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Chapter 14: Q. 20 (page 1119)

Give a formula for a normal vector to the surface S determined by y = g(x,z), where g(x,z) is a function with continuous partial derivatives.

Short Answer

Expert verified

$Therefore the required normal vector is = - g_{x} i + j - g_{z} k .$

Step by step solution

01

Step 1. Given

g(x, z) is a function with continuous partial derivatives.

02

Step 2. Find rx and rz

$B y a b o v e s t e p s, t h e s u r f a c e S i s p a r a m e t r i z e d b y a s f o l l o w s r (x, z) = <g (x, z), x, z> T h e n \begin{aligned} r_{x} = \frac{\partial}{\partial y} r (x, z) \\ = \frac{\partial}{\partial y} <g (x, z), x, z> \\ = <\frac{\partial g}{\partial y}, 1, 0> \\ = <g_{y}, 1, 0> \end{aligned} a n d \begin{aligned} r_{z} = \frac{\partial}{\partial z} r (y, z) \\ = \frac{\partial}{\partial z} <g (x, z), x, z> \\ = <\frac{\partial g}{\partial z}, 0, 1> \\ = <g_{z}, 0, 1> \end{aligned}$

03

Step 3. Find n=rz×rx.

$\begin{aligned} n = r_{z} \times r_{x} \\ = <0, g_{z}, 1> \times <1, g_{x}, 0> \\ = |\begin{array}{ccc} i & j & k \\ 0 & g_{z} & 1 \\ 1 & g_{x} & 0 \end{array}| \\ = [g_{z} \cdot 0 - 1 \cdot g_{x}] i - [0 \cdot 0 - 1 \cdot 1] j + [0 \cdot g_{x} - g_{z} \cdot 1] k \\ = - g_{x} i + j - g_{z} k . \end{aligned} Therefore the required normal vector is = - g_{x} i + j - g_{z} k .$

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Most popular questions from this chapter

Calculus of vector-valued functions: Calculate each of the following.

$\frac{d}{d t} (r (t)), where r (t) = 3 \cos^{2} t i + 5 t j + \frac{t}{t^{2} + 1} k$

Suppose that an electric field is given by

$E = 2 y i + 2 x y j + y z k$

Compute the flux $\int_{S} E \cdot n d A$ of the field through the unit cube $[0, 1] \times [0, 1] \times [0, 1]$ .

Compute dS for your parametrization in Exercise 7.

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$Use the vector field F (x, y) = x^{2} e^{y} i + \cos (x) \sin (y) j and Green ’ s Theorem to write the line integral of F (x, y) about the unit circle, traversed counterclockwise, as a double integral . Do not evaluate the integral .$

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