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Apply Green's Theoren to evaluate the integral. C6x+y) dx + (3xy"+4y) dy ter

Apply Green's Theoren to evaluate the integral. C6x+y) dx + (3xy"+4y) dy ter

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Applying Green's Theorem to Evaluate an Integral** Task: Apply Green's Theorem to evaluate the integral. \[ \oint_{C} (6x + y^3) \, dx + (3xy^2 + 4y) \, dy \] **C**: Any simple closed curve in the plane for which Green's Theorem holds. Answers: - (a) 2 - (b) 0 - (c) -2 - (d) There is not enough information **Explanation:** Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the plane region \( D \) bounded by \( C \). The theorem states: \[ \oint_{C} (L \, dx + M \, dy) = \iint_{D} \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) \, dA \] Where \( L = 6x + y^3 \) and \( M = 3xy^2 + 4y \). To solve the problem, compute: \[ \frac{\partial M}{\partial x} = \frac{\partial}{\partial x}(3xy^2 + 4y) = 3y^2 \] \[ \frac{\partial L}{\partial y} = \frac{\partial}{\partial y}(6x + y^3) = 3y^2 \] Thus, \[ \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} = 3y^2 - 3y^2 = 0 \] Therefore, the double integral over \( D \) is zero: \[ \iint_{D} \left( 0 \right) \, dA = 0 \] Hence, the value of the line integral is **0**, corresponding to answer (b).
Transcribed Image Text:**Applying Green's Theorem to Evaluate an Integral** Task: Apply Green's Theorem to evaluate the integral. \[ \oint_{C} (6x + y^3) \, dx + (3xy^2 + 4y) \, dy \] **C**: Any simple closed curve in the plane for which Green's Theorem holds. Answers: - (a) 2 - (b) 0 - (c) -2 - (d) There is not enough information **Explanation:** Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the plane region \( D \) bounded by \( C \). The theorem states: \[ \oint_{C} (L \, dx + M \, dy) = \iint_{D} \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) \, dA \] Where \( L = 6x + y^3 \) and \( M = 3xy^2 + 4y \). To solve the problem, compute: \[ \frac{\partial M}{\partial x} = \frac{\partial}{\partial x}(3xy^2 + 4y) = 3y^2 \] \[ \frac{\partial L}{\partial y} = \frac{\partial}{\partial y}(6x + y^3) = 3y^2 \] Thus, \[ \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} = 3y^2 - 3y^2 = 0 \] Therefore, the double integral over \( D \) is zero: \[ \iint_{D} \left( 0 \right) \, dA = 0 \] Hence, the value of the line integral is **0**, corresponding to answer (b).
Expert Solution
Step 1

Given integration is

∮c(6x+y3)dx+(3xy2+4y)dy

C : Any simple curve in the plane for which Green's theorem holds.

Green's theorem states that,

∫CP dx + Q dy = ∫∫R∂Q∂x-∂P∂ydx dy --- (1)

Step 2

Comparing the given integral with the Green theorem we get,

P=6x+3y3Q=3xy2+4y--- (2)

Differentiating P partially with respect to y we get,

∂P∂y=3y2 --- (3)

Differentiating Q partially with respect to x we get,

∂Q∂x=3y2 ---- (4)

 

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