r\times \sqrt{h^2+r^2}+2\times \pi \times r^2$ ………… (vi) Using the given values $h=8\text{ units}$ and $r=6\text{ units }$ in equation (vi)

  $\begin{array}{l}A=\left(\pi \times 6\times \sqrt{8^2+6^2}\right){\text{ unit}}^{\text{2}}\text{ + }\left(2\times \pi \times 6^2\right){\text{ unit}}^{\text{2}}\\ A=\left(\pi \times 6\times 10\right){\text{ unit}}^{\text{2}}\text{ + }\left(2\times \pi \times 36\right){\text{ unit}}^{\text{2}}\\ A=60\pi {\text{ unit}}^{\text{2}}\text{ + 72}\pi {\text{ unit}}^{\text{2}}\\ A=132\pi {\text{ unit}}^{\text{2}}\\ A=414.690{\text{ unit}}^{\text{2}}\end{array}$

 Hence, the total volume of the given solid is $\text{414}{\text{.690 unit}}^{\text{2}}.$