Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

$F (x, y, z) = i + j + k$ , where S is the lower half of the unit sphere, with n pointing outwards.

The required flux of the vector field through the surface S is $2 π$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

Consider the given question,

$F (x, y, z) = i + j + k$

Step 2. Find the Flux of Fx,y,zthough an oriented surface S.

If a surface S is the graph of $z = z (x, y)$ , then the Flux of $F (x, y, z)$ through S is given below,

$\begin{aligned} \int_{S} F (x, y, z) \cdot n d S = \iint_{D} (F (x, y, z) \cdot n) ∥z_{x} \times z_{y}∥ d A \\ = \iint_{D} (F (x, y, z) \cdot n) \sqrt{{(\frac{\partial z}{\partial x})}^{2} + {(\frac{\partial z}{\partial y})}^{2} + 1} d A \dots . . . (i) \end{aligned}$

Here, the surface S is the lower half of the unit sphere, so its equation will be,

$z = - \sqrt{1 - x^{2} - y^{2}}$

Now first find $\frac{\partial z}{\partial x}$ . The first partial derivates of z are given below,

$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} (- \sqrt{1 - x^{2} - y^{2}}) = \frac{x}{\sqrt{1 - x^{2} - y^{2}}}$

Step 3. Find the value of ∂z∂x2+∂z∂y2+1.

On finding the value of $\frac{\partial z}{\partial y}$ ,

role="math" localid="1650342662824" $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y} (- \sqrt{1 - x^{2} - y^{2}}) = \frac{y}{\sqrt{1 - x^{2} - y^{2}}}$

Then,

$\begin{aligned} \sqrt{{(\frac{\partial z}{\partial x})}^{2} + {(\frac{\partial z}{\partial y})}^{2} + 1} = \sqrt{{(\frac{x}{\sqrt{1 - x^{2} - y^{2}}})}^{2} + {(\frac{y}{\sqrt{1 - x^{2} - y^{2}}})}^{2} + 1} \\ = \sqrt{\frac{x^{2}}{1 - x^{2} - y^{2}} + \frac{y^{2}}{1 - x^{2} - y^{2}} + 1} \\ = \frac{\sqrt{1 - x^{2} - y^{2}}}{\sqrt{1}} \end{aligned}$

Step 4. Find the choice of n.

The choice of n should have pointing upwards. Then the following vector, $v = <- \frac{\partial z}{\partial x}, - \frac{\partial z}{\partial y}, 1>$ is normal to the surface.

Now, for $z = - \sqrt{1 - x^{2} - y^{2}}$ , gives the following vector perpendicular to this surface,

$\begin{aligned} v = <- \frac{\hat{\partial z}}{\partial x}, - \frac{\partial z}{\partial y}, 1> \\ = <- \frac{y}{\sqrt{1 - x^{2} - y^{2}}}, - \frac{y}{\sqrt{1 - x^{2} - y^{2}}}, 1 \end{aligned}$

Then the desired normal vector will be,

$\begin{aligned} n = \frac{1}{∥ <- \frac{x}{\sqrt{1 - x^{2} - y^{2}}}, - \frac{y}{\sqrt{1 - x^{2} - y^{2}}}, 1}> <- \frac{x}{\sqrt{1 - x^{2} - y^{2}}}, - \frac{y}{\sqrt{1 - x^{2} - y^{2}}}, 1> \\ = \frac{1}{\sqrt{{(- \frac{x}{\sqrt{1 - x^{2} - y^{2}}})}^{2} + {(- \frac{y}{\sqrt{1 - x^{2} - y^{2}}})}^{2} + 1^{2}}} <- \frac{x}{\sqrt{1 - x^{2} - y^{2}}}, - \frac{y}{\sqrt{1 - x^{2} - y^{2}}}, 1> \\ = \sqrt{1 - x^{2} - y^{2}} <- \frac{x}{\sqrt{1 - x^{2} - y^{2}}}, - \frac{y}{\sqrt{1 - x^{2} - y^{2}}}, 1> \end{aligned}$

Step 5. Find the value of F(x,y,z)⋅n.

The value of $F (x, y, z) \cdot n$ will be,

$\begin{aligned} F (x, y, z) \cdot n = ⟨ 1, 1, 1 ⟩ \cdot \sqrt{1 - x^{2} - y^{2}} <- \frac{x}{\sqrt{1 - x^{2} - y^{2}}}, - \frac{y}{\sqrt{1 - x^{2} - y^{2}}}, 1> \\ = \sqrt{1 - x^{2} - y^{2}} ⟨ 1, 1, 1 ⟩ \cdot <- \frac{x}{\sqrt{1 - x^{2} - y^{2}}}, - \frac{y}{\sqrt{1 - x^{2} - y^{2}}}, 1> \\ = \sqrt{1 - x^{2} - y^{2}} (1 - \frac{x}{\sqrt{1 - x^{2} - y^{2}}}) + 1 (- \frac{y}{\sqrt{1 - x^{2} - y^{2}}}) + 1 \cdot 1) \\ = \sqrt{1 - x^{2} - y^{2}} (- \frac{x}{\sqrt{1 - x^{2} - y^{2}}} - \frac{y}{\sqrt{1 - x^{2} - y^{2}}} + 1) \\ = \sqrt{1 - x^{2} - y^{2}} (1 - \frac{x + y}{\sqrt{1 - x^{2} - y^{2}}}) \end{aligned}$

Step 6. Find the polar coordinates.

Substituting these values in equation (i),

$\begin{aligned} \int_{S} F (x, y, z) \cdot n d S = \iint_{D} (F (x, y, z) \cdot n) (\sqrt{{(\frac{\partial z}{\partial x})}^{2} + {(\frac{\partial z}{\partial y})}^{2} + 1}) d A \\ = \iint_{D} (\sqrt{1 - x^{2} - y^{2}} (1 - \frac{x + y}{\sqrt{1 - x^{2} - y^{2}}})) (\frac{1}{\sqrt{1 - x^{2} - y^{2}}}) d A \\ = \iint_{D} (1 - \frac{x + y}{\sqrt{1 - x^{2} - y^{2}}}) d A \dots \dots . (i i) \end{aligned}$

The surface S is the lower half of the unit sphere, so the region of integration D is the unit disk in the xy-plane and centered at the origin.

Hence, in polar coordinates, the region of integration will be,

$D = {(r, θ) ∣ 0 \leq r \leq 1, 0 \leq θ \leq 2 π}$

In this case,

localid="1650343222895" $\begin{aligned} x = r \cos θ, \\ y = r \sin θ \end{aligned}$

Step 7. Using equation (ii), write the flux of the vector field through the surface S.

Using equation (ii), the flux of the vector field through the surface S,

$\begin{aligned} \int_{S} F (x, y, z) \cdot n d S = \iint_{D} (1 - \frac{x + y}{\sqrt{1 - x^{2} - y^{2}}}) d A \\ = \int_{0}^{2 π} \int_{0}^{1} (1 - \frac{r \cos θ + r \sin θ}{\sqrt{1 - r^{2}}}) r d r d θ \\ = \int_{0}^{2 π} [\int_{0}^{1} 1 d r - (\cos θ + \sin θ) \int_{0}^{1} \frac{r^{2}}{\sqrt{1 - r^{2}}} d r] d θ \\ = \int_{0}^{2 π} [[1 - 0] - (\cos θ + \sin θ) [\frac{1}{2} (- 1 \sqrt{1 - 1^{2}} + \sin^{- 1} 1) - \frac{1}{2} (- 0 \sqrt{1 - 0^{2}} + \sin^{- 1} 0)]] d θ \\ \begin{aligned} \begin{array}{r} = \int_{0}^{2 π} [1 - \frac{π}{4} (\cos θ + \sin θ)] d θ \\ = {[θ - \frac{π}{4} (\sin θ - \cos θ)]}_{0}^{2 π} \\ = (2 π - \frac{π}{4} (\sin 2 π - \cos 2 π) - (0 - \frac{π}{4} (\sin 0 - \cos 0)) \end{array} \\ = 2 π \end{aligned} \end{aligned}$

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Calculus

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Short Answer

$F (x, y, z) = i + j + k$ , where S is the lower half of the unit sphere, with n pointing outwards.

Step by Step Solution

Step 1. Given information.

Step 2. Find the Flux of Fx,y,zthough an oriented surface S.

Step 3. Find the value of ∂z∂x2+∂z∂y2+1.

Step 4. Find the choice of n.

Step 5. Find the value of F(x,y,z)⋅n.

Step 6. Find the polar coordinates.

Step 7. Using equation (ii), write the flux of the vector field through the surface S.

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Calculus

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Short Answer

F(x,y,z)=i+j+k, where S is the lower half of the unit sphere, with n pointing outwards.

Step by Step Solution

Step 1. Given information.

Step 2. Find the Flux of Fx,y,zthough an oriented surface S.

Step 3. Find the value of ∂z∂x2+∂z∂y2+1.

Step 4. Find the choice of n.

Step 5. Find the value of F(x,y,z)⋅n.

Step 6. Find the polar coordinates.

Step 7. Using equation (ii), write the flux of the vector field through the surface S.

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$F (x, y, z) = i + j + k$ , where S is the lower half of the unit sphere, with n pointing outwards.