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

  • #1
runinfang
4
0
Homework Statement
Calculate the line integral ##\oint_C (2y \, dx + 3x \, dy)##, where ##C## is the circle ##x^2 + y^2 = 1##, using the Green's Theorem. Please state the theorem and describe the resolution step by step.
Relevant Equations
Green's Theorem: Let ##M## and ##N## be functions of two variables ##x## and ##y##, such that they have continuous first partial derivatives on an open disk ##B## in ##\mathbb{R}^2##. 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:

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

$$
\iint_R 1 \, dA = \int_{0}^{2\pi} \int_{0}^{1} r \, dr \, d\theta
= \int_{0}^{2\pi} \left[\frac{r^2}{2}\right]_{0}^{1} \, d\theta
= \int_{0}^{2\pi} \frac{1}{2} \, d\theta
= \frac{1}{2} \times 2\pi
= \pi
$$
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
(MENTOR) Note: We use MathJax on-site for our Latex notation.

Please consider using it in the future. There is a tutorial below in my signature for reference.
 
  • #3
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.
 
  • Like
Likes runinfang and jedishrfu
  • #4
It's true. Thank you. I was getting confused about it.
 

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

What is Green's Theorem?

Green's Theorem relates the line integral around a simple, closed curve C to a double integral over the plane region D bounded by C. It states that if L and M are functions of (x, y) with continuous partial derivatives on an open region that contains D, then the line integral around C is equal to the double integral over D of the partial derivatives: ∮C (L dx + M dy) = ∬D ( (∂M/∂x) - (∂L/∂y) ) dA.

How do you set up an integral using Green's Theorem for a circular path?

To set up an integral using Green's Theorem for a circular path, you need to define the vector field F = (L, M) and ensure that the curve C is positively oriented (counterclockwise). Then, convert the line integral ∮C F · dr into the double integral ∬D ( (∂M/∂x) - (∂L/∂y) ) dA over the region D enclosed by the circle.

What are the conditions for applying Green's Theorem?

The conditions for applying Green's Theorem are: the curve C must be a positively oriented, simple, closed curve; the region D enclosed by C must be simply connected; and the functions L and M must have continuous partial derivatives on an open region containing D.

Can Green's Theorem be used for any vector field?

Green's Theorem can be used for any vector field as long as the conditions for the theorem are met. The vector field components L and M must have continuous partial derivatives, and the curve C and region D must satisfy the requirements of the theorem.

How do you handle singularities when using Green's Theorem?

When dealing with singularities, Green's Theorem can still be applied by excluding the singular points from the region D and adjusting the curve C accordingly. This often involves breaking the region into smaller, non-singular regions and applying the theorem to each part, or using a limiting process to handle the singularity.

Similar threads

Replies
8
Views
1K
Replies
4
Views
1K
Replies
10
Views
1K
Replies
3
Views
1K
Replies
1
Views
700
Replies
8
Views
2K
Replies
4
Views
1K
Back
Top