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Chapter 6: Q14P (page 335)

$∮ (yi - xj + zk) - dr$ around the circumference of the circle of radius $2$ , center at the origin, in the $(x, y)$ plane.

Short Answer

Expert verified

The solution derived is $I = - 8 π .$

Step by step solution

01

Given Information.

The given expression is $∮ (yi - xj + zk) - dr$

02

Definition of vector.

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines its magnitude.

03

Apply Stokes’s theorem.

Apply Stokes' theorem and the fact that $\nabla \times V = - 2 \hat{k}, Z = 0$ and then the question can be solved as shown below.

$I = - 2 \int_{σ} dσ$

$= 2$ (Area of circle)

$= - 2 (4) π$

$= - 8 π$

Hence, the solution derived is $I = - 8 π$ .

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Most popular questions from this chapter

Question: $\int \int curl (x^{2} yi - xzk) \cdot ndσ$ over the closed surface of the ellipsoid

. $\frac{x^{2}}{4} + \frac{y^{2}}{9} + \frac{z^{2}}{16} = 1$

Warning: Stokes’ theorem applies only to an open surface. Hints: Could you cut the given surface into two halves? Also see (d) in the table of vector identities (page 339).

Find vector fields $A$ such that role="math" localid="1657346627450" $V = curl A$ for each givenrole="math" localid="1657346639484" $V .$

$\iint V \times ndσ$ over the entire surface of the volume in the first octant bounded by $x^{2} + y^{2} + z^{2} = 16$ and the coordinate planes, where $V = (x + x^{2} - y^{2}) i + (2 xyz - 2 xy) j - {xz}^{2} k$

If A and B are the diagonals of a parallelogram, find a vector formula for the area of the parallelogram.

$H i n t : I n t e g r a t e (g)$ Derive the following vector integral theorems

(a) $\int_{v o l u m e τ} \nabla ϕ d τ = \oint_{\underset{i n c l o s i n g τ}{s u r f a c e}} ϕ n d σ$

Hint: In the divergence theorem (10.17), substitute $V = ϕC$ where is an arbitrary constant vector, to obtain $C \cdot \int \nabla ϕ d τ = C \cdot \oint ϕ n d σ$ Since C is arbitrary, let C=i to show that the x components of the two integrals are equal; similarly, let C=j and C=k to show that the y components are equal and the z components are equal.

(b) $\int_{volume τ} \nabla \times V d τ = \oint_{\underset{inclosing τ}{surface}} n \times V d σ$

Hint: Replace in the divergence theorem by where is an arbitrary constant vector. Follow the last part of the hint in (a).

(c) localid="1659323284980" $\int_{\underset{b o u n d i n g σ}{c u r v e}} ϕ d r = \oint_{s u r f a c e} σ (n \times \nabla ϕ) d σ .$

(d) $\int_{\underset{boundingσ}{curve}} ϕ d r \times V = \oint_{surface} (n \times \nabla) \times V d σ$

Hints for (c) and (d): Use the substitutions suggested in (a) and (b) but in Stokes' theorem (11.9) instead of the divergence theorem.

(e) $\int_{volume τ} \nabla ϕ d τ = \oint_{\underset{inclosing τ}{surface}} ϕ V \cdot n d σ - \oint_{\underset{inclosing τ}{surface}} ϕ V \cdot \nabla ϕ n d τ .$

Hint: Integrate (7.6) over volume and use the divergence theorem.

(f) localid="1659324199695" $\int_{volume τ} V \cdot (\nabla \times \cup) d τ = \int_{volume τ} V \cdot (\nabla \times \cup) d τ + \oint_{\underset{inclosing τ}{surface}} (\cup \times V) \cdot n d σ$

Hint: Integrate (h) in the Table of Vector Identities (page 339) and use the divergence theorem.

(g) $\int_{s u r f a c e o f σ} ϕ (\nabla \times V) \cdot n d σ = \int_{s u r f a c e o f σ} (\nabla \times \nabla ϕ) \cdot n d σ + \oint_{\underset{b o u n d i n g}{c u r v e}} ϕ V \cdot d r$

$H i n t : I n t e g r a t e (g)$ in the Table of Vector Identities (page 339) and use Stokes' Theorem.

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