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Essential Calculus: Early Transcendentals
Found in: Page 730
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Illustration

Short Answer

Set up, but do not evaluate, integral expressions for

  1. The mass
  2. The centre of mass, and
  3. The moment of inertia about the z-axis.

The solid of exercise 19; \({\rm{\rho (x,y,z) = }}\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \).

  1. The mass of the solid is \(m = \int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - y}}} {\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} } } } {\rm{dzdydx}}\).
  2. Therefore, the centres of the mass are:

\(\begin{aligned}\rm m &= \int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - y}}} {\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} } } } {\rm{\;dz\;dy\;dx}}\\\rm\bar x &= \frac{{\rm{1}}}{{\rm{m}}}\int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - y}}} {\rm{x}} } } \sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} {\rm{\;dz\;dy\;dx}}\\\rm\bar y &= \frac{{\rm{1}}}{{\rm{m}}}\int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - y}}} {\rm{y}} } } \sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} {\rm{\;dz\;dy\;dx}}\\\rm\bar z &= \frac{{\rm{1}}}{{\rm{m}}}\int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - y}}} {\rm{z}} } } \sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} {\rm{\;dz\;dy\;dx}}\end{aligned}\)

  1. The moment of inertia about the z-axis is \({{\rm{I}}_{\rm{z}}}{\rm{ = }}\int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - y}}} {{{\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right)}^{{\rm{3/2}}}}} } } {\rm{dzdydx}}\).
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Concept Introduction

Triple integrals are the three-dimensional equivalents of double integrals. They're a way to add up an unlimited number of minuscule quantities connected with points in a three-dimensional space.

Step 2: Find the mass

Exercising is a great way to become in shape.

The region bounded by \({\rm{z = 0,y = }}{{\rm{x}}^{\rm{2}}}{\rm{ and y + z = 1}}\) is called the solid.

Along the line \({\rm{y = 1}}\), the planes \({\rm{y + z = 1and z = 0}}\) intersect at the \({\rm{xy}}\)plane.

We can define the region by looking at the diagram. E

\(\left\{ {{\rm{(x,y,z)}} \in {\rm{E}}\mid \;\;\;{\rm{0}} \le {\rm{z}} \le {\rm{1 - y,}}\;\;\;{{\rm{x}}^{\rm{2}}} \le {\rm{y}} \le {\rm{1,}}\;{\rm{ - 1}} \le {\rm{x}} \le {\rm{1}}} \right\}\)

The solid's mass is

\(\text{m=}\iiint_{\text{E}}{\text{ }\!\!\rho\!\!\text{ }}\text{(x,y,z)dV=}\int_{\text{-1}}^{\text{1}}{\int_{{{\text{x}}^{\text{2}}}}^{\text{1}}{\int_{\text{0}}^{\text{1-y}}{\sqrt{{{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}}}}}\text{dzdydx}\)

We are not to judge the situation because it wants us not to.

Therefore, the mass of the solid is \(m = \int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - y}}} {\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} } } } {\rm{dzdydx}}\).

Step 3: Find the centre of mass

Exercising is a great way to become in shape.

The region bounded by \({\rm{z = 0,y = }}{{\rm{x}}^{\rm{2}}}{\rm{ and y + z = 1}}\) is called the solid.

Along the line\({\rm{y = 1}}\), the planes \({\rm{y + z = 1and z = 0}}\)intersect at the \({\rm{xy}}\)plane.

We can define the region by looking at the diagram. E

The coordinates of the centre of mass are calculated as follows:

\(\begin{aligned}& \bar{x}=\frac{1}{m}\iiint_{E}{x}\rho (x,y,z)dV=\frac{1}{m}\int_{-1}^{1}{\int_{{{x}^{2}}}^{1}{\int_{0}^{1-y}{x}}}\sqrt{{{x}^{2}}+{{y}^{2}}}dzdydx \\ & \bar{y}=\frac{1}{m}\iiint_{E}{y}\rho (x,y,z)dV=\frac{1}{m}\int_{1}^{1}{\int_{{{x}^{2}}}^{1}{\int_{0}^{1-y}{y}}}\sqrt{{{x}^{2}}+{{y}^{2}}}dzdydx \\ & \bar{z}=\frac{1}{m}\iiint_{E}{z}\rho (x,y,z)dV=\frac{1}{m}\int_{-1}^{1}{\int_{{{x}^{2}}}^{1}{\int_{0}^{1-y}{z}}}\sqrt{{{x}^{2}}+{{y}^{2}}}dzdydx \\\end{aligned}\)

Where

\(\text{m=}\iiint_{\text{E}}{\text{ }\!\!\rho\!\!\text{ }}\text{(x,y,z)dV=}\int_{\text{-1}}^{\text{1}}{\int_{{{\text{x}}^{\text{2}}}}^{\text{1}}{\int_{\text{0}}^{\text{1-y}}{\sqrt{{{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}}}}}\text{dzdydx}\)

The problem instructs us to refrain from evaluating the integrals.

Therefore, the centres of the mass are

\(\begin{aligned}\rm m &= \int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - y}}} {\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} } } } {\rm{\;dz\;dy\;dx}}\\\rm\bar x &= \frac{{\rm{1}}}{{\rm{m}}}\int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - y}}} {\rm{x}} } } \sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} {\rm{\;dz\;dy\;dx}}\\\rm\bar y &= \frac{{\rm{1}}}{{\rm{m}}}\int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - y}}} {\rm{y}} } } \sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} {\rm{\;dz\;dy\;dx}}\\\rm\bar z &= \frac{{\rm{1}}}{{\rm{m}}}\int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - y}}} {\rm{z}} } } \sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} {\rm{\;dz\;dy\;dx}}\end{aligned}\)

Step 4: Find the moment of inertia about the z-axis

Exercising is a great way to become in shape.

The region bounded by \({\rm{z = 0,y = }}{{\rm{x}}^{\rm{2}}}{\rm{ and y + z = 1}}\) is called the solid.

Along the line\({\rm{y = 1}}\), the planes \({\rm{y + z = 1and z = 0}}\) intersect at the \({\rm{xy}}\) plane.

We can define the region by looking at the diagram. E

The moment of inertia about the z-axis is

\(\begin{aligned} {I_z} &= \iiint_E {\left( {{x^2} + {y^2}} \right)}\rho (x,y,z)dV \\ &= \int_{ - 1}^1 {\int_{{x^2}}^1 {\int_0^{1 - y} {\left( {{x^2} + {y^2}} \right)} } } \sqrt {{x^2} + {y^2}} dzdydx{\text{ }} \\ &= \int_{ - 1}^1 {\int_{{x^2}}^1 {\int_0^{1 - y} {{{\left( {{x^2} + {y^2}} \right)}^{3/2}}} } } dzdydx \\ \end{aligned} \)

We are not to judge the situation because it wants us not to.

Therefore, the moment of inertia about the z-axis is \({{\rm{I}}_{\rm{z}}}{\rm{ = }}\int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{{\rm{x}}^{\rm{2}}}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{1 - y}}} {{{\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right)}^{{\rm{3/2}}}}} } } {\rm{dzdydx}}\).

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