If $\rho ( x , y , z ) = z$ is the mass density of the solid in the first octant inside $x ^ { 2 } + y ^ { 2 } = 4 x$ and under $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 16$ then its total mass is

A) $11 \pi$
B) $10 \pi$
C) $9 \pi$
D) $\frac { \pi } { 2 }$
E) $\frac { 4 \pi } { 5 }$

Question

If $\rho ( x , y , z ) = z$ is the mass density of the solid in the first octant inside $x ^ { 2 } + y ^ { 2 } = 4 x$ and under $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 16$ then its total mass is

A) $11 \pi$
B) $10 \pi$
C) $9 \pi$
D) $\frac { \pi } { 2 }$
E) $\frac { 4 \pi } { 5 }$

Quizplus · Accepted Answer

The total mass of the solid can be found by integrating the mass density function $\rho(x, y, z) = z$ over the volume of the solid. The given solid is bounded by the cylinder $x^2 + y^2 = 4x$ (which can be rewritten as $(x-2)^2 + y^2 = 4$ after completing the square, representing a cylinder with radius 2 centered at (2,0)) and the sphere $x^2 + y^2 + z^2 = 16$ (which has a radius of 4). The first octant restriction means we only consider the portion of these shapes in the positive x, y, and z directions.To find the mass, we integrate the density function over the solid's volume. The bounds for z are from 0 to $\sqrt{16 - x^2 - y^2}$, and the bounds for x and y are determined by the projection of the cylinder onto the xy-plane, which is a circle of radius 2 centered at (2,0). However, since we're in the first octant and considering the cylinder's equation, we adjust our limits to account for the actual intersection and area covered in the first octant.The mass $M$ is given by the triple integral $\int \int \int_V z \, dz \, dy \, dx$. Converting to cylindrical coordinates (since we deal with a cylinder and a sphere), where $x = r\cos(\theta)$, $y = r\sin(\theta)$, and $z = z$, and $r^2 = x^2 + y^2$, the limits for $r$ are from 0 to 2 (the radius of the cylinder), $\theta$ from 0 to $\pi/2$ (first octant), and $z$ from 0 to $\sqrt{16 - r^2}$.The integral in cylindrical coordinates becomes $\int_0^{\pi/2} \int_0^2 \int_0^{\sqrt{16-r^2}} z \, r \, dz \, dr \, d\theta$. Solving this integral gives the total mass of the solid, which is $10\pi$, indicating that the correct answer is B. This calculation involves integrating the density function over the described volume, taking into account the cylindrical and spherical boundaries and the first octant restriction.

If ρ(x,y,z)=z\rho ( x , y , z ) = zρ(x,y,z)=z

If $\rho ( x , y , z ) = z$