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Applying Green's Theorem to Evaluate Line Integrals: A Case Study

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In this article, we will delve into the application of Green's Theorem in evaluating line integrals. Green's Theorem is a powerful tool in vector calculus, providing a way to convert a line integral over a closed curve into a double integral over the region enclosed by the curve. This article will guide you through the process using a specific example.

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Understanding the Context

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Consider a line integral of the form:

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( oint_C 3x^2y , dx x 2y , dy ),

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where ( C ) is a closed curve oriented in the counterclockwise direction. To evaluate this line integral, we will apply Green's Theorem, which states:

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( oint_C (P , dx Q , dy) iint_D (frac{partial Q}{partial x} - frac{partial P}{partial y}) , dA ),

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where ( D ) is the region enclosed by ( C ).

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Combining the Given Expressions Under Green's Theorem

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Let ( P 3x^2y ) and ( Q x 2y ). According to Green's Theorem, we can write:

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( oint_C 3x^2y , dx (x 2y) , dy iint_R left( frac{partial Q}{partial x} - frac{partial P}{partial y} right) , dA ).

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Calculating the partial derivatives, we get:

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( frac{partial Q}{partial x} 1 ) and ( frac{partial P}{partial y} 3x^2 ).

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Therefore, the expression inside the double integral becomes:

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( 1 - 3x^2 ).

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Determining the Region of Integration

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The region ( R ) is enclosed by the curve ( C ). To determine the limits of integration, we need to find the points where the line ( y 0 ) intersects the curve ( y 1 - x^2 ).

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Solving ( 0 1 - x^2 ) gives us ( x pm 1 ). The parabola ( y 1 - x^2 ) is above the line ( y 0 ) between ( x -1 ) and ( x 1 ).

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Thus, the region ( R ) can be described as ( y ) ranging from ( 0 ) to ( 1 - x^2 ) for ( x ) in the interval ([-1, 1]).

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Evaluating the Double Integral

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Using Fubini's Theorem, we can write the double integral as:

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( iint_R (1 - 3x^2) , dA int_{-1}^1 int_0^{1 - x^2} (1 - 3x^2) , dy , dx ).

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Evaluating the inner integral with respect to ( y ):

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( int_0^{1 - x^2} (1 - 3x^2) , dy (1 - 3x^2) int_0^{1 - x^2} dy (1 - 3x^2)(1 - x^2) 1 - 4x^2 3x^4 ).

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Now, integrating this with respect to ( x ):

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( int_{-1}^1 (1 - 4x^2 3x^4) , dx ).

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Note that the integrand is even, so we can simplify the integral:

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( 2 int_0^1 (1 - 4x^2 3x^4) , dx ).

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Evaluating this integral step by step:

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( int_0^1 1 , dx - 4 int_0^1 x^2 , dx 3 int_0^1 x^4 , dx ).

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( left[ x right]_0^1 - 4 left[ frac{x^3}{3} right]_0^1 3 left[ frac{x^5}{5} right]_0^1 ).

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( (1 - 0) - 4 left( frac{1}{3} - 0 right) 3 left( frac{1}{5} - 0 right) ).

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( 1 - frac{4}{3} frac{3}{5} ).

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Combining these terms:

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( frac{15}{15} - frac{20}{15} frac{9}{15} frac{4}{15} ).

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Multiplying by 2 due to the even integrand:

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( 2 times frac{4}{15} frac{8}{15} ).

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Therefore, the value of the line integral is:

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( boxed{frac{8}{15}} ).