Circulation Workshop on Piecewise-Bounded Regions

Circulation on Regions with Several Boundary Pieces

A circulation problem can look complicated when the boundary has several straight segments or a curved edge. The calculation becomes more manageable when the boundary is treated as a collection of pieces and the region is described with bounds that fit its shape.

For a planar vector field $\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle$, circulation around a closed curve $C$ is the line integral $\oint_C P\,dx+Q\,dy$. Green’s theorem can replace that boundary integral with an area integral of the scalar curl density. The theorem is especially useful when the curl density and the region’s bounds make the area integral straightforward; a direct boundary calculation then provides an independent check.

Green’s theorem for circulation: If $P$ and $Q$ have continuous first partial derivatives on an open set containing a region $D$, and $C$ is the positively oriented, piecewise-smooth, simple closed boundary of $D$, then $$ \oint_C P\,dx+Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA. $$ Positive orientation means that $D$ stays to the left as $C$ is traversed; for a simple boundary without holes, this is counterclockwise.

The expression $Q_x-P_y$ is the planar curl density. A region may have corners, and its boundary may consist of lines and arcs, provided the boundary is piecewise smooth and the other hypotheses hold. The key tasks are to preserve the orientation and describe all of $D$ exactly once in the area integral.

Why the Area Integral Gives Circulation

The theorem is the general result used in the examples below. Its mechanism can be seen directly on a rectangle, where each side’s contribution combines with the opposite side to produce a partial derivative across the interior.

Derivation on a rectangle: Let $D=[a,b]\times[c,d]$, with its boundary traversed counterclockwise, and suppose $P,Q$ have continuous first partial derivatives on a neighborhood of $D$. Add the four oriented side integrals: $$ \begin{aligned} \oint_C P\,dx+Q\,dy ={}&\int_a^b P(x,c)\,dx+\int_c^d Q(b,y)\,dy\\ &-\int_a^b P(x,d)\,dx-\int_c^d Q(a,y)\,dy. \end{aligned} $$ Group the terms involving $P$ and those involving $Q$. By the fundamental theorem of calculus, $$ \begin{aligned} \oint_C P\,dx+Q\,dy ={}&\int_a^b\bigl(P(x,c)-P(x,d)\bigr)\,dx\\ &+\int_c^d\bigl(Q(b,y)-Q(a,y)\bigr)\,dy\\ ={}&-\int_a^b\int_c^d P_y(x,y)\,dy\,dx +\int_c^d\int_a^b Q_x(x,y)\,dx\,dy\\ ={}&\iint_D (Q_x-P_y)\,dA. \end{aligned} $$ The last equality uses equality of iterated integrals for continuous functions. Opposite boundary pieces measure changes across the rectangle; the fundamental theorem turns those changes into the interior derivatives.

For triangles and curved regions, Green’s theorem gives the corresponding relation under its stated hypotheses. The practical challenge is usually not the theorem itself, but choosing bounds and tracking the orientation of each boundary piece.

Worked Example: A Triangular Region

Worked Example: Circulation Around a Triangle

Let $\mathbf{F}(x,y)=\langle -y,x\rangle$, and let $D$ be the triangle with vertices $(0,0)$, $(2,0)$, and $(0,1)$. Find the counterclockwise circulation using Green’s theorem, then check by direct boundary integration.

Here $P=-y$ and $Q=x$, so the curl density is $$ Q_x-P_y=1-(-1)=2. $$ The sloping edge has equation $y=1-\frac{x}{2}$. A vertical slice runs from $y=0$ to that edge, while $x$ runs from $0$ to $2$. Therefore, $$ \begin{aligned} \oint_C -y\,dx+x\,dy &=\int_0^2\int_0^{1-x/2}2\,dy\,dx\\ &=\int_0^2 2\left(1-\frac{x}{2}\right)\,dx\\ &=\int_0^2(2-x)\,dx\\ &=\left[2x-\frac{x^2}{2}\right]_0^2 =4-2=2. \end{aligned} $$

For the direct check, traverse the bottom edge from $(0,0)$ to $(2,0)$, the diagonal to $(0,1)$, and the vertical edge back to $(0,0)$. On the bottom edge, $y=0$ and $dy=0$, so its contribution is zero. Parametrize the diagonal by $x=2-2t,\ y=t$, for $0\le t\le1$. Then $dx=-2\,dt,\ dy=dt$, and $$ P\,dx+Q\,dy=(-t)(-2\,dt)+(2-2t)\,dt=2\,dt. $$ Its contribution is $\int_0^1 2\,dt=2$. On the final edge $x=0$, so $dx=0$ and $Q=x=0$; its contribution is zero. The boundary total is $0+2+0=2$, agreeing with the area integral.

Choosing Bounds for Straight-Sided Regions

For a triangle, one convenient description is often a vertical slice: identify the left and right $x$-values, then write the lower and upper $y$-boundaries as functions of $x$. A horizontal-slice description may be simpler if the relevant boundaries are easier to express as $x$-values depending on $y$. Either order is valid when the bounds cover the region exactly once.

For a rectangle, constant bounds usually make the area integral especially direct. The next example uses a nonconstant curl density, so it also illustrates why the density must be integrated over the whole region rather than evaluated at one convenient point.

Worked Example: A Rectangular Region

Let $\mathbf{F}(x,y)=\langle xy,x^2+y\rangle$, and let $D=[0,2]\times[1,3]$. Compute the counterclockwise circulation using Green’s theorem and verify it around the four edges.

We have $P=xy$, $Q=x^2+y$, $Q_x=2x$, and $P_y=x$. Thus the curl density is $x$, and the rectangle has bounds $0\le x\le2$, $1\le y\le3$: $$ \begin{aligned} \oint_C xy\,dx+(x^2+y)\,dy &=\int_0^2\int_1^3 x\,dy\,dx\\ &=\int_0^2 x(3-1)\,dx\\ &=\int_0^2 2x\,dx =\left[x^2\right]_0^2=4. \end{aligned} $$

For a direct check, add the four sides in counterclockwise order. On the bottom, $y=1$, $x$ goes from $0$ to $2$, and $dy=0$: $$ \int_0^2 x(1)\,dx=\left[\frac{x^2}{2}\right]_0^2=2. $$ On the right, $x=2$, $y$ goes from $1$ to $3$, and $dx=0$: $$ \int_1^3(4+y)\,dy =\left[4y+\frac{y^2}{2}\right]_1^3 =\left(12+\frac92\right)-\left(4+\frac12\right)=12. $$ On the top, $y=3$ and $x$ goes from $2$ to $0$: $$ \int_2^0 3x\,dx =\left[\frac{3x^2}{2}\right]_2^0=-6. $$ On the left, $x=0$ and $y$ goes from $3$ to $1$: $$ \int_3^1 y\,dy =\left[\frac{y^2}{2}\right]_3^1=-4. $$ The sum is $2+12-6-4=4$, which matches Green’s theorem.

Worked Example: A Region with a Circular Arc

A curved boundary does not prevent using Green’s theorem. It does mean that the bounds or the boundary parametrization must describe the arc accurately. In the first-quadrant quarter disk below, vertical slices have a simple upper boundary, and the circular arc has a simple trigonometric parametrization.

Worked Example: Circulation Around a Quarter Disk

Let $\mathbf{F}(x,y)=\langle 0,x^2\rangle$, and let $D$ be the quarter disk $x^2+y^2\le4$, $x\ge0$, $y\ge0$. Find its counterclockwise circulation and check it directly.

Here $P=0$, $Q=x^2$, so $Q_x-P_y=2x$. For each $x$ between $0$ and $2$, $y$ ranges from $0$ to $\sqrt{4-x^2}$. Green’s theorem gives $$ \begin{aligned} \oint_C x^2\,dy &=\int_0^2\int_0^{\sqrt{4-x^2}}2x\,dy\,dx\\ &=\int_0^2 2x\sqrt{4-x^2}\,dx. \end{aligned} $$ Set $u=4-x^2$, so $du=-2x\,dx$. When $x=0$, $u=4$; when $x=2$, $u=0$. Hence $$ \int_0^2 2x\sqrt{4-x^2}\,dx =\int_4^0-\sqrt{u}\,du =\int_0^4\sqrt{u}\,du =\left[\frac{2}{3}u^{3/2}\right]_0^4 =\frac{16}{3}. $$

For the boundary check, traverse the bottom radius from $(0,0)$ to $(2,0)$, the circular arc to $(0,2)$, then the left radius back to $(0,0)$. On the bottom radius $dy=0$, so the contribution is zero. On the arc, use $x=2\cos t,\ y=2\sin t$, with $0\le t\le\frac{\pi}{2}$; then $dy=2\cos t\,dt$, and $$ x^2\,dy=(2\cos t)^2(2\cos t\,dt)=8\cos^3 t\,dt. $$ Therefore the arc contributes $$ \int_0^{\pi/2}8\cos^3t\,dt =8\int_0^{\pi/2}\cos t(1-\sin^2t)\,dt =8\left[\sin t-\frac{\sin^3t}{3}\right]_0^{\pi/2} =8\left(1-\frac13\right)=\frac{16}{3}. $$ On the left radius $x=0$, so $x^2\,dy=0$. The total boundary integral is $0+\frac{16}{3}+0=\frac{16}{3}$, as required.

Orientation and a Common Sign Error

Orientation is part of the circulation, not a cosmetic choice. Reversing the direction of a closed curve changes the sign of its line integral. Green’s theorem in the form stated above requires positive orientation, so a clockwise traversal must either be reversed or given a minus sign.

For example, the triangle calculation above gives $2$ counterclockwise. If its boundary is traversed clockwise, the circulation is $-2$, not $2$. A common incorrect calculation is to keep the positive curl density and write $$ \oint_{C_{\mathrm{clockwise}}} P\,dx+Q\,dy =\iint_D (Q_x-P_y)\,dA=2. $$ This applies the positively oriented version of Green’s theorem to a negatively oriented boundary. The correction is $$ \oint_{C_{\mathrm{clockwise}}} P\,dx+Q\,dy =-\iint_D (Q_x-P_y)\,dA=-2. $$ The direct boundary check makes the same sign change because reversing any parametrized path reverses the direction of integration.

Practical check: Before integrating, write the boundary pieces in traversal order and confirm that the region remains on the left. In an area integral, also compute $Q_x-P_y$ in that order; using $P_y-Q_x$ changes the sign.

A Reliable Workflow

For a circulation problem on a piecewise-bounded region, use the following sequence. The final direct check is especially valuable when the boundary has multiple pieces or a curved segment.

Identify the field and the orientation.
Write $P$ and $Q$, and confirm that the boundary is traversed counterclockwise for the standard Green’s theorem formula.

Compute the curl density.
Differentiate to obtain $Q_x-P_y$, keeping the order of the terms explicit.

Describe the entire region.
Choose bounds that cover the region once. For curved edges, solve the boundary equation for the upper or lower slice as appropriate.

Evaluate the area integral.
Integrate the density over the stated bounds, showing how each limit is used.

Check along the boundary.
Parametrize or describe every edge and arc in its actual traversal direction, compute its contribution, and add the pieces.

Why These Checks Matter

The area and boundary calculations organize the same circulation in different ways. Green’s theorem is often more efficient when the curl density is simple, while direct integration exposes mistakes in edge direction, parametrization, or omitted boundary pieces. Agreement between the two methods is strong evidence that the setup is consistent; disagreement is a prompt to check the hypotheses, orientation, derivatives, and bounds.

In particular, the region must be the one actually enclosed by the chosen boundary, and the field must have the required continuous first partial derivatives on a neighborhood of that region. If the field is not defined or smooth enough there, the stated theorem cannot simply be applied. When the hypotheses hold, corners and circular arcs are allowed: piecewise smoothness accommodates finitely many smooth pieces joined at endpoints.

Check Your Understanding

Use the stated orientation and show enough work to make your bounds and signs clear.

For $\mathbf{F}(x,y)=\langle -y,x\rangle$ on the counterclockwise triangle with vertices $(0,0),(3,0),(0,2)$, find the curl density and circulation using Green’s theorem.
For $D=[1,3]\times[0,2]$, write an iterated-integral description of the circulation in terms of $Q_x-P_y$.
For a clockwise traversal of a region whose positively oriented circulation is $7$, what is the circulation? Explain the sign.
For the first-quadrant quarter disk $x^2+y^2\le9$, give vertical-slice bounds for an area integral over the region.
State two things to check if a direct boundary calculation disagrees with the Green’s theorem calculation.