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Q. 13

Expert-verified

Calculus

Found in: Page 1140

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Why is $d A$ in Green’s Theorem replaced by $d S$ in Stokes’ Theorem?

$d A$ refers to ordinary area while $d S$ refers to the surface area. So, $d A$ in Green's Theorem is replaced by $d S$ in Stokes' Theorem because .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information

$d A$ in Green’s Theorem replaced by $d S$ in Stokes’ Theorem.

Step 2. Understanding both the Theorems

Green's Theorem states that;"Let $R$ be a region in the plane with smooth boundary curve $C$ oriented counterclockwise by $r (t) = ⟨ (x (t), y (t)) ⟩$ for $a \leq t \leq b$ .If a vector field $F (x, y) = <F_{1} (x, y), F_{2} (x, y)>$ is defined by $R$ , then, $\int_{C} F \cdot d r = \iint_{R} (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) d A . "$ Stokes' Theorem states that;"Let be $S$ an oriented, smooth or piecewise-smooth surface bounded by a curve $C$ . Suppose that $n$ is an oriented unit normal vector of $S$ and $C$ has a parametrization that traverses $C$ in the counterclockwise direction with respect to $n$ .If a vector field $F (x, y, z) = F_{1} (x, y, z) i + F_{2} (x, y, z) j + F_{3} (x, y, z) k$ is defined on $S$ , then, $\int_{C} F (x, y, z) \cdot d r = \iint_{S} curl F (x, y, z) \cdot n d S "$ .

Step 3. Understanding the notations

Both the notations $d A$ and $d S$ refer to area.In the case of Green's Theorem, $d A$ refers to the ordinary area.In the case of Stokes' Theorem, $d S$ refers to the surface area. In this case, $d S$ keeps track of any distortion to the area of the surface $S$ caused by transforming a patch of the $u v -$ plane to parametrize $S$ . Hence, $d A$ in Green's Theorem can be replaced by $d S$ in Stokes' Theorem.

Most popular questions for Math Textbooks

The current through a certain passage of the San Juan Islands in Washington State is given by

$F = <F_{1} (x, y), F_{2} (x, y)> = <0, 1.152 - 0.8 x^{2}> .$

Consider a disk R of radius 1 mile and centered on this region. Denote the boundary of the disk by localid="1650455155947" $\partial R$ .

(a) Compute localid="1650455166518" $\iint_{R} (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) d A$ .
(b) Show that $\int_{\partial R} F \cdot n d s = \int_{0}^{2 π} (1.152 - 0.8 \cos^{2} θ) \sin θ d θ .$ Conclude that Green's Theorem is valid for the current in this area of the San Juan Islands.
(c) What do the integrals from Green's Theorem tell us about this region of the San Juan Islands?

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

Find the area of S is the portion of the plane with equation y−z = $\frac{π}{2}$

that lies above the rectangle determined by 0 ≤ x ≤ 4 and 3 ≤ y ≤ 6.

Find the work done by the vector field

$F (x, y) = x^{3} y^{2} i + (y - x) j$
in moving an object around the triangle with vertices $(1, 1), (2, 2)$ , and $(3, 1)$ , starting and ending at $(2, 2)$ .

Let R be a simply connected region in the xy-plane. Show that the portion of the paraboloid with equation $z = x^{2} + y^{2}$ determined by R has the same area as the portion of the saddle with equation $z = x^{2} - y^{2}$ determined by R.

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