Which weighs more? For 0 ≤ r ≤ 1, the solid bounded bythe cone z = 4 - 4r and the solid bounded by the paraboloidz = 4 - 4r2 have the same base in the xy-plane and the sameheight. Which object has the greater mass if the density of bothobjects is ρ(r, θ, z) = 10 - 2z?

Answered: Which weighs more? For 0 ≤ r ≤ 1, the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Which weighs more? For 0 ≤ r ≤ 1, the solid bounded by
the cone z = 4 - 4r and the solid bounded by the paraboloid
z = 4 - 4r² have the same base in the xy-plane and the same
height. Which object has the greater mass if the density of both
objects is ρ(r, θ, z) = 10 - 2z?

Expert Solution

Step 1

Given: Solid bounded by cone $z = 4 - 4 r$ and the solid bounded by the paraboloid $z = 4 - 4 r^{2}$ have the same base and same height and density is $ρ (r, θ, z) = 10 - 2 z$ .

To Find: Which object has greater mass?

Step 2

Given the solid $E$ in cylindrical coordinate with desity $ρ (r, θ, z)$ defined by the inequalities.

$E = \{(r, θ, z) 0 \leq r \leq 1, 0 \leq θ \leq 2 π, 0 \leq z \leq F (r, θ)\}$

The mass of the solid is given by the triple integral.

$M = \int \int \int d m = \int \int \int ρ (r, θ, z) d V = \int_{θ = 0}^{θ = 2 π} \int_{r = 0}^{r = 1} \int_{0}^{F (r, θ)} ρ (r, θ, z) r d z d r d θ$

The mass of the solid under the cone is calculated as follows:

$\begin{array}{rcl} M_{c} & = & \int_{θ = 0}^{θ = 2 π} \int_{r = 0}^{r = 1} \int_{0}^{F (r, θ)} ρ (r, θ, z) r d z d r d θ \\ = & \int_{θ = 0}^{θ = 2 π} \int_{r = 0}^{r = 1} r \int_{0}^{4 - 4 r} (10 - 2 z) d z d r d θ \\ = & \int_{θ = 0}^{θ = 2 π} \int_{r = 0}^{r = 1} r {[10 z - z^{2}]}_{0}^{4 - 4 r} d r d θ \\ = & \int_{θ = 0}^{θ = 2 π} \int_{r = 0}^{r = 1} r [10 (4 - 4 r) - {(4 - 4 r)}^{2} - 0] d r d θ \\ = & \int_{θ = 0}^{θ = 2 π} \int_{r = 0}^{r = 1} [- 16 r^{3} - 8 r^{2} + 24 r] d r d θ \\ = & \int_{θ = 0}^{θ = 2 π} {[- 4 r^{4} - \frac{8}{3} r^{3} + 12 r]}_{0}^{1} d θ \\ = & \int_{θ = 0}^{θ = 2 π} [- 4 - \frac{8}{3} + 12 - 0] d θ \\ = & \int_{θ = 0}^{θ = 2 π} \frac{16}{3} d θ \\ = & \frac{16}{3} {[θ]}_{0}^{2 π} \\ = & \frac{32 π}{3} \end{array}$

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