Calculate this Integral around the Circular Path using Green's Theorem

Is my resolution correct? I can't identify.

Calculate the line integral ∮C(2ydx+3xdy), where C is the circle x2+y2=1, using the Green's Theorem.

Green's Theorem:

Let M and N be functions of two variables x and y, such that they have continuous first partial derivatives in an open disk B in R2. If C is a simple closed curve that is piecewise smooth and entirely contained in B, and if R is the region bounded by C, then:

$$
\oint_C (M \, dx + N \, dy) = \iint_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) \, dA
$$

Conditions of Green's Theorem:

Condition 1: M and N have continuous partial derivatives.

For the function M=2y and N=3x, their partial derivatives are continuous everywhere. This is because these functions are polynomials, and polynomials have continuous partial derivatives at all points in their domain.

Condition 2: C is a simple closed curve and is contained in a domain where M and N have continuous partial derivatives.

The circle x2+y2=1 is a simple closed curve that is smooth. Furthermore, it is entirely contained in the plane R2, where M and N have continuous partial derivatives. Therefore, this condition is satisfied.

Condition 3: R is the region bounded by C.

The region R is the unit disk x2+y2≤1, which is closed and bounded by the circle x2+y2=1. Thus, R is indeed the region bounded by C, satisfying the third condition.

Therefore, all conditions of Green's Theorem are satisfied, allowing us to apply the theorem to solve the line integral along the circle x2+y2=1.

Step 1: Identify M and N and verify if the conditions of Green's Theorem are satisfied:

For the line integral ∮C(2ydx+3xdy), we have:

M=2y and N=3x

Both functions have continuous partial derivatives everywhere, as they are polynomials. Therefore, the first condition of Green's Theorem is satisfied.

Step 2: Calculate the partial derivatives of(M and N:

$$
\frac{\partial M}{\partial y} = 2 \quad \text{and} \quad \frac{\partial N}{\partial x} = 3
$$

Step 3: Apply Green's Theorem:

According to Green's Theorem:

$$
\oint_C (2y \, dx + 3x \, dy) = \iint_R (3 - 2) \, dA = \iint_R 1 \, dA
$$

Step 4: Simplify the double integral:

$$
\iint_R 1 \, dA = \iint_R 1 \, dA
$$

Step 5: Calculate the Double Integral over R using polar coordinates:

To calculate the double integral, we use polar coordinates since R is the unit disk.

$$
\iint_R 1 \, dA = \int_{0}^{2\pi} \int_{0}^{1} r \, dr \, d\theta
= \int_{0}^{2\pi} \left[\frac{r^2}{2}\right]_{0}^{1} \, d\theta
= \int_{0}^{2\pi} \frac{1}{2} \, d\theta
= \frac{1}{2} \times 2\pi
= \pi
$$

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Did you have a question about this?

I always find it impossible to remember where the minus sign goes unless I use the general Stokes’ theorem instead (of which Green’s theorem is a special case). Stokes’ makes it easy:
$$
\oint_C (2y \, dx + 3x\, dy) = \int_\Omega d(2y \, dx + 3x\, dy) = \int_S (2\,dy\wedge dx + 3\, dx\wedge dy) = \int_S (-2 + 3) dx\wedge dy = \int_S dx\wedge dy = \pi
$$
where the last step is just using the area of the unit circle.

It's true. Thank you. I was getting confused about it.