Verifying Greens Theorem in the Plane for a Specific Line Integral

Green's theorem is a fundamental result in vector calculus which relates a line integral around a simple closed curve to a double integral over the plane region bounded by that curve. This article will walk through the steps to verify Green's theorem for the given line integral and demonstrate how to compute it using this theorem.

Introduction

Green's theorem allows us to convert a line integral around a simple closed curve C into a double integral over a planar region. The line integral we will verify is:

$$int_{C} left(x^2 - 2y^3 , dx 3xy , dyright)$$

The curve C is the boundary of the region enclosed by the circles x2 y2 1 and x2 y2 4. This circle problem is a great example to illustrate the power of Green's theorem in simplifying computations.

Main Steps

Step 1: Identify the Functions P and Q

First, we identify the functions P and Q from the line integral:

$$P x^2 - 2y^3, quad Q 3xy$$

Step 2: Compute the Partial Derivatives

Next, we compute the partial derivatives ?Q/?x and ?P/?y:

For Q 3xy:

$$frac{partial Q}{partial x} 3y$$

For P x^2 - 2y^3:

$$frac{partial P}{partial y} -6y^2$$

Step 3: Compute the Difference of the Derivatives

We then compute the difference:

$$frac{partial Q}{partial x} - frac{partial P}{partial y} 3y - (-6y^2) 3y 6y^2$$

Step 4: Set Up the Double Integral

We need to integrate this expression over the region R bounded by the circles x2 y2 1 and x2 y2 4. It is convenient to use polar coordinates for this region.

Using the polar coordinate transformation, where x rcos(θ) and y rsin(θ), the region R is defined by:

$$1 leq r leq 2, quad 0 leq θ leq 2π$$

Step 5: Rewrite the Integrand in Polar Coordinates

Substituting y rsin(θ) into the expression:

$$3y 6y^2 3rsin(θ) 6r^2sin^2(θ)$$

Step 6: Set Up the Double Integral in Polar Coordinates

The double integral becomes:

$$iint_{R} 3rsin(θ) 6r^2sin^2(θ) , dA int_{0}^{2π} int_{1}^{2} 3rsin(θ) 6r^2sin^2(θ) cdot r , dr , dθ$$

This simplifies to:

$$int_{0}^{2π} int_{1}^{2} 3r^2sin(θ) 6r^3sin^2(θ) , dr , dθ$$

Step 7: Evaluate the Inner Integral

First, evaluate the inner integral with respect to r from 1 to 2:

For 3r2sin(θ):

$$int_{1}^{2} 3r^2 , dr 3 left[ frac{r^3}{3} right]_{1}^{2} [r^3]_{1}^{2} 8 - 1 7$$

For 6r3sin2(θ):

$$int_{1}^{2} 6r^3 , dr 6 left[ frac{r^4}{4} right]_{1}^{2} 6 left[ frac{16}{4} - frac{1}{4} right] 6 left[ 4 - frac{1}{4} right] 6 left[ frac{15}{4} right] frac{90}{4} 22.5$$

Step 8: Combine and Evaluate the Outer Integral

Now, substitute back into the integral and evaluate the outer integral with respect to θ from 0 to 2π:

$$int_{0}^{2π} left( 7sin(θ) - 22.5sin^2(θ) right) , dθ$$

The integral of sin(θ) over a full period is zero:

$$int_{0}^{2π} sin(θ) , dθ 0$$

For the second term, use the identity sin2(θ) (1 - cos(2θ))/2:

$$int_{0}^{2π} sin^2(θ) , dθ frac{1}{2} int_{0}^{2π} 1 - cos(2θ) , dθ frac{1}{2} left[ 2π - 0 right] π$$

Hence:

$$-22.5 int_{0}^{2π} sin^2(θ) , dθ -22.5 cdot π -22.5π$$

Conclusion

Thus, the value of the double integral computed using Green's theorem is:

$$iint_{R} left(frac{partial Q}{partial x} - frac{partial P}{partial y}right) , dA -22.5π$$

According to Green's theorem, this should equal the value of the line integral:

$$int_{C} left(x^2 - 2y^3 , dx 3xy , dyright) -22.5π$$

This verifies Green's theorem for this specific case.