Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

To sketch the solid whose volume is given by the iterated integral and evaluate it.

The value of the given iterated integral is \(\frac{{16\pi }}{3}\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The region \(D\) is \(\left\{ {(r,\theta ,z)\mid 0 \le r \le 2,0 \le \theta \le 2\pi ,0 \le z \le r} \right\}\).

Step 2: Formula used of Integration

Integration \(\mathop \smallint \nolimits^b f(x)dx = f(b) - f(a)\)

Step 3: Draw the sketch

Since \(r\) varies between \(0\)and \(2\) and \(\theta \) varies between \(0\) and \(2\pi \), the region is the circle in the \(xy\)-plane with radius \(2\) . \(z\)varies from 0 to \(r\). So, the outline sketch of the given region \(D\) is given below in the Figure 1.

Step 4: Integrate the integral with respect to \(z\)

Integrate the given iterated integral with respect to \(z\) and apply the limit of it. \(\begin{array}{c}\int_0^2 {\int_0^{2\pi } {\int_0^r r } } dzd\theta dr &=& \int_0^2 {\int_0^{2\pi } {(rz)} _0^rd\theta } dr\\ &=& \int_0^2 {\int_0^{2\pi } {(r(r) - r(0))} _0^rd\theta } dr\\ &=& \int_0^2 {\int_0^{2\pi } {\left( {{r^2} - 0} \right)} } d\theta dr\\ &=& \int_0^{2\pi } {\int_0^{{r^2}} d } \theta dr\end{array}\)

Step 5: Integrate the integral with respect to \(r\)

Integrate the given iterated integral with respect to \(r\) and apply the limit of it.

\(\begin{array}{c}\int_0^2 {\int_0^{2\pi } {\int_0^r r } } dzd\theta dr &=& \int_0^2 {\left( {{r^2}\theta } \right)_0^{2\pi }} dr\\ &=& \int_0^2 {\left( {{r^2}(2\pi ) - {r^2}(0)} \right)} dr\\ &=& \int_0^2 {\left( {2\pi {r^2} - 0} \right)} dr\\ &=& \int_0^2 2 \pi {r^2}dr\end{array}\)

Step 6: Integrate the integral with respect to \(\theta \)

Integrate the given iterated integral with respect to \(\theta \) and apply the limit of it.

\(\begin{array}{c}\int_0^2 {\int_0^{2\pi } {\int_0^r r } } dzd\theta dr &=& \left( {2\pi \frac{{{r^3}}}{3}} \right)_0^2\\ &=& 2\pi \left( {\frac{{{2^3}}}{3} - \frac{{{0^3}}}{3}} \right)\\ &=& 2\pi \left( {\frac{8}{3} - 0} \right)\\ &=& \frac{{16\pi }}{3}\end{array}\)

Thus, the value of the given iterated integral is \(\frac{{16\pi }}{3}\).

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

To sketch the solid whose volume is given by the iterated integral and evaluate it.

Step by Step Solution

Step 1: Given data

Step 2: Formula used of Integration

Step 3: Draw the sketch

Step 4: Integrate the integral with respect to \(z\)

Step 5: Integrate the integral with respect to \(r\)

Step 6: Integrate the integral with respect to \(\theta \)

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

To sketch the solid whose volume is given by the iterated integral and evaluate it.

Step by Step Solution

Step 1: Given data

Step 2: Formula used of Integration

Step 3: Draw the sketch

Step 4: Integrate the integral with respect to \(z\)

Step 5: Integrate the integral with respect to \(r\)

Step 6: Integrate the integral with respect to \(\theta \)

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