A tetrahedron solid is formed by x + 2y + z = 4 in the first octant. This solid has a density...

Question:

A tetrahedron solid is formed by {eq}x + 2y + z = 4{/eq} in the first octant. This solid has a density function {eq}\rho (x,y,z) = y +1{/eq}. Set up the tiple integral for its MASS using the differential {eq}dV{/eq} given below:

{eq}\iiint \rho (x,y,z) dydxdz{/eq}

Find the center of mass {eq}(\overline{x} , \overline{y}, \overline{z}){/eq} using this order of differentials for each moment of inertia.

Center of Mass:

The center of mass of any solid {eq}E {/eq} in {eq}\mathbb{R}^3 {/eq} is denoted by the point {eq}(\bar{x}, \bar{y}, \bar{z}) {/eq}.

The coordinates of its center of mass are related to its mass {eq}M {/eq} and its density function {eq}\rho(x,y,z) {/eq} such that:

$$M = \iiint \limits_E \rho(x,y,z) \,dV \\ \bar{x} = \frac{1}{M} \iiint \limits_E x \,\rho(x,y,z) \,dV \\ \bar{y} = \frac{1}{M} \iiint \limits_E y \,\rho(x,y,z) \,dV \\ \bar{z} = \frac{1}{M} \iiint \limits_E z \,\rho(x,y,z) \,dV $$

Answer and Explanation:

Given a tetrahedron solid formed by the plane {eq}x + 2y + z = 4 {/eq} in the first octant. The solid's density function is given as {eq}\rho(x,y,z) = y + 1 {/eq}.

Let {eq}D {/eq} be the region on the xz-plane which is the projection of the tetrahedron solid onto the said plane such that it is bounded by the surface {eq}y = \frac{4 - x - z}{2} {/eq}.

Graph of the Tetrahedron

Graph of the Tetrahedron

We let the region {eq}D {/eq} to be on the xz-plane so that the order of integration will first be with respect to y.

As such, determine the equation of the intersection of the plane with the xz-plane {eq}(y = 0) {/eq}:

{eq}\begin{align*} &y = \frac{4 - x - z}{2} = 0 \\ &\Rightarrow 4 - x - z = 0 \\ &\Rightarrow x + z = 4 \end{align*} {/eq}

Graph of the Region D on the XZ-Plane

Graph of the Region D on the XZ-Plane

For the order of integration to be with respect to z the last, the region {eq}D {/eq} must be defined in terms of z. If so, based on the graph, the region can be described by the inequalities {eq}0 \le z \le 4 \text{ and } 0 \le x \le 4 - z {/eq}.

Therefore, the tetrahedron can be defined as {eq}\left\{\, (x,y) \,|\, 0 \le z \le 4 \,,\, 0 \le x \le 4 - z \text{ and } 0 \le y \le \frac{4 - x - z}{2} \,\right\} {/eq}.


MASS OF THE TETRAHEDRON

{eq}\begin{align*} M &= \boxed{ \int_0^4 \int_0^{4-z} \int_0^\frac{4-x-z}{2} (y + 1) \,dy \,dx \,dz } & \text{[Integrate with respect to } y \text{]} \\ &= \int_0^4 \int_0^{4-z} \left. \left( \frac{y^2}{2} + y \right) \right|_0^\frac{4-x-z}{2} \,dx \,dz & \text{[Fundamental Theorem of Calculus]} \\ &= \int_0^4 \int_0^{4-z} \left( \left[ \frac{1}{2} \left( \frac{4-x-z}{2} \right)^2 + \frac{4-x-z}{2} \right] - 0 \right) \,dx \,dz \\ &= \int_0^4 \int_0^{4-z} \left( \frac{2xz - 8x + x^2 - 8z + z^2 + 16}{8} + \frac{16 - 4x - 4z}{8} \right) \,dx \,dz \\ &= \int_0^4 \int_0^{4-z} \left[ \frac{x^2}{8} + \frac{x}{4} ( z - 6 ) + \left( \frac{z^2}{8} - \frac{3z}{2} + 4 \right) \right] \,dx \,dz & \text{[Integrate with respect to } x \text{]} \\ &= \int_0^4 \left. \left[ \frac{x^3}{24} + \frac{x^2}{8} (z - 6) + x \left( \frac{z^2}{8} - \frac{3z}{2} + 4 \right) \right] \right|_0^{4-z} \,dz & \text{[Fundamental Theorem of Calculus]} \\ &= \int_0^4 \left( \left[ \frac{(4-z)^3}{24} + \frac{(4-z)^2}{8} (z - 6) + (4-z) \left( \frac{z^2}{8} - \frac{3z}{2} + 4 \right) \right] - 0 \right) \,dz \\ &= \int_0^4 \left( - \frac{z^3}{24} + \frac{3z^2}{4} - 4z + \frac{20}{3} \right) \,dz & \text{[Integrate with respect to } z \text{]} \\ &= \left. \left( - \frac{z^4}{96} + \frac{z^3}{4} - 2z^2 + \frac{20z}{3} \right) \right|_0^4 & \text{[Fundamental Theorem of Calculus]} \\ &= \left[ - \frac{4^4}{96} + \frac{4^3}{4} - 2(4)^2 + \frac{20(4)}{3} \right] - 0 & \text{[Simplify]} \\ &= - \frac{8}{3} + 16 - 32 + \frac{80}{3} \\ M &= \boxed{ 8 } & \boxed{ \text{Mass of the Tetrahedron} } \end{align*} {/eq}


CENTER OF MASS

Using its mass, the coordinates of the center of mass of the tetrahedron are:

{eq}\begin{align*} \bar{x} &= \frac{1}{M} \int_0^4 \int_0^{4-z} \int_0^\frac{4-x-z}{2} x (y + 1) \,dy \,dx \,dz & \left[ \text{For all constant } k \,,\, \int k \cdot f(x) \,dx = k \int f(x)\,dx \right] \\ &= \frac{1}{8} \int_0^4 \int_0^{4-z} x \int_0^\frac{4-x-z}{2} (y + 1) \,dy \,dx \,dz & \left[ \text{Recall that } \int_0^\frac{4-x-z}{2} (y + 1) \,dy = \frac{x^2}{8} + \frac{x}{4} ( z - 6 ) + \left( \frac{z^2}{8} - \frac{3z}{2} + 4 \right) \right] \\ &= \frac{1}{8} \int_0^4 \int_0^{4-z} x \left[ \frac{x^2}{8} + \frac{x}{4} ( z - 6 ) + \left( \frac{z^2}{8} - \frac{3z}{2} + 4 \right) \right] \,dx \,dz \\ &= \frac{1}{8} \int_0^4 \int_0^{4-z} \left[ \frac{x^3}{8} + \frac{x^2}{4} ( z - 6 ) + x \left( \frac{z^2}{8} - \frac{3z}{2} + 4 \right) \right] \,dx \,dz & \text{[Integrate with respect to } x \text{]} \\ &= \frac{1}{8} \int_0^4 \left. \left[ \frac{x^4}{32} + \frac{x^3}{12} (z - 6) + \frac{x^2}{2} \left( \frac{z^2}{8} - \frac{3z}{2} + 4 \right) \right] \right|_0^{4-z} \,dz & \text{[Fundamental Theorem of Calculus]} \\ &= \frac{1}{8} \int_0^4 \left( \left[ \frac{(4-z)^4}{32} + \frac{(4-z)^3}{12} (z - 6) + \frac{(4-z)^2}{2} \left( \frac{z^2}{8} - \frac{3z}{2} + 4 \right) \right] - 0 \right) \,dz \\ &= \frac{1}{8} \int_0^4 \left( \frac{z^4}{96} - \frac{z^3}{4} + 2z^2 - \frac{20z}{3} + 8 \right) \,dz & \text{[Integrate with respect to } z \text{]} \\ &= \frac{1}{8} \left. \left( \frac{z^5}{480} - \frac{z^4}{16} + \frac{2z^3}{3} - \frac{10z^2}{3} + 8z \right) \right|_0^4 & \text{[Fundamental Theorem of Calculus]} \\ &= \frac{1}{8} \left( \left[ \frac{4^5}{480} - \frac{4^4}{16} + \frac{2(4)^3}{3} - \frac{10(4)^2}{3} + 8(4) \right] - 0 \right) & \text{[Simplify]} \\ &= \frac{1}{8} \left( \frac{32}{15} - 16 + \frac{128}{3} - \frac{160}{3} + 32 \right) \\ &= \frac{1}{8} \left( \frac{112}{15} \right) \\ \bar{x} &= \boxed{ \frac{14}{15} } & \boxed{ X \text{-Coordinate of the Tetrahedron's Center of Mass} } \\ \\ \bar{y} &= \frac{1}{M} \int_0^4 \int_0^{4-z} \int_0^\frac{4-x-z}{2} y (y + 1) \,dy \,dx \,dz \\ &= \frac{1}{8} \int_0^4 \int_0^{4-z} \int_0^\frac{4-x-z}{2} (y^2 + y) \,dy \,dx \,dz & \text{[Integrate with respect to } y \text{]} \\ &= \frac{1}{8} \int_0^4 \int_0^{4-z} \left. \left( \frac{y^3}{3} + \frac{y^2}{2} \right) \right|_0^\frac{4-x-z}{2} \,dx \,dz & \text{[Fundamental Theorem of Calculus]} \\ &= \frac{1}{8} \int_0^4 \int_0^{4-z} \left( \left[ \frac{1}{3} \left( \frac{4-x-z}{2} \right)^3 + \frac{1}{2} \left( \frac{4-x-z}{2} \right)^2 \right] - 0 \right) \,dx \,dz \\ &= \frac{1}{8} \int_0^4 \int_0^{4-z} \left[ -\frac{x^3}{24} + \frac{x^2}{8} (5 - z) + \frac{x}{8} (10z - z^2 - 24) + \left( - \frac{z^3}{24} + \frac{5z^2}{8} - 3z + \frac{14}{3} \right) \right] \,dx \,dz & \text{[Integrate with respect to } x \text{]} \\ &= \frac{1}{8} \int_0^4 \left. \left[ -\frac{x^4}{96} + \frac{x^3}{24} (5 - z) + \frac{x^2}{16} (10z - z^2 - 24) + x \left( - \frac{z^3}{24} + \frac{5z^2}{8} - 3z + \frac{14}{3} \right) \right] \right|_0^{4-z} \,dz & \text{[Fundamental Theorem of Calculus]} \\ &= \frac{1}{8} \int_0^4 \left( \left[ -\frac{(4-z)^4}{96} + \frac{(4-z)^3}{24} (5 - z) + \frac{(4-z)^2}{16} (10z - z^2 - 24) + (4-z) \left( - \frac{z^3}{24} + \frac{5z^2}{8} - 3z + \frac{14}{3} \right) \right] - 0 \right) \,dz \\ &= \frac{1}{8} \int_0^4 \left( \frac{z^4}{96} - \frac{5z^3}{24} + \frac{3z^2}{2} - \frac{14z}{3} + \frac{16}{3} \right) \,dz & \text{[Integrate with respect to } z \text{]} \\ &= \frac{1}{8} \left. \left( \frac{z^5}{480} - \frac{5z^4}{96} + \frac{z^3}{2} - \frac{7z^2}{3} + \frac{16z}{3} \right) \right|_0^4 & \text{[Fundamental Theorem of Calculus]} \\ &= \frac{1}{8} \left( \left[ \frac{4^5}{480} - \frac{5(4)^4}{96} + \frac{4^3}{2} - \frac{7(4)^2}{3} + \frac{16(4)}{3} \right] - 0 \right) & \text{[Simplify]} \\ &= \frac{1}{8} \left( \frac{32}{15} - \frac{40}{3} + 32 - \frac{112}{3} + \frac{64}{3} \right) \\ &= \frac{1}{8} \left( \frac{24}{5} \right) \\ \bar{y} &= \boxed{ \frac{3}{5} } & \boxed{ Y \text{-Coordinate of the Tetrahedron's Center of Mass} } \\ \\ \bar{z} &= \frac{1}{M} \int_0^4 \int_0^{4-z} \int_0^\frac{4-x-z}{2} z (y + 1) \,dy \,dx \,dz & \left[ \text{For all constant } k \,,\, \int k \cdot f(x) \,dx = k \int f(x)\,dx \right] \\ &= \frac{1}{8} \int_0^4 z \int_0^{4-z} \int_0^\frac{4-x-z}{2} (y + 1) \,dy \,dx \,dz & \left[ \text{Recall that } \int_0^{4-z} \int_0^\frac{4-x-z}{2} (y + 1) \,dy \,dx = - \frac{z^3}{24} + \frac{3z^2}{4} - 4z + \frac{20}{3} \right] \\ &= \frac{1}{8} \int_0^4 z \left( - \frac{z^3}{24} + \frac{3z^2}{4} - 4z + \frac{20}{3} \right) \,dz \\ &= \frac{1}{8} \int_0^4 \left( - \frac{z^4}{24} + \frac{3z^3}{4} - 4z^2 + \frac{20z}{3} \right) \,dz & \text{[Integrate with respect to } z \text{]} \\ &= \frac{1}{8} \left. \left( - \frac{z^5}{120} + \frac{3z^4}{16} - \frac{4z^3}{3} + \frac{10z^2}{3} \right) \right|_0^4 & \text{[Fundamental Theorem of Calculus]} \\ &= \frac{1}{8} \left( \left[ - \frac{4^5}{120} + \frac{3(4)^4}{16} - \frac{4(4)^3}{3} + \frac{10(4)^2}{3} \right] - 0 \right) & \text{[Simplify]} \\ &= \frac{1}{8} \left( - \frac{128}{15} + 48 - \frac{256}{3} + \frac{160}{3} \right) \\ &= \frac{1}{8} \left( \frac{112}{15} \right) \\ \bar{z} &= \boxed{ \frac{14}{15} } & \boxed{ Z \text{-Coordinate of the Tetrahedron's Center of Mass} } \end{align*} {/eq}

The center of mass of the given tetrahedron is located at the point {eq}\boxed{ (\bar{x}, \bar{y}, \bar{z} ) = \left( \frac{14}{15}, \frac{3}{5}, \frac{14}{15} \right) } {/eq}.


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Double Integrals & Evaluation by Iterated Integrals

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4
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