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Chapter 14: Q. 19 (page 1119)

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

Short Answer

Expert verified

$Therefor the required normal vector is = i - f_{y} j - f_{z} k .$

Step by step solution

01

Step 1. Given

f(y, z) is a function with continuous partial derivatives.

02

Step 2. Find ry and rz

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

03

Step 3. Find n=rn×ry

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

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Most popular questions from this chapter

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

$\int r (t) d t, where r (t) = e^{t} i + t^{3} j - 4 k$

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.

Find

$\begin{aligned} \int_{S} F (x, y, z) \cdot n d S if \\ F (x, y, z) = \frac{\ln (x^{2} + y^{2} + 1)}{z + 3} i + \frac{y}{y + 1} j + e^{z^{2}} k \end{aligned}$

Where S is the portion of the sphere with radius 2, centered at the origin, and that lies below the plane with equation $z = - \sqrt{2}$ , with n pointing outwards.

$F = <x \ln (x z), 5 z, \frac{1}{y^{2} + 1}>$ , where S is the region of the plane with equation $12 x - 9 y + 3 z = 10$ , where $2 \leq x \leq 3$ and $5 \leq y \leq 10$ , with n pointing upwards.

Make a chart of all the new notation, definitions, and theorems in this section, including what each new item means in terms you already understand.

See all solutions

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