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Q. 34

Expert-verified

Calculus

Found in: Page 1119

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

$f (x, y, z) = <\frac{1}{x \ln (z + 2 y)}>$ , where S is the surface given by $r (u, v) = 2 u i + (u + v) j + (2 u - 2 v) k$ for $3 \leq u \leq 7$ and $2 \leq v \leq 10$ .

The required integral over the surface S is $24 \ln (\frac{\ln (28)}{\ln (12)})$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

Consider the given question,

$f (x, y, z) = <\frac{1}{x \ln (z + 2 y)}>, r (u, v) = 2 u i + (u + v) j + (2 u - 2 v) k$

Step 2. Find the integral of fx,y,z over a smooth surface S.

If a surface S is parametrized by $r (u, v)$ for $(u, v) \in D$ , then the integral of $f (x, y, z)$ over a smooth surface S is given below,

$\int_{S} f (x, y, z) d S = \iint {}_{D}f (x (u, v), y (u, v), z (u, v)) ||r_{u} \times r_{w}|| d A . . . . . . (i)$

Now, find $r_{u}, r_{v}$ ,

role="math" localid="1650303986567" $r_{u} = \frac{\partial}{\partial u} r (u, v) = \frac{\partial}{\partial u} (2 u i + (u + v) j + (2 u - 2 v) k) = 2 i + j + 2 k r_{v} = \frac{\partial}{\partial v} r (u, v) = \frac{\partial}{\partial v} (2 u i + (u + v) j + (2 u - 2 v) k) = 0 i + j - 2 k$

Step 3. Find ru×rv.

On solving $||r_{u} \times r_{v}||$ ,

$r_{u} \times r_{v} = (2 i + j + 2 k) \times (0 i + j - 2 k) = |\begin{matrix} i & j & k \\ 2 & 1 & 2 \\ 0 & 1 & - 2 \end{matrix}| = - 4 i + 4 j + 2 k$

Then, $||r_{u} \times r_{v}|| = ||- 4 i + 4 j + 2 k|| = \sqrt{{(- 4)}^{2} + 4^{2} + 2^{2}} = \sqrt{36} = 6$

Consider the following function, localid="1650304882776" $f (x, y, z)$ ,

localid="1650304885354" $f (x (u, v), y (u, v), z (u, v)) = \frac{1}{2 u \ln ((2 u - 2 v) + 2 (u + v))} = \frac{1}{2 u \ln (4 u)}$

Step 4. Consider the given function.

Substitute the values in equation (i),

$D = \{(u, v) 3 \leq u \leq 7, 2 \leq u \leq 10\}$

Take the integral of f over the surface S,

$\begin{array}{r} \int_{S} f (x, y, z) d S = \iint_{D} f (x (u, v), y (u, v), z (u, v)) ∥r_{u} \times r_{v}∥ d A \\ = \int_{2}^{10} \int_{3}^{7} [\frac{1}{2 u \ln (4 u)}) (6) d u d v \\ = 3 \int_{2}^{10} \int_{3}^{7} \frac{1}{u \ln (4 u)} d u d v \\ = 3 \int_{2}^{10} [\int_{3}^{7} \frac{1}{u \ln (4 u)} d u] d v \\ = 3 \int_{2}^{10} [\ln (\ln (4 u))]_{3}^{7} d v \end{array}$

Step 5. Continue solving the above equation.

On solving the above equation,

$\begin{aligned} \int_{S} f (x, y, z) d S = 3 \int_{2}^{10} [\ln (\ln (28)) - \ln (\ln (12))] d v \\ = 3 \int_{2}^{10} \ln (\frac{\ln (28)}{\ln (12)}) d v \\ = 3 \ln (\frac{\ln (28)}{\ln (12)}) \int_{2}^{10} d v \\ = 3 \ln (\frac{\ln (28)}{\ln (12)}) [v]_{2}^{10} \\ = 3 \ln (\frac{\ln (28)}{\ln (12)}) [10 - 2] \\ = 24 \ln (\frac{\ln (28)}{\ln (12)}) \end{aligned}$

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