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Found in: Page 730

Essential Calculus: Early Transcendentals

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

Set up, but do not evaluate, integral expressions for The mass The center of mass, and 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 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$${{\rm{I}}_{\rm{z}}}{\rm{ = }}\int_{{\rm{ - 1}}}^{\rm{1}} {\int_{{\rm{ - }}\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} }^{\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}} } {\int_{\rm{0}}^{\sqrt {{\rm{1 - }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}} } {\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right)} } } \sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}} {\rm{dzdydx}}$$.