The volume of a right circular cone with radius r and height h is V = pir^2h/3.
a. Approximate the change in the volume of the cone when the radius changes from r to r and the height changes from h = 4.00 to h = 3.96.
b. Approximate the change in the volume of the cone when the radius changes from r to r and the height changes from h = 10.0 to h = 9.92.
a. The approximate change in volume is dV = _______. (Type an integer or decimal rounded to two decimal places as needed.)
b. The approximate change in volume is dV = ___________ (Type an integer or decimal rounded to two decimal places as needed.)

The question is incomplete. The complete question is :

The volume of a right circular cone with radius r and height h is V = pir^2h/3. a. Approximate the change in the volume of the cone when the radius changes from r = 5.9 to r = 6.8 and the height changes from h = 4.00 to h = 3.96.

b. Approximate the change in the volume of the cone when the radius changes from r = 6.47 to r = 6.45 and the height changes from h = 10.0 to h = 9.92.

a. The approximate change in volume is dV = _______. (Type an integer or decimal rounded to two decimal places as needed.)

b. The approximate change in volume is dV = ___________ (Type an integer or decimal rounded to two decimal places as needed.)

Solution :

Given :

The volume of the right circular cone with a radius r and height h is

[tex]$V=\frac{1}{3} \pi r^2 h$[/tex]

[tex]$dV = d\left(\frac{1}{3} \pi r^2 h\right)$[/tex]

[tex]$dV = \frac{1}{3} \pi h \times d(r^2)+\frac{1}{3} \pi r^2 dh$[/tex]

[tex]$dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$[/tex]

a). The radius is changed from r = 5.9 to r = 6.8 and the height is changed from h = 4 to h = 3.96

So, r = 5.9  and dr = 6.8 - 5.9 = 0.9

     h = 4  and dh = 3.96 - 4 = -0.04

Now, [tex]$dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$[/tex]

[tex]$dV = \frac{2}{3} \pi (5.9)(4)(0.9)+\frac{1}{3} \pi (5.9)^2 (-0.04)$[/tex]

[tex]$dV=44.484951 - 1.458117$[/tex]

[tex]$dV=43.03$[/tex]

Therefore, the approximate change in volume is dV = 43.03 cubic units.

b).  The radius is changed from r = 6.47 to r = 6.45 and the height is changed from h = 10 to h = 9.92

So, r = 6.47  and dr = 6.45 - 6.47 = -0.02

     h = 10  and dh = 9.92 - 10 = -0.08

Now, [tex]$dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$[/tex]

[tex]$dV = \frac{2}{3} \pi (6.47)(10)(-0.02)+\frac{1}{3} \pi (6.47)^2 (-0.08)$[/tex]

[tex]$dV=-2.710147-3.506930$[/tex]

[tex]$dV= -6.22$[/tex]

Hence, the approximate change in volume is dV = -6.22 cubic units

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