The excitement of learning separates youth from old age. As long as you’re learning, you’re not old, Rosalyn S.Yalow.
Silence is a source of great strength, Lao Tzu.
A vector field is an assignment of a vector $\vec{F}$ to each point (x, y) in a space, i.e., $\vec{F} = M\vec{i}+N\vec{j}$ where M and N are functions of x and y.
A vector field on a plane can be visualized as a collection of arrows, each attached to a point on the plane. These arrows represent vectors with specific magnitudes and directions.
Work is defined as the energy transferred when a force acts on an object and displaces it along a path. In the context of vector fields, we calculate the work done by a force field along a curve or trajectory C using a line integral. The work done by a force field $\vec{F}$ along a curve C is: W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy = \int_{C} \vec{F}·\hat{\mathbf{T}}ds$, where $\hat{\mathbf{T}}$ is the unit tangent vector.
A vector field is conservative if there exist a scalar function such that $\vec{F}$ = ∇f (the vector field is its gradient). This scalar function is known or referred to as the potential function associated with the vector field.
Theorem. Fundamental theorem of calculus for line integral. If $\vec{F}$ is a conservative vector field in a simply connected region of space (i.e., a region with no holes), and if f is a scalar potential function for $\vec{F}$ in that region, then $\int_{C} \vec{F}·d\vec{r} = \int_{C} ∇f·d\vec{r} = f(P_1)-f(P_0)$ where P_{0} and P_{1} are the initial and final points of the curve C, respectively.
Conservative force. A force $\vec{F}$ is considered conservative if the work done by the force around any closed curve C is zero. Mathematically, this is expressed as $\int_{C} \vec{F}·d\vec{r} = 0$.
Path independence. A force field is path-independent, meaning the work done by the force in moving an object from one point to another is the same, regardless of the path taken between the two points.
Gradient Field. A vector field $\vec{F}$ is a gradient field if it can be expressed as the gradient of a scalar potential function f. In mathematical terms, this means $\vec{F} = ∇f$, where f is a scalar function and the vector field $\vec{F}$ has components $\vec{F} = ⟨M, N⟩ = ⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨f_x, f_y⟩$. Here, f is the potential function associated with the vector field $\vec{F}$. The lineal integral of $\vec{F}$ along a path C measures the work done by the vector field in moving an object along the path C. If $\vec{F}$ is a gradient field, then: $\int_{C} \vec{F}·d\vec{r} = f(P_1)-f(P_0).$
Exact differential. In the context of differential forms, a differential expression Mdx+Ndy is called an exact differential if there exist a scalar function f(x, y) such that df = $\frac{∂f}{∂x}dx+\frac{∂f}{∂y}dy$ = Mdx + Ndy. This implies that the vector field $\vec{F} = ⟨M, N⟩$ is conservative, and there exists a potential function f such that $\vec{F} = ∇f$.
If a differential is exact, then the line integral of $\vec{F}$ over any path C can be evaluated by simply finding the difference in the potential function values at the endpoints of the path.
These four properties —conservative force, path independence, gradient field, and exact differential— are different perspectives of the same fundamental concept: the conservativeness of a vector field.
The criterion for checking whether a vector field $\vec{F}$ is conservative can be summarized as follows: If $\frac{∂M}{∂y} = \frac{∂N}{∂x}$, and if the functions M and N have continuous first partial derivatives across the entire domain D, and the domain D is open and simply connected, then $\vec{F}$ is a gradient field, meaning it is conservative.
Definition. The positive orientation of a simple closed curve C refers to a single counterclockwise traversal of C. This keeps the region R enclosed by C always on the left as $\vec{r(t)}$ traverses C.
Notation. An integral around a positive oriented, closed curve C can be represented as $ \oint_C Mdx + N dy$.
Green’s theorem is a fundamental result in vector calculus that establishes a relationship between a line integral around a simple, closed curve C and a double integral over the region R enclosed by C.
Green’s Theorem. If C is a positively oriented (counterclockwise) simple closed curve enclosing a region R, and $\vec{F} = ⟨M(x, y), N(x,y)$ is a vector field that is defined and has continuous partial derivatives on an open region containing R, then Green's Theorem states that: $\oint_C \vec{F} \cdot{} d\vec{r} = \iint_R curl(\vec{F}) dA ↭ \oint_C Mdx + N dy = \iint_R (N_x-M_y)dA = \iint_R (\frac{∂N}{∂x}-\frac{∂M}{∂y})dA$ where
Let $\vec{F} = ⟨x^2+y^2, -2xy⟩$. Evaluate $\oint_C \vec{F} \cdot{} d\vec{r}$ where C and D are the curve and region below respectively (Refer to Figure i for a visual representation and aid in understanding it)
First, let’s check if the vector field $\vec{F}$. A vector field is conservative if there exists a scalar potential function f such that $\vec{F}$ = = ∇f. A necessary condition for this is that the curl of $\vec{F}$ should be zero, or equivalently: $\frac{∂M}{∂y} = \frac{∂N}{∂x}$.
However, $\vec{F}$ is not conservative since $\frac{∂M}{∂y} = \frac{∂}{∂y}(x^2+y^2) = 2y ≠ \frac{∂N}{∂x} = \frac{∂}{∂x}(-2xy) = -2y$. This implies that the line integral $\oint_C \vec{F} \cdot{} d\vec{r}$ depends on the path C and cannot be evaluated using a potential function.
The curve C is composed of two segments:
$\oint_C \vec{F} \cdot{} d\vec{r} = \oint_{C_1} \vec{F} \cdot{} d\vec{r} + \oint_{C_2} \vec{F} \cdot{} d\vec{r} = \int_{-2}^{2} (t^2+0)(-dt)+0 + \int_{π}^{2π} 4(-2sin(t)dt)-2(2cos(t))(2sin(t))(2cos(t)dt) = -\int_{-2}^{2} t^2dt + \int_{π}^{2π} (-8sin(t)-16cos^2(t)sin(t))dt$
$\int_{-2}^{2} t^2dt = \frac{t^3}{3}\bigg|_{-2}^{2} = \frac{8}{3}-(\frac{-8}{3}) = \frac{16}{3}$
$\int_{π}^{2π} (-8sin(t)-16cos^2(t)sin(t))dt = \int_{π}^{2π} -8sin(t)dt -\int_{π}^{2π} 16cos^2(t)sin(t)dt$
$\int_{π}^{2π} -8sin(t)dt = -8·[-cos(t)]\bigg|_{π}^{2π} = 8[cos(2π)-cos(π)] = 8·[1-(-1)] = 16.$
$\int_{π}^{2π} -16cos^2(t)sin(t)dt =[\text{Using substitution: } u = cos(t), du = -sin(t)dt.\text{ Limits of integration: } t = π→ u = cos(π) = -1, t = 2π→ u = cos(2π) = 1] 16\int_{-1}^{1} u^2 = \frac{u^3}{3}\bigg|_{-1}^{1} = \frac{16}{3}[1^3-(-1)^3] = \frac{16}{3}·2 = \frac{32}{3}$
$\int_{π}^{2π} (-8sin(t)-16cos^2(t)sin(t))dt = \int_{π}^{2π} -8sin(t)dt -\int_{π}^{2π} 16cos^2(t)sin(t)dt = 16 + \frac{32}{3} = \frac{48}{3} + \frac{32}{3} = \frac{80}{3}$
$\oint_C \vec{F} \cdot{} d\vec{r} = -\int_{-2}^{2} t^2dt + \int_{π}^{2π} (-8sin(t)-16cos^2(t)sin(t))dt = \frac{-16}{3}+ \frac{80}{3} = \frac{64}{3}$
Green’s Theorem states that for a vector field $\vec{F} = ⟨M, N⟩$, the line integral around a simple, closed curve C can be related to a double integral over the region R enclosed by C:
$\oint_C \vec{F} \cdot{} d\vec{r} = \iint_R (\frac{∂N}{∂x}-\frac{∂M}{∂y})dA$
$\iint_R (\frac{∂N}{∂x}-\frac{∂M}{∂y})dA = \iint_R (-2y-2x)dA = -4\iint_R ydA =[\text{Polar coordinates}] -4\int_{π}^{2π}\int_{0}^{2} (rsin(θ))rdrdθ = -4\int_{π}^{2π}sin(θ)dθ\int_{0}^{2}r^2dr$
Inner integral: $\int_{0}^{2}r^2dr = \frac{r^3}{3}\bigg|_{0}^{2} = \frac{8}{3}$
Outer integral: $4·\frac{8}{3}\int_{π}^{2π}sin(θ)dθ = 4·\frac{8}{3}(-cos(θ))\bigg|_{π}^{2π} = -4·\frac{8}{3}[-cos(2π)-(-cos(π))] = -4·\frac{8}{3}[-1-(1)] = -4·\frac{-8}{3}·2 = \frac{64}{3}$
$\oint_C \vec{F} \cdot{} d\vec{r} = \iint_R (\frac{∂N}{∂x}-\frac{∂M}{∂y})dA = \frac{64}{3},$ consistent with our earlier calculation.
Identify the Vector Field $\vec{F}$. The given line integral is $\oint_C xydx + x^2y^3dy$. We can identify the components M(x, y) and N(x, y) of the vector field $\vec{F}$: $\vec{F} = ⟨M, N⟩ = ⟨xy, x^2y^3⟩$
$\oint_C xydx + x^2y^3dy = $[Using Green’s Theorem (the curve does satisfy the required conditions), we can convert the line integral around the closed curve C to a double integral over the region R enclosed by C. Green’s Theorem states: $\oint_C Mdx + Ndy = \iint_R (N_x-M_y) dA$] = $\iint_R (N_x-M_y) dA$.
We first compute the partial derivatives: $N_x = \frac{∂}{∂x}(x^2y^3) = 2xy^3, M_y = \frac{∂}{∂y}(xy)=x$
Thus, the integrand for the double integral is: $N_x-M_y = 2xy^3-x$
$\iint_R (N_x-M_y) dA = \iint_R (2xy^3-x)dA$
The region R is the triangle with vertices (0, 0), (1,0), and (1,2). The boundaries of this region can be described by the inequalities: 0 ≤ x ≤ 1, 0 ≤ y ≤ 2x.
Now, we set up the double integral:
$\iint_R (2xy^3-x)dA = \int_{0}^{1} (\int_{0}^{2x} (2xy^3-x)dy)dx$
We evaluate the inner integral first: $\int_{0}^{2x} (2xy^3-x)dy = 2x\frac{y^4}{4}-xy = (\frac{xy^4}{2}-xy)\bigg|_{0}^{2x} = 8x^5-2x^2$
Now, evaluate the outer integral:
$\iint_R (2xy^3-x)dA = \int_{0}^{1} (\int_{0}^{2x} (2xy^3-x)dy)dx = \int_{0}^{1} (8x^5-2x^2)dx = \frac{8x^6}{6}-\frac{2x^3}{3} = (\frac{4x^6}{3}-\frac{2x^3}{3})\bigg|_{0}^{1} = \frac{4}{3}-\frac{2}{3} = \frac{2}{3}.$
The integral involves the vector field $\vec{F}$ = ⟨M, N⟩ where $M=ye^{-x}, N=(\frac{1}{2}x^2-e^{-x})$
Apply Green’s Theorem. Green’s Theorem relates the line integral around the closed curve C to a double integral over the region R enclosed by C:
$\oint_C Mdx + Ndy = \iint_R (\frac{∂N}{∂x}-\frac{∂M}{∂y})dA$
Calculate the partial derivatives:
$\oint_C Mdx + Ndy = \oint_C ye^{-x}dx + (\frac{1}{2}x^2-e^{-x})dy =$[Using Green’s Theorem, we can convert the line integral to a double integral. $N_x = \frac{∂}{∂x}(\frac{1}{2}x^2-e^{-x}) = x+e^{-x}, M_y = \frac{∂}{∂y}(ye^{-x})=e^{-x}$] =$\iint_R curl(\vec{F}) dA = \iint_R (N_x-M_y) dA.$
Set Up the Double Integral. Using Green’s Theorem: $\oint_C ye^{-x}dx + (\frac{1}{2}x^2-e^{-x})dy = \iint_R (x+e^{-x}-e^{-x}) dA =[\text{This simplifies to:}] \iint_R x dA = Area(R)\overline {x} =[1] (π·1^2)·2=2π$
[1] Recall that the center of mass of a distribution of mass in space is the unique point at any given time where the weighted relative position of the distributed mass sums to zero. It is the point where the entire mass of an object may be assumed to be concentrated to visualize its motion. Mathematically, for a region R with constant density δ, the x-coordinate of the center of mass $\overline {x}$ is given by: $\overline {x} = \frac{1}{Mass} \iint_R\ x·δ·dA$ =[If the density δ is uniform, and particularly δ=1, then:] $\frac{1}{Area(R)} \iint_R xdA$.
Since the circle is symmetric around the x-axis, its center of mass lies along the x-axis at the x-coordinate of the circle’s center, which is 2. Hence, $\overline {x} = 2.$
The vector field $\vec{F}$ associated with this line integral can be expressed as: $\vec{F} = ⟨M, N⟩ = ⟨y^3, - x^3⟩$.
Apply Green’s Theorem. Green’s Theorem relates the line integral around the closed curve C to a double integral over the region R enclosed by C:
$\oint_C Mdx + Ndy = \iint_R (\frac{∂N}{∂x}-\frac{∂M}{∂y})dA$. Here, the curve C is a circle, which is a simple closed curve, so Green’s Theorem applies.
$\oint_C y^3dx - x^3dy = $[Using Green’s Theorem, we can convert the line integral around the closed curve C into a double integral over the region R enclosed by C. Green’s Theorem states: $\oint_C Mdx + Ndy = \iint_R (N_x-M_y) dA$] = $\iint_R (N_x-M_y) dA$ where $M=y^3, N=-x^3$.
We first compute the partial derivatives: $N_x = \frac{∂}{∂x}(-x^3) = -3x^2, M_y = \frac{∂}{∂y}(y^3)=3y^2$
Thus, the integrand for the double integral is: $N_x-M_y = -3x^2-3y^2$
Set up the Double Integral: $\oint_C y^3dx - x^3dy = \iint_R curl(\vec{F}) dA = \iint_R (N_x-M_y) dA = \iint_R (-3x^2-3y^2)dA$ where the region R is the disk of radius 2 centered at the origin.
Since R is a disk, it’s convenient to switch to polar coordinates. In polar coordinates, we can describe the disk as: x = rcos(θ), y = rsin(θ), x^{2}+y^{2} = r^{2}, dA = rdrdθ.
The limits for r are from 0 to 2, and for θ are from 0 to 2π. The integral becomes:
$-3\iint_R (x^2+y^2)dA = -3\int_{0}^{2π}\int_{0}^{2} r^3drdθ$
We evaluate the inner integral first: $\int_{0}^{2} r^3dr = \frac{r^4}{4}\bigg|_{0}^{2} = \frac{2^4}{4} = 4$
Now, evaluate the outer integral:
$-3 \int_{0}^{2π} \int_{0}^{2} r^3drdθ = -3\int_{0}^{2π} 4dθ = -3·4θ\bigg|_{0}^{2π} = -3·(8π-0) = -24π$
Final Answer: $\oint_C y^3dx - x^3dy = -24π$
Here, C is a closed curve representing the boundary of the rectangle, and it is oriented counterclockwise. The vector field $\vec{F} = ⟨M, N ⟩ = ⟨x^2y, y-3⟩$.
Apply Green’s Theorem. Green’s Theorem relates the line integral around the closed curve C to a double integral over the region R enclosed by C:
$\oint_C Mdx + Ndy = \iint_R (\frac{∂N}{∂x}-\frac{∂M}{∂y})dA$. The region R is the interior of the rectangle C encloses.
$\oint_C x^2ydx + (y-3)dy = $[Using Green’s Theorem (the curve does satisfy the required conditions), we can convert the line integral around the closed curve C to a double integral over the region R enclosed by C. Green’s Theorem states: $\oint_C Mdx + Ndy = \iint_R (N_x-M_y) dA$] = $\iint_R (N_x-M_y) dA$ where $M=x^2y, N=y-3$.
We first compute the partial derivatives: $N_x = \frac{∂}{∂x}(y-3) = 0, M_y = \frac{∂}{∂y}(x^2y)=x^2$
Thus, the integrand for the double integral is: $N_x-M_y = 0 - x^2 = -x^2.$
The boundaries of this rectangular region can be described by the inequalities: 1 ≤ x ≤ 4 and 1 ≤ y ≤ 5. The double integral over the region R becomes:
$\iint_R (N_x-M_y) dA = \iint_R -x^2dA = \int_{1}^{5}(\int_{1}^{4} -x^2dx)dy.$
Evaluate the inner integral first:
$\int_{1}^{4} -x^2dx = \frac{-x^3}{3}\bigg|_{1}^{4} = -(\frac{4^3}{3}-\frac{1^3}{3}) = -(\frac{64}{3}-\frac{1}{3}) = \frac{-63}{3} = -21$
Now, evaluate the outer integral:
$\iint_R (N_x-M_y) dA = \iint_R -x^2dA = \int_{1}^{5}(\int_{1}^{4} -x^2dx)dy = \int_{1}^{5} -21dy = -21y\bigg|_{1}^{5} = -21(5-1) = -21·4 = -84.$
$\oint_C x^2ydx + (y-3)dy= -84.$
One significant implication of Green’s theorem is that if $\vec{F}$ is defined everywhere in the plane and the curl of the vector field is zero everywhere in a region containing R, $curl(\vec{F})=0$, then the line integral of $\vec{F}$ around any close curve within this region will also be zero $\oint_C \vec{F} \cdot{} d\vec{r} = \iint_R curl(\vec{F}) dA = \iint_R 0·dA = 0$, indicating that the vector field is conservative.
Apply Green’s Theorem. Green’s Theorem relates the line integral around the closed curve C to a double integral over the region R enclosed by C:
$\oint_{-C} Mdx + Ndy = -\iint_R (\frac{∂N}{∂x}-\frac{∂M}{∂y})dA =[\frac{∂M}{∂y}=3, \frac{∂N}{∂x}=2] -\iint_R(2 -3)dA = \iint_R dA = $ area of R =[The area of the circle with radius 2, since (x-1)^{2}+(y-5)^{2}=4 is a circle of radius 2] π·2^{2} = 4π.
The region R is the annular region (a ring-shaped region) between the two circles x^{2}+ y^{2} = 1 (a smaller circle with radius 1) and x^{2}+ y^{2} = 9 (a larger circle with radius 3). The curve C is the boundary of this region, consisting of two circular paths: one for the outer boundary (radius 3) and one for the inner boundary (radius 1) (Refer to Figure ii for a visual representation and aid in understanding it).
Outer circle x^{2}+ y^{2} = 9: The curve is oriented counterclockwise, which is positive. Inner circle x^{2}+ y^{2} = 1: The curve is oriented clockwise, which is negative. Effectively, when applying Green’s Theorem, the integral over the region R should consider the area of the region enclosed by the outer circle minus the area of the region enclosed by the inner circle.
Apply Green’s Theorem. Green’s Theorem relates the line integral around a positively oriented (counterclockwise) curve C to a double integral over the region R enclosed by C:
$\oint_{C} Mdx + Ndy = \iint_R (\frac{∂N}{∂x}-\frac{∂M}{∂y})dA = \iint_R (3x^2+3y^2)dA$
Set Up the Double Integral in Polar Coordinates, where x = rcos(θ), y = rsin(θ), we have x^{2} + y^{2} = r^{2}. Therefore, the integral becomes,
$\iint_R 3(x^2+y^2)dA = \int_{0}^{2π} \int_{1}^{3} 3r^2·rdrdθ = 3\int_{0}^{2π} \int_{1}^{3} r^3drdθ$
Evaluate the integral:
Inner (or radial) integral: $\int_{1}^{3} r^3dr = \frac{r^4}{4}\bigg|_{1}^{3} = \frac{3^4}{4}-\frac{1}{4} = \frac{81}{4}-\frac{1}{4} = \frac{80}{4} = 20$
Outer integral with respect to theta: $\iint_R 3(x^2+y^2)dA = 3\int_{0}^{2π} 20dθ = 3·20θ\bigg|_{0}^{2π} = 3·20·2π = 120π$. Thus, the value of the line integral is $\oint_C (x^3 - y^3)dx + (x^3+y^3)dy = 120π$
Green’s Theorem is a fundamental result in vector calculus that establishes a relationship between a line integral around a simple closed curve C and a double integral over the region R enclosed by C. Specifically, Green’s Theorem states that: $\oint_C Mdx + Ndy = \iint_R (\frac{∂N}{∂x}-\frac{∂M}{∂y})dA$
First, we will prove that $\oint_C Mdx = \iint_R -M_ydA$, the reader could observe that this is just a special case when N = 0. Similarly we can argue that $\oint_C Ndy = \iint_R N_xdA$. Adding these two results together, we obtain Green’s Theorem: $\oint_C Mdx + Ndy = \iint_R (N_x-M_y)dA$
Futhermore, we can decompose or break down the region R into simpler, smaller, and more manageable regions (Figure A). If we can prove that $\oint_{C_1} Mdx = \iint_{R_1} -M_ydA, \oint_{C_2} Mdx = \iint_{R_2} -M_ydA$, the we can add this two statements together to prove the general case, and that’s true [1] because the boundary between R_{1} and R_{2} is traversed twice, but in opposite directions, canceling each other out.
Thus, we have:
$\oint_C Mdx =[1] \oint_{C_1} Mdx + \oint_{C_2} Mdx = \iint_{R_1} -M_ydA + \iint_{R_2} -M_ydA = \iint_{R} -M_ydA$
Futhermore, we are going to cut or split the regions in vertically simple regions, which are regions bounded between two curves, f_{1} and f_{2}, ∀x: a < x < b, f_{1}(x) < y < f_{2}(x).
Therefore, we only need to prove $\oint_C Mdx = \iint_R -M_ydA$ if R is vertically simple and C is a boundary of R taken counterclockwise.
We split the integral into four parts (Figure C). Let’s consider R as a vertically simple region bounded by the curves y = f_{1}(x) and y = f_{2}(x) from x = a to x = b. The boundary C can be divided into four segments:
Integral along C_{1}. The path from (a, f_{1}(a)) to (b, f_{1}(b)): $\oint_{C_1} Mdx =$[Notice that M = M(x, y), y = f_{1}(x), a ≤ x ≤ b] =$\int_{a}^{b} M(x, f_1(x))dx$
Integral along C_{2}. The path from (b, f_{1}(b)) to (b, f_{2}(b)): $\oint_{C_2} Mdx =$[Since x = b and dx = 0, this integral contributes 0] 0.
Integral along C_{3}. The path from (b, f_{2}(b)) to (a, f_{2}(a)): $\oint_{C_3} Mdx =$[y = f_{2}(x), b ≤ x ≤ a] =$\int_{b}^{a} M(x, f_2(x))dx = -\int_{a}^{b} M(x, f_2(x))dx$
Integral along C_{4}. The path from (a, f_{2}(a)) to (a, f_{1}(a)): $\oint_{C_4} Mdx =$[Since x = a and dx = 0, this integral also contributes 0] 0.
Putting all together, $\oint_C Mdx = \oint_{C_1} Mdx + \oint_{C_2} Mdx + \oint_{C_3} Mdx + \oint_{C_4} Mdx = \int_{a}^{b} M(x, f_1(x))dx -\int_{a}^{b} M(x, f_2(x))dx = \int_{a}^{b} [M(x, f_1(x))-M(x, f_2(x))]dx$
On the other side of the equation, $\iint_{R} -M_ydA = -\int_{a}^{b}(\int_{f_1(x)}^{f_2(x)} \frac{∂M}{∂y}dy)dx$
The inner integral is $\int_{f_1(x)}^{f_2(x)} \frac{∂M}{∂y}dy = M(x, f_2(x))-M(x, f_1(x))$
Therefore: $\iint_{R} -M_ydA = -\int_{a}^{b}(\int_{f_1(x)}^{f_2(x)} \frac{∂M}{∂y}dy)dx = -\int_{a}^{b} (M(x, f_2(x))-M(x, f_1(x)))dx$ ⇒Thus, we have shown: $\oint_C Mdx = \iint_R -M_ydA$ ∎
The area of a region R in the plane can be calculated using a double integral over that region. Mathematically, the area is expressed as: Area(R) = $\iint_{R} dA$. Here, dA represents a differential element of area in the region R.
However, we can also use Green’s Theorem to find the area of R by converting a double integral over R into a line integral around the boundary C of R.
Green’s Theorem states that for a positively oriented, simple, closed curve C enclosing a region R: $\oint_{C} Mdx + Ndy = \iint_R (\frac{∂N}{∂x}-\frac{∂M}{∂y})dA$.
To calculate the area of R, we want the double integral on the right-hand side of Green’s Theorem to simply become the area Area(R) = $\iint_{R} dA$. This requires that the expression inside the double integral, $\frac{∂N}{∂x}-\frac{∂M}{∂y}$, equals 1.
There are several ways to define M and N to obtain this.
Example: Area of an Ellipse (Refer to Figure iii for a visual representation and aid in understanding it).
Use a line integral to find the area enclosed by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$.
To apply Green’s Theorem, we parametrize the ellipse using: x = acos(t), y = bsin(t), 0 ≤ t ≤ 2π. The differentials are: dx = -asin(t)dt, dy = bcos(t)dt.
Set Up the Line Integral
Using the third option, M = -y, N = x, the area enclosed by the ellipse can be computed as:
Area(R)= $\frac{1}{2}\oint_{C} xdy-ydx = \frac{1}{2}\int_{0}^{2π} (acos(t))(bcos(t)dt)-(bsin(t))(-asin(t)dt) =[\text{Simplifying}] \frac{1}{2}\int_{0}^{2π}(abcos^2(t)+absin^2(t))dt =[\text{Since: }cos^2(t)+sin^2(t)=1] \frac{1}{2}\int_{0}^{2π} abdt = \frac{1}{2}abt\bigg|_{0}^{2π} = πab$. For a circle of radius r, a = b = r, Area = πr^{2}.