Vector calculus. Green's theorem

Statement

Let $D$ be a bounded simply connected domain in the plane whose boundary $\partial D$ is a piecewise smooth closed contour traversed anticlockwise (the domain stays on the left), and let $P$ and $Q$ be continuously differentiable on the closure of $D$ . Then

\partial D \oint P d x + Q d y = D \iint (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d x d y .

The theorem turns a contour integral into a double integral over the domain. This pays off when the integrands themselves have no elementary antiderivative but the difference of the derivatives is simple — the typical case in the problems below.

Hypotheses to check

Three of them, and all three really are violated in problems:

The contour is closed. The theorem does not apply to an open arc; sometimes the arc is closed with an auxiliary segment whose contribution is then subtracted.
The orientation is anticlockwise. Traversed clockwise, the answer changes sign.
Smoothness in the whole domain, not just on the contour. A singular point inside $D$ (for instance, a denominator $x^{2} + y^{2}$ at the origin) breaks the theorem; such points are cut out with a small circle.

Two common uses

Area as a contour integral. Taking $P = - y$ , $Q = x$ gives $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2$ and

S (D) = \frac{1}{2} \partial D \oint x d y - y d x .

Vanishing. If $P$ depends only on $x$ and $Q$ only on $y$ , both derivatives are zero and the integral over any closed contour is zero — however frightening $P$ and $Q$ look.

Common mistakes

Computing $\partial P / \partial x - \partial Q / \partial y$ . The order is exactly this: the derivative of the second function with respect to the first variable, minus the derivative of the first with respect to the second.
Forgetting to check the orientation and losing the sign.
Applying the theorem to a contour that encloses a singularity of the field.

Problems

Problem 1 (A contour integral of functions with no elementary antiderivative). Compute the integral around a closed contour

Γ \oint (x^{3} e^{x^{4}} d x + sin (e^{y}) d y),

where $Γ$ is the circle $x^{2} + y^{2} = 4$ , traversed anticlockwise.

Solution.

Direct computation is hopeless: after parametrisation the integrals of $e^{16 c o s^{4} θ}$ and $sin (e^{2 s i n θ})$ appear, and these have no elementary antiderivatives. That is exactly the signal to use Green's theorem — it replaces the functions themselves with their derivatives.

Green's theorem.

Γ \oint P d x + Q d y = D \iint (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d x d y,

where $D$ is the disc $x^{2} + y^{2} ⩽ 4$ .

Checking the hypotheses. The contour is closed and traversed anticlockwise, and the functions $P = x^{3} e^{x^{4}}$ and $Q = sin (e^{y})$ are continuously differentiable everywhere in the plane — there are no singular points inside $D$ . The theorem applies.

The derivatives. The function $P$ depends only on $x$ and the function $Q$ only on $y$ , so

\frac{\partial Q}{\partial x} = 0, \frac{\partial P}{\partial y} = 0.

Result. The integrand of the double integral is identically zero:

D \iint 0 d x d y = 0.

This is a general fact, not a coincidence: if $P = P (x)$ and $Q = Q (y)$ , the field is potential (the potential is the sum of the antiderivatives), and its circulation around any closed contour is zero, however complicated the functions themselves are.

Answer: $0$ .

Problem 2 (The area of an ellipse as a contour integral). Compute

Γ \oint (- y d x + x d y),

where $Γ$ is the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ , traversed anticlockwise.

Solution.

By Green's theorem. Here $P = - y$ , $Q = x$ , so

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (- 1) = 2,

and the integral reduces to the area:

Γ \oint (- y d x + x d y) = D \iint 2 d x d y = 2 S (D) .

This is the standard use of the theorem: the combination $- y d x + x d y$ exists precisely for computing areas.

The area of an ellipse with semi-axes $a = 3$ and $b = 2$ is $S = π ab = 6 π$ , so the integral is $12 π$ .

A check by direct computation. Parametrise: $x = 3 cos t$ , $y = 2 sin t$ , $t \in [0; 2 π]$ . Then $d x = - 3 sin t d t$ , $d y = 2 cos t d t$ and

- y d x + x d y = 6 sin^{2} t d t + 6 cos^{2} t d t = 6 d t .

The integrand turned out to be constant, so

\int_{0}^{2 π} 6 d t = 12 π .

It agrees.

Answer: $12 π$ .

Problem 3 (Green's theorem for a triangular contour). Compute

Γ \oint (x y d x + x^{2} d y),

where $Γ$ is the boundary of the triangle with vertices $O (0; 0)$ , $A (1; 0)$ , $B (1; 1)$ , traversed anticlockwise.

Solution.

Direct computation would take three integrals along the three sides. Green's theorem reduces everything to one double integral over the triangle $D$ .

The integrand. Here $P = x y$ , $Q = x^{2}$ :

\frac{\partial Q}{\partial x} = 2 x, \frac{\partial P}{\partial y} = x, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x .

The domain. The vertices $O (0; 0)$ , $A (1; 0)$ , $B (1; 1)$ give a triangle bounded by the lines $y = 0$ below, $x = 1$ on the right and $y = x$ above. It is more convenient to describe it as $0 ⩽ x ⩽ 1$ , $0 ⩽ y ⩽ x$ .

Checking the orientation. The traversal $O \to A \to B \to O$ is anticlockwise: the domain stays on the left. No sign change is needed.

Integration.

D \iint x d x d y = \int_{0}^{1} \int_{0}^{x} x d y d x = \int_{0}^{1} x \cdot x d x = \int_{0}^{1} x^{2} d x = \frac{1}{3} .

Answer: $\frac{1}{3}$ .

Problem 4 (An open arc: close it and subtract). Compute

\int_{Γ} (e^{x} sin y - 3 y) d x + (e^{x} cos y - 3) d y,

where $Γ$ is the upper semicircle $x^{2} + y^{2} = 2 x$ , $y ⩾ 0$ , traversed from the point $A (2; 0)$ to the point $O (0; 0)$ .

Solution.

The contour is not closed, so Green's theorem cannot be applied directly. Direct computation along the semicircle leads to integrals of $e^{1 + c o s t} sin (sin t)$ — hopeless. The trick: close the arc with a segment along which the integral is easy, and subtract its contribution.

Closing. Add the diameter $O A$ — the segment of the $O x$ axis from $O (0; 0)$ to $A (2; 0)$ . The semicircle from $A$ to $O$ followed by the diameter from $O$ to $A$ goes round the upper half-disc $D$ anticlockwise, so

\int_{Γ} + \int_{O A} = \oint_{\partial D} = D \iint (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d x d y .

The integrand.

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = e^{x} cos y - (e^{x} cos y - 3) = 3.

The exponentials cancelled — that is what the field was chosen for.

The double integral. The circle $x^{2} + y^{2} = 2 x$ is $(x - 1)^{2} + y^{2} = 1$ , of radius $1$ , so $S (D) = π /2$ and $\oint_{\partial D} = 3 \cdot \frac{π}{2} = \frac{3 π}{2}$ .

The integral along the diameter. On the $O x$ axis: $y = 0$ , $d y = 0$ , and $P = e^{x} sin 0 - 0 = 0$ . The whole integral along the segment is zero.

Result. $\int_{Γ} = \frac{3 π}{2} - 0 = \frac{3 π}{2}$ .

The closing segment is chosen so that the integral along it is simple; here it vanished entirely, but that is not always so — and then it must be subtracted.

Answer: $\frac{3 π}{2}$ .

Problem 5 (The area under an arch of a cycloid). Find the area of the figure bounded by one arch of the cycloid $x = 2 (t - sin t)$ , $y = 2 (1 - cos t)$ , $t \in [0; 2 π]$ , and the $O x$ axis.

Solution.

Area as a contour integral is convenient where the boundary is given parametrically and $y$ cannot be expressed through $x$ — as for a cycloid.

Which formula to take. Any of $S = \oint x d y = - \oint y d x = \frac{1}{2} \oint (x d y - y d x)$ will do, traversed anticlockwise. The boundary has two parts: a segment of the $O x$ axis and the arch. On the segment $y = 0$ and $d y = 0$ , so for the formula $S = - \oint y d x$ it contributes nothing — that is the one to choose.

Orientation. Anticlockwise means first along the $O x$ axis from left to right, from $(0; 0)$ to $(4 π; 0)$ , and then along the arch from right to left, that is with $t$ decreasing from $2 π$ to $0$ . The cycloid's parameter runs the other way, and this has to be taken into account.

Computation.

S = - \int_{2 π}^{0} y x^{'} (t) d t = \int_{0}^{2 π} 2 (1 - cos t) \cdot 2 (1 - cos t) d t = 4 \int_{0}^{2 π} (1 - cos t)^{2} d t .

Expand: $(1 - cos t)^{2} = 1 - 2 cos t + cos^{2} t$ . Over a full period $\int cos t = 0$ and $\int cos^{2} t = π$ , so the integral is $2 π + π = 3 π$ and

S = 4 \cdot 3 π = 12 π .

This is a classical result: the area under an arch of a cycloid is three times the area of the rolling circle ( $π \cdot 2^{2} = 4 π$ ).

Answer: $S = 12 π$ .

Problem 6 (Green's theorem for an annulus). Compute $\oint_{\partial D} (- x^{2} y d x + x y^{2} d y)$ , where $D$ is the annulus $1 ⩽ x^{2} + y^{2} ⩽ 4$ and its boundary $\partial D$ is traversed in the positive direction.

Solution.

The positive direction for an annulus. The boundary of an annulus consists of two circles, and "positive" means so that the region stays on the left: the outer circle $x^{2} + y^{2} = 4$ is traversed anticlockwise, and the inner one $x^{2} + y^{2} = 1$ clockwise. In this form Green's theorem holds for a region that is not simply connected as well.

The integrand. $P = - x^{2} y$ , $Q = x y^{2}$ :

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = y^{2} + x^{2} .

The double integral in polar coordinates.

D \iint (x^{2} + y^{2}) d x d y = \int_{0}^{2 π} \int_{1}^{2} r^{2} \cdot r d r d φ = 2 π \cdot \frac{2 ^{4} - 1 ^{4}}{4} = \frac{15 π}{2} .

A check by direct computation. On a circle of radius $R$ , anticlockwise: $x = R cos t$ , $y = R sin t$ , and

- x^{2} y d x + x y^{2} d y = R^{4} (cos^{2} t sin^{2} t + cos^{2} t sin^{2} t) d t = 2 R^{4} cos^{2} t sin^{2} t d t,

while $\int_{0}^{2 π} 2 cos^{2} t sin^{2} t d t = π /2$ . The outer circle gives $16 \cdot π /2 = 8 π$ , the inner one (clockwise — with a minus sign) $- π /2$ . In total $8 π - π /2 = 15 π /2$ — it agrees.

If the inner circle is mistakenly traversed anticlockwise too, the result is $8 π + π /2$ — the answer for "a disc plus another disc", not for the annulus.

Answer: $\frac{15 π}{2}$ .

Problem 7* (A singular point inside the contour). Compute

\oint_{Γ} \frac{( x - y ) d x + ( x + y ) d y}{x ^{2} + y ^{2}},

where $Γ$ is the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1$ , traversed anticlockwise. What is the same integral around any closed contour that does not enclose the origin?

Solution.

Parametrising the ellipse gives an unwieldy fraction with $4 cos^{2} t + 9 sin^{2} t$ in the denominator. Green's theorem cannot be applied to the ellipse itself: at the origin, inside the contour, the field is not defined. The trick is to cut the singular point out with a small circle.

The integrand in Green's theorem. $P = \frac{x - y}{x ^{2} + y ^{2}}$ , $Q = \frac{x + y}{x ^{2} + y ^{2}}$ . Direct differentiation gives

\frac{\partial Q}{\partial x} = \frac{y ^{2} - x ^{2} - 2 x y}{( x ^{2} + y ^{2} ) ^{2}} = \frac{\partial P}{\partial y},

that is, $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$ everywhere except the origin.

Cutting out the singularity. Let $γ_{ε}$ be the circle $x^{2} + y^{2} = ε^{2}$ , with $ε$ small enough for it to lie inside the ellipse. Between the ellipse and $γ_{ε}$ lies a region with no singular points. Its boundary is the ellipse anticlockwise and $γ_{ε}$ clockwise, and by Green's theorem the integral around this boundary equals the integral of zero:

\oint_{Γ} - \oint_{γ_{ε}} = 0 ⟹ \oint_{Γ} = \oint_{γ_{ε}},

where both curves are now traversed anticlockwise. The integral around the ellipse has been reduced to one around a circle of our own choosing.

The integral around the circle. $x = ε cos t$ , $y = ε sin t$ :

(x - y) d x + (x + y) d y = ε^{2} (- (cos t - sin t) sin t + (cos t + sin t) cos t) d t = ε^{2} d t .

Dividing by $x^{2} + y^{2} = ε^{2}$ gives $\oint_{γ_{ε}} = \int_{0}^{2 π} d t = 2 π$ .

A contour that does not enclose the origin. Then the field is smooth inside it, Green's theorem applies directly, and the integral is $0$ .

The shape of the contour did not enter the answer at all: all that matters is whether it encloses the singular point. The field here is the sum of $\frac{x d x + y d y}{x ^{2} + y ^{2}} = d ln x^{2} + y^{2}$ , whose circulation is always zero, and the polar-angle field from the related problem on line integrals, which gives the $2 π$ .

Answer: $2 π$ ; around a contour that does not enclose the origin, $0$ .

Statement

Hypotheses to check

Two common uses

Common mistakes

Problems

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