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Chapter 6: Q1P (page 322)

Evaluate each of the following integrals in the easiest way you can.
$\oint (2 y d x - 3 x d y)$ around the square bounded by $x = 3, x = 5, y = 1 andy = 3$

Short Answer

Expert verified

The solution to this problem is /=-20.

Step by step solution

01

Given Information.

The given information is $\oint (2 y d x - 3 x d y)$

02

Definition of Green’s Theorem.

The Green's theorem connects a line integral around a simple closed curve C to a double integral over the plane region D circumscribed by C in vector calculus. Stokes' theorem has a two-dimensional special case.

03

Find the solution.

Write the values of P and Q.

P = 2y

Q = - 3x

Find the difference of their differentiation

$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = - 3 - 2 = - 5$

The integral is in the form mentioned below.

$I = - 5 \int_{1}^{3} \int_{3}^{5} d x d y = - 5 \times 4 = - 20$

Hence, the solution to this problem is / = - 20.

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Most popular questions from this chapter

$H i n t : I n t e g r a t e (g)$ 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.

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).

(a) Suppose that a hill (as in Fig. 5.1) has the equation $32 - x^{2} - 4 y^{2}$ , where $z = height measured from some refrence level$ (in hundreds of feet). Sketch acontour map (that is, draw on one graph a set of curves $z =$ const.); use the contours $z = 32, 19, 12, 7, 0$ (b) If you start at the point $(3, 2)$ and in the direction $i + j$ , are you going up hillor downhill, and how fast?

Evaluate $\iint V \cdot n d σ$ over the curved surface of the hemisphere $x^{2} + y^{2} + z^{2} = 9, z \geq 0$ , if $V = yi + xzj + (2 z - 1) k$ .Careful: See Problem 9.

$\iint r \times ndσ$ over the entire surface of the hemisphere $x^{2} + y^{2} + z^{2} = 9$ , $z \geq 0$

where $r = xi + yj + zk$ .

See all solutions

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