Solve the problem.
-Find the center of mass of the region of constant density bounded by the paraboloid $z = 49 - x ^ { 2 } - y ^ { 2 }$ and the $x y$-plane.
A) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 49 } { 3 }$
B) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 49 } { 2 }$
C) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 7 } { 3 }$
D) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 7 } { 2 }$

Question

Quizplus · Accepted Answer

The center of mass $\bar{x}$ and $\bar{y}$ are 0 due to symmetry. For $\bar{z}$, use the volume integral of $z$ over the solid divided by the volume of the solid. The volume is found by integrating the area of circular slices, leading to $\bar{z} = \frac{1}{V} \int z dV = \frac{49}{3}$.

Solve the Problem z=49−x2−y2z = 49 - x ^ { 2 } - y ^ { 2 }z=49−x2−y2

Solve the Problem $z = 49 - x ^ { 2 } - y ^ { 2 }$