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




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, 



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, 



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