Solve the problem.
-The radius of a ball is claimed to be 4.5 inches, with a possible error of 0.05 inch. Use differentials to approximate the maximum possible error in calculating the volume of the sphere and the surface area of the
Sphere.
A) $8.1 \pi \mathrm { in } ^ { 3 } ; 3.6 \pi \mathrm { in } ^ { 2 }$
B) $0.9 \pi \mathrm { in } ^ { 3 } ; 18 \pi \mathrm { in. } ^ { 2 }$
C) $4.05 \pi$ in. $^ { 3 } ; 1.8 \pi$ in. $^ { 2 }$
D) $1.8 \pi$ in. $^ { 3 } ; 4.05 \pi$ in. $^ { 2 }$

Question

Solve the problem.
-The radius of a ball is claimed to be 4.5 inches, with a possible error of 0.05 inch. Use differentials to approximate the maximum possible error in calculating the volume of the sphere and the surface area of the
Sphere. 
A) $8.1 \pi \mathrm { in } ^ { 3 } ; 3.6 \pi \mathrm { in } ^ { 2 }$
B) $0.9 \pi \mathrm { in } ^ { 3 } ; 18 \pi \mathrm { in. } ^ { 2 }$
C) $4.05 \pi$ in. $^ { 3 } ; 1.8 \pi$ in. $^ { 2 }$
D) $1.8 \pi$ in. $^ { 3 } ; 4.05 \pi$ in. $^ { 2 }$

Quizplus · Accepted Answer

The differential for volume $dV = 4\pi r^2 dr$ gives $dV = 4\pi (4.5)^2 (0.05) = 4.05\pi$ in$^3$. The differential for surface area $dA = 8\pi r dr$ gives $dA = 8\pi (4.5)(0.05) = 1.8\pi$ in$^2$.

Solve the Problem 8.1πin3;3.6πin28.1 \pi \mathrm { in } ^ { 3 } ; 3.6 \pi \mathrm { in } ^ { 2 }8.1πin3;3.6πin2

Solve the Problem $8.1 \pi \mathrm { in } ^ { 3 } ; 3.6 \pi \mathrm { in } ^ { 2 }$