Chapter 12.4, Problem 26PPS

(a)

To determine

Describe if height is tripled, how it affects the volume of the cylinder.

Answer to Problem 26PPS

Tripled

Explanation of Solution

Given :

A cylinder has a radius of 5 centimeters and a height of 8 centimeters.

Calculation:

The volume of a cylinder with radius r and height h is :

  $V=\pi r^{2}h$

The height is tripled , $h_1=3h$ , so , the new volume:

  $V_1=\pi r^2{h}_1=\pi r^2(3h)=3\pi r^{2}h=3V$

So, the volume becomes three times , if the height is tripled.

(b)

To determine

Describe if radius is tripled, how it affects the volume of the cylinder.

Answer to Problem 26PPS

Nine times

Explanation of Solution

Given :

A cylinder has a radius of 5 centimeters and a height of 8 centimeters.

Calculation:

The volume of a cylinder with radius r and height h is :

  $V=\pi r^{2}h$

The radius is tripled , $r_1=3r$ , so , the new volume:

  $V_1=\pi r_1{}^{2}h=\pi {(3r)}^{2}h=9\pi r^{2}h=9V$

So, the volume becomes nine times , if the radius is tripled.

(c)

To determine

Describe if both the radius and height are tripled, how it affects the volume of the cylinder.

Answer to Problem 26PPS

Twenty-seven times

Explanation of Solution

Given :

A cylinder has a radius of 5 centimeters and a height of 8 centimeters.

Calculation:

The volume of a cylinder with radius r and height h is :

  $V=\pi r^{2}h$

The radius and height are tripled , $r_1=3r$ , $h_1=3h$ , so , the new volume:

  $V_1=\pi r_1{}^2{h}_1=\pi {(3r)}^2(3h)=27\pi r^{2}h=27V$

So, the volume becomes twenty-seven times , if the radius and height are tripled.

(d)

To determine

Describe if the dimensions are exchanged , how it affects the volume of the cylinder.

Answer to Problem 26PPS

Increase by 60%

Explanation of Solution

Given :

A cylinder has a radius of 5 centimeters and a height of 8 centimeters.

Calculation:

The volume of a cylinder with radius r and height h is :

  $V=\pi r^{2}h=\pi 5^{2}8=200\pi$

The dimensions are exchanged , $r_1=h$ , $h_1=r$ , so , the new volume:

  $V_1=\pi r_1{}^2{h}_1=\pi h^{2}r=\pi 8^{2}5=320\pi$

The volume is increased by :

  $\Delta V=\frac{320\pi -200\pi }{200\pi }\times 100\%=\frac{120}{2}\%=60\%$

So, the volume is increased by 60% , if the dimensions are exchanged.