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Q. 46

Expert-verified

Calculus

Found in: Page 1120

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Find $\int_{S} 1 d S$ , where S is the portion of the surface with equation $x = e^{y z} - e^{- y z}$ that lies on the positive side of the circle of radius 3 and centered at the origin in the yz-plane.

Ans: The required surface area is $\int_{0}^{2 π} \int_{0}^{3} (\sqrt{r^{2} (e^{r^{2} \sin 2 θ} + e^{- r^{2} \sin 2 θ} + 2) + 1}) r d r d θ$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

given,

$x = e^{y z} - e^{- y z}$

Step 2. Consider the following surface:

The surface S is the portion of the surface determined by $y x = e^{y z} - e^{- y z}$ that lies on the positive side of the circle of radius 3 and is centered at the origin in the yz-plane.
The objective is to find the value of $\int_{S} 1 d S$ . The formula for Finding the Area of a Surface $x = f (y, z)$ : If a surface S is given by $x = f (y, z) for f (y, z) \in D \subseteq ℝ^{2}$ , then the surface area of the smooth surface is,

role="math" localid="1650475395157" $\begin{aligned} \int_{S} 1 d S = \int_{S} \sqrt{{(\frac{\partial x}{\partial y})}^{2} + {(\frac{\partial x}{\partial z})}^{2} + 1} d A \\ = \iint_{D} \sqrt{{(\frac{\partial x}{\partial y})}^{2} + {(\frac{\partial x}{\partial z})}^{2} + 1 d A} . . . (1) \end{aligned}$

Step 3. Here, the surface S is the portion of the following surface;

$x = e^{y z} - e^{- y z}$

Now, first, find $\frac{\partial x}{\partial y} and \frac{\partial x}{\partial z}$ . The partial derivative of x is,

$\begin{aligned} \frac{\partial x}{\partial y} = \frac{\partial}{\partial y} (e^{y z} - e^{- y z}) \\ = z (e^{y z} + e^{- y z}) \end{aligned}$

and,

$\begin{aligned} \frac{\partial x}{\partial z} = \frac{\partial}{\partial z} (e^{y z} - e^{- y z}) \\ = y (e^{y z} + e^{- y z}) \end{aligned}$

Step 4. Now,

The surface S lies on the positive side of the circle of radius 3 and is centered at the origin in the -plane. The region of integration D is the disk $y^{2} + z^{2} = 9$ is shown in the following figure.
In polar coordinates, the region of integration is described as follows,

$D = {(r, θ) ∣ 0 \leq r \leq 3, 0 \leq θ \leq 2 π} .$

In this case,

$y = r \cos θ, z = r \sin θ, y^{2} + z^{2} = r^{2}, a n d d A = r d r d θ .$

Step 5. Then,

Substitute $\frac{\partial x}{\partial y} = z (e^{y z} + e^{- y z}) and \frac{\partial x}{\partial z} = y (e^{j z} + e^{- y z})$ into formula (1), then the area of the surface is,

localid="1650476425232" $\begin{aligned} \int_{S} d S = \iint_{D} (\sqrt{{(\frac{\partial x}{\partial y})}^{2} + {(\frac{\partial x}{\partial z})}^{2} + 1}) d A \\ = \iint_{D} (\sqrt{{(z (e^{y z} + e^{- y z}))}^{2} + {(y (e^{y z} + e^{- y z}))}^{2} + 1}) d A \\ = \iint_{D} (\sqrt{z^{2} {(e^{y z} + e^{- y z})}^{2} + y^{2} {(e^{y z} + e^{- y z})}^{2} + 1}) d A \\ = \iint_{D} (\sqrt{(y^{2} + z^{2}) {(e^{y z} + e^{- y z})}^{2} + 1}) d A \\ = \int_{0}^{2 π} \int_{0}^{3} (\sqrt{r^{2} {(e^{r \cos θ \sin θ} + e^{- r \cos θ r \sin θ})}^{2} + 1}) r d r d θ \\ = \int_{0}^{2 z} \int_{0}^{3} (\sqrt{r^{2} {(e^{\frac{1}{2} r^{2} \sin 2 θ} + e^{- \frac{1}{2} r^{2} \sin 2 θ})}^{2} + 1}) r d r d θ \\ = \int_{0}^{2 π} \int_{0}^{3} (\sqrt{r^{2} (e^{r^{2} \sin 2 θ} + e^{- r^{2} \sin 2 θ} + 2) + 1}) r d r d θ \end{aligned}$

Therefore, the required surface area is the above integral.

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