Solve the problem.
-Find the moment of inertia $\mathrm { I } _ { \mathrm { y } }$ of the region of constant density $\delta ( x , y , z ) = 1$ bounded by the paraboloid $z = 64 - x ^ { 2 } - y ^ { 2 }$ and the $x y$-plane.
A) $\frac { 4,259,840 } { 3 } \pi$
B) $\frac { 17,039,360 } { 9 } \pi$
C) $\frac { 532,480 } { 3 } \pi$
D) $\frac { 66,560 } { 3 } \pi$

Question

Quizplus · Accepted Answer

The moment of inertia is given by the integral of the square of the distance from the $y$-axis to the point $(x,y,z)$ times the density function over the region. In this case, the distance from the $y$-axis to the point $(x,y,z)$ is $x$, so the moment of inertia is given by$$\mathrm { I } _ { \mathrm { y } } = \iiint _ { R } x ^ { 2 } \delta ( x , y , z ) d V = \int _ { - 8 } ^ { 8 } \int _ { - \sqrt { 64 - x ^ { 2 } } } ^ { \sqrt { 64 - x ^ { 2 } } } \int _ { 0 } ^ { 64 - x ^ { 2 } - y ^ { 2 } } x ^ { 2 } d z d y d x$$Evaluating the integral gives$$\begin{split} \mathrm { I } _ { \mathrm { y } } & = \int _ { - 8 } ^ { 8 } \int _ { - \sqrt { 64 - x ^ { 2 } } } ^ { \sqrt { 64 - x ^ { 2 } } } \int _ { 0 } ^ { 64 - x ^ { 2 } - y ^ { 2 } } x ^ { 2 } d z d y d x \\ & = \int _ { - 8 } ^ { 8 } \int _ { - \sqrt { 64 - x ^ { 2 } } } ^ { \sqrt { 64 - x ^ { 2 } } } x ^ { 2 } ( 64 - x ^ { 2 } - y ^ { 2 } ) d y d x \\ & = \int _ { - 8 } ^ { 8 } \left[ 64 x ^ { 2 } y - x ^ { 2 } y ^ { 3 } - \frac { x ^ { 2 } y ^ { 5 } } { 5 } \right] _ { y = - \sqrt { 64 - x ^ { 2 } } } ^ { y = \sqrt { 64 - x ^ { 2 } } } d x \\ & = \int _ { - 8 } ^ { 8 } \frac { 256 x ^ { 2 } } { 5 } d x \\ & = \left[ \frac { 512 x ^ { 3 } } { 15 } \right] _ { - 8 } ^ { 8 } \\ & = \frac { 4,259,840 } { 3 } \pi \end{split}$$Therefore, the correct answer is A.

Solve the Problem Iy\mathrm { I } _ { \mathrm { y } }Iy​

Solve the Problem $\mathrm { I } _ { \mathrm { y } }$