Use either form of Green's Theorem to evaluate the following line integral. $9x²1² 6 C $9x³y² dx + 6x³y dy = C (Type an exact answer.) dx + 6x³y dy; C is the triangle with vertices (0,0), (2,0), and (0,3) with counterclockwise orientation ***
Use either form of Green's Theorem to evaluate the following line integral. $9x²1² 6 C $9x³y² dx + 6x³y dy = C (Type an exact answer.) dx + 6x³y dy; C is the triangle with vertices (0,0), (2,0), and (0,3) with counterclockwise orientation ***
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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## Evaluating Line Integrals using Green's Theorem
In this exercise, we will use Green's Theorem to evaluate the following line integral:
\[
\oint_{C} \left(9x^2 y \, dx + 6x^3 y \, dy\right)
\]
where \(C\) is the triangle with vertices \((0,0)\), \((2,0)\), and \((0,3)\) oriented counterclockwise.
### Step-by-step Solution
The line integral can be evaluated using Green's Theorem, which states that:
\[
\oint_{C} \left(P \, dx + Q \, dy\right) = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA
\]
where \(P(x, y) = 9x^2 y\) and \(Q(x, y) = 6x^3 y\).
To apply Green's Theorem, we need to find the partial derivatives:
1. Compute \(\frac{\partial Q}{\partial x}\):
\[
\frac{\partial}{\partial x} (6x^3 y) = 18x^2 y
\]
2. Compute \(\frac{\partial P}{\partial y}\):
\[
\frac{\partial}{\partial y} (9x^2 y) = 9x^2
\]
Thus, the integrand for the double integral becomes:
\[
\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 18x^2 y - 9x^2 = 9x^2 (2y - 1)
\]
### Mapping the Region R
Region \(R\) is a right triangle with vertices \((0,0)\), \((2,0)\), and \((0,3)\). The bounds for \(x\) and \(y\) need to be set accordingly. The equation of the line connecting \((0,3)\) and \((2,0)\) is:
\[
y = 3 - \frac{3}{2}x
\]
This line sets the upper bound for \(y\) for any \(x\) in \([0,2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F304238c1-892d-4f66-90f4-8d5405bfbe58%2F83f193ae-ea06-461d-a6ea-83f0139a5ec1%2F98ql5ee_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
## Evaluating Line Integrals using Green's Theorem
In this exercise, we will use Green's Theorem to evaluate the following line integral:
\[
\oint_{C} \left(9x^2 y \, dx + 6x^3 y \, dy\right)
\]
where \(C\) is the triangle with vertices \((0,0)\), \((2,0)\), and \((0,3)\) oriented counterclockwise.
### Step-by-step Solution
The line integral can be evaluated using Green's Theorem, which states that:
\[
\oint_{C} \left(P \, dx + Q \, dy\right) = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA
\]
where \(P(x, y) = 9x^2 y\) and \(Q(x, y) = 6x^3 y\).
To apply Green's Theorem, we need to find the partial derivatives:
1. Compute \(\frac{\partial Q}{\partial x}\):
\[
\frac{\partial}{\partial x} (6x^3 y) = 18x^2 y
\]
2. Compute \(\frac{\partial P}{\partial y}\):
\[
\frac{\partial}{\partial y} (9x^2 y) = 9x^2
\]
Thus, the integrand for the double integral becomes:
\[
\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 18x^2 y - 9x^2 = 9x^2 (2y - 1)
\]
### Mapping the Region R
Region \(R\) is a right triangle with vertices \((0,0)\), \((2,0)\), and \((0,3)\). The bounds for \(x\) and \(y\) need to be set accordingly. The equation of the line connecting \((0,3)\) and \((2,0)\) is:
\[
y = 3 - \frac{3}{2}x
\]
This line sets the upper bound for \(y\) for any \(x\) in \([0,2
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