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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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Short Answer

Set up, but do not evaluate, integral expressions for

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

The hemisphere .

  1. The mass of the solid is\(\int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{\rm{ - }}\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} }^{\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} } {\int_{\rm{0}}^{\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}} } {\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}} } } } {\rm{dzdydx}}\).

  2. The centres of the mass are:

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

  3. The moment of inertia about the z-axis is\(\int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{\rm{ - }}\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} }^{\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} } {\int_{\rm{0}}^{\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}} } {\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right)} } } \sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}} {\rm{dzdydx}}\).

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 

  1. For the upper hemisphere, the equation is

\({\rm{z = }}\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}} \)

The region inside the sphere can be defined as

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

The solid's mass is

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

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

Step 3: Find the centre of mass

  1. For the upper hemisphere, the equation is

\({\rm{z = }}\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}} \)

The region inside the sphere can be defined as

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

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

Where

The problem instructs us to refrain from evaluating the integrals.

Therefore, the centres of the mass are

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

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

  1. For the upper hemisphere, the equation is

\({\rm{z = }}\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}} \)

The region inside the sphere can be defined as

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

The moment of inertia about the z-axis is

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

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

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