Let W be the region between the spheres $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 1$ and $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4$ .Given that $\int _ { W } \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { 1 / 2 } d V = 15 \pi$ , evaluate the integral $\int _ { w } \left( 64 x ^ { 2 } + 36 y ^ { 2 } + 144 z ^ { 2 } \right) ^ { 1 / 2 } d V$ , where $\bar{W}$ is the region between the ellipsoids $\frac { x ^ { 2 } } { 3 ^ { 2 } } + \frac { y ^ { 2 } } { 4 ^ { 2 } } + \frac { z ^ { 2 } } { 2 ^ { 2 } } = 1$ and $\frac { x ^ { 2 } } { 3 ^ { 2 } } + \frac { y ^ { 2 } } { 4 ^ { 2 } } + \frac { z ^ { 2 } } { 2 ^ { 2 } } = 4$ .
A) $8640 \pi$
B) $2160 \pi$
C) $24 \pi$
D) $360 \pi$
E) $360 \pi ^ { 3 }$

Question

Quizplus · Accepted Answer

The given integral over the region $\bar{W}$ can be transformed using the scaling factors from the ellipsoids. The transformation involves scaling $x$, $y$, and $z$ by 3, 4, and 2 respectively. The integral $\int _ { W } \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) ^ { 1 / 2 } d V = 15 \pi$ is scaled by the volume factor $3 \times 4 \times 2 = 24$ and the integrand scales by the factor $\sqrt{64 \times 36 \times 144} = 864$. Thus, the integral evaluates to $15 \pi \times 24 \times 864 = 8640 \pi$.

Let W Be the Region Between the Spheres x2+y2+z2=1x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 1x2+y2+z2=1

Let W Be the Region Between the Spheres $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 1$