Green's Theorem Practice Problems

Green’s Theorem gives you a relationship between the line integral of a 2D vector field over a closed path in a plane and the double integral over the region that it encloses. However, the integral of a 2D conservative field over a closed path is zero, which is a type of special case in Green’s Theorem. 

What is Green's Theorem?

Green's Theorem is one of the four fundamental theorems of calculus, in which all four are closely related. Once you learn about the line integral and surface integral concepts, you will learn how Stokes's theorem is based on the principle of linking the macroscopic and microscopic circulations.

Let C be the positively oriented, smooth, and simple closed curve in a plane, and D be the region bounded by the C. If L and M are the functions of (x, y) defined on the open region, containing D and have continuous partial derivatives, then Green’s theorem is stated as:

ϕ(Ldx + Mdy) = ∫∫_D(∂M/∂x-∂L/∂y)dxdy

Where the path integral is traversed counterclockwise along with C.

Important Formulas

Using this formula, we can write Green's Theorem as

ϕ(Ldx + Mdy) = ∫∫_D(dM/∂x-∂L/dy)dxdy

The line integral

∫_C F.ds = 0

Green's Theorem: Practice Questions with Solution

Problem 1: Use Green’s Theorem to Solve the given function F = (x-xy)+ y² when a particle moves counterclockwise along the rectangle whose vertices are given as (0,0),  (4,0),  (4,6), and  (0,6).

Solution:

Using Green’s Theorem, we know rhat

Given F = (P, Q) with P = x-xy and Q = y², let's compute the partial derivatives needed for Green's Theorem.

First, compute the partial derivatives:

\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(y^2) = 0

\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x - xy) = -x

Now, apply Green's Theorem:

 \oint_C \mathbf{F} \cdot d\mathbf{s} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \iint_R \left( 0 - (-x) \right) \, dA = \iint_R x \,

Region R is a rectangle with vertices (0,0), (4,0), (4,6), and (0,6)). To evaluate the double integral over this rectangular region, we set up the integral:

\iint_R x \, dA = \int_{0}^{6} \int_{0}^{4} x \, dx \,

Evaluate the inner integral with respect to x:

\int_{0}^{4} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{4}

= \frac{4^2}{2} - \frac{0^2}{2}

= \frac{16}{2}

= 8

Now, evaluate the outer integral with respect to y:

\int_{0}^{6} 8 \, dy = 8y \int_{0}^{6} = 8 \times 6 - 8 \times 0 = 48

Therefore, value of line integral is:

\oint_C \mathbf{F} \cdot d\mathbf{s} = 48

Problem 2: Calculate the line integral of ∫_c (3 - x) dx + (x - y) dy in which C is the union of a unit circle which is centred at the origin and is oriented negatively, and the circle that has radius 2 and is centred at the origin and is oriented positively.

Solution:

Green's Theorem states:

\oint_C \mathbf{F} \cdot d\mathbf{s} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \,

where P = 3 - x and Q = x - y.

First, we compute the partial derivatives:

\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x - y) = 1

\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(3 - x) = 0

Now, apply Green's Theorem:

\oint_C \mathbf{F} \cdot d\mathbf{s} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \iint_R (1 - 0) \, dA = \iint_R 1 \, dA

Region R is the annular region between the unit circle (radius 1, oriented negatively) and the circle with radius 2 (oriented positively). The area of this region can be calculated as the difference between the areas of the two circles.

Area of larger circle (radius 2) is:

Area larger circle =\text{Area}_{\text{larger circle}} = \pi (2^2) = 4

Area of smaller circle (radius 1) is:

Area smaller circle =\text{Area}_{\text{smaller circle}} = \pi (1^2) = \pi

Therefore, area of annular region R is:

Area R=\text{Area}_R = 4\pi - \pi = 3\pi

Thus, value of line integral is:

\oint_C \mathbf{F} \cdot d\mathbf{s} = \iint_R 1 \, dA = \text{Area}_R = 3\pi

So, the line integral \int_C (3 - x) \, dx + (x - y) \, dy is:

3\pi

Problem 3: Problem: Evaluate the line integral \oint_C (x^3 - y) \, dx + (x + y^3) \, dy, where C is the triangle with vertices (0,0), (1,0), and (1,1), oriented counterclockwise.

Solution:

Identify P(x, y) = x^3 - y and Q(x, y) = x + y^3.

Compute the partial derivatives: \frac{\partial Q}{\partial x} = 1, \quad \frac{\partial P}{\partial y} = -1​

Apply Green's Theorem: \oint_C (x^3 - y) \, dx + (x + y^3) \, dy = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \iint_R (1 - (-1)) \, dA = \iint_R 2 \, dA

Area A of the triangle with vertices (0,0), (1,0), and (1,1) is \frac{1}{2}​: \iint_R 2 \, dA = 2 \cdot \frac{1}{2} = 1

Hence, the line integral is: \oint_C (x^3 - y) \, dx + (x + y^3) \, dy = 1.

Problem 4: Problem: Evaluate the line integral \oint_C y \, dx - x \, dy, where C is the positively oriented ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.

Solution:

Identify P(x,y)=yand Q(x,y)=-x.

Compute the partial derivatives: \frac{\partial Q}{\partial x} = -1, \quad \frac{\partial P}{\partial y} = 1

Apply Green's Theorem: \oint_C y \, dx - x \, dy = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \iint_R (-1 - 1) \, dA = \iint_R -2 \, dA

Area A of the ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 is \pi ab: \iint_R -2 \, dA = -2 \cdot \pi ab.

Hence, the line integral is: \oint_C y \, dx - x \, dy = -2\pi ab

Problem 5: Problem: Evaluate the line integral \oint_C (x^2 - y^2) \, dx + (2xy) \, dy, where C is the positively oriented circle x^2 + y^2 = 1.

Solution:

Identify P(x, y) = x^2 - y^2 and Q(x,y)=2xy.

Compute the partial derivatives: \frac{\partial Q}{\partial x} = 2y, \quad \frac{\partial P}{\partial y} = -2y

Apply Green's Theorem: \oint_C (x^2 - y^2) \, dx + (2xy) \, dy = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \iint_R (2y + 2y) \, dA = \iint_R 4y \, dA

Since R is a symmetric region around the origin (the unit circle), the integral of 4yover this region is zero because y is odd and the region is symmetric: \iint_R 4y \, dA = 0

Hence, line integral is: \oint_C (x^2 - y^2) \, dx + (2xy) \, dy = 0

Worksheet: Green's Theorem

Problem 1: Evaluate the line integral \oint_C (2x - y) \, dx + (x + 3y) \, dy, where C is the circle x² + y² = 1.

Problem 2: Evaluate the line integral \oint_C (x^2 - y^2) \, dx + (2xy) \, dy, where C is the square with vertices (1, 1), (1, -1), (-1, -1), and (-1, 1).

Problem 3: Use Green's Theorem to find the area enclosed by the ellipse x²/4+y²=1.

Problem 4: Evaluate the line integral \oint_C (e^x \cos(y)) \, dx + (e^x \sin(y)) \, dy, where C is the boundary of the region bounded by x = 0, x = 1, y = 0, and y = 1.

Problem 5: Compute the line integral \oint_C \left(y^2 \sin(x)\right) \, dx + \left(x^2 \cos(y)\right) \, dy, where C is the boundary of the triangle with vertices (0, 0), (1, 0), and (0, 1).

Problem 6: Evaluate the line integral \oint_C (xy) \, dx + (x^2) \, dy, where C is the boundary of the region bounded by x = 0, y = 0, and x + y = 1.

Problem 7: Evaluate the line integral \oint_C (y^3) \, dx + (x^3) \, dy, where C is the circle x² + y² = 4.

Problem 8: Use Green's Theorem to find the work done by the force field F = (x², y² )in moving a particle once counterclockwise around the square with vertices (0, 0), (1, 0), (1, 1), and (0, 1).

Problem 9: Let C be the boundary of the region enclosed by y = x² and y = 2x. Use Green's Theorem to compute the line integral of \mathbf{F} = (x^2 - y^2, xy) around C.

Problem 10: Find the flux of the vector field \mathbf{F} = (x^2 y, xy^2) across the boundary of the region enclosed by the curve C, where C is the circle x² + y² = 4, using Green’s Theorem.

Answer Key:

1. Ans . 2π
2. Ans. 0
3. Ans. 2π
4. Ans. e-1
5. Ans. 0
6. Ans. 1/3
7. Ans. 0
8. Ans. 1
9. Ans. 0
10. Ans. -1/2​

Related Articles:

• Stoke's Theorem
• Divergence Theorem
• Vector Calculus

• Ensure the curve C is positively oriented.
• Verify that the vector field F has continuous partial derivatives in the region.
• Ensure that the region D is properly defined and that C is a simple closed curve.