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Green’s Theorem in the Plane · Tutorial 807 of 999

Setting Double-Integral Bounds for Green’s Theorem

Learn to translate planar regions into correct Cartesian bounds and use those bounds to set up both forms of Green’s theorem.

Advanced 9 min read

What You'll Learn

  • Describe regions using vertical or horizontal slices and choose the more useful order of integration.
  • Find Cartesian limits for a disk from its circle equation.
  • Set up triangle bounds when the upper boundary changes between edges.
  • Split piecewise-bounded regions into intervals with consistent limits.
  • Set up Green’s theorem circulation and flux double integrals over the same region.
  • Check that bounds include the whole region without adding or omitting pieces.

Why Bounds Matter in Green’s Theorem

Green’s theorem relates a line integral around a positively oriented, simple closed curve to a double integral over the region it encloses. Once the appropriate Green’s theorem integrand is identified, the area integral still needs correct limits. Those limits must describe the entire region: a bound that uses the wrong edge, covers only part of the region, or extends beyond it produces an incorrect setup.

This tutorial focuses on expressing disks, triangles, and piecewise-bounded regions with Cartesian limits. The essential task is to imagine slicing the region. A vertical slice leads to an integral with \(dy\) first; a horizontal slice leads to an integral with \(dx\) first. If the slice’s endpoints change from one part of the region to another, the bounds must be split into pieces.

Definition: A region is vertically simple if each vertical line through it meets the region in one interval, and the region can be written as \(a\le x\le b\), \(g_1(x)\le y\le g_2(x)\). It is horizontally simple if each horizontal line through it meets the region in one interval, and it can be written as \(c\le y\le d\), \(h_1(y)\le x\le h_2(y)\). A region may have either description or both.

The Green’s Theorem Integrals to Set Up

Let \(\mathbf{F}=(P,Q)\) be a vector field whose first partial derivatives are continuous on an open set containing a simple planar region \(D\) and its boundary \(C\). With \(C\) positively oriented (counterclockwise), the circulation form and outward-flux form of Green’s theorem are:

$$ \oint_C P\,dx+Q\,dy = \iint_D (Q_x-P_y)\,dA $$
$$ \oint_C P\,dy-Q\,dx = \iint_D (P_x+Q_y)\,dA. $$

Here \(Q_x\) denotes \(\frac{\partial Q}{\partial x}\), and similarly for the other partial derivatives. The two area integrals use the same bounds for \(D\); only their integrands differ. If \(D\) is vertically simple, for example, the circulation integral is set up as an iterated integral with \(Q_x-P_y\) between the lower and upper \(y\)-limits. The flux integral uses those same limits with \(P_x+Q_y\). Green’s theorem itself requires the stated smoothness and boundary conditions; having plausible-looking bounds does not establish those hypotheses.

How to Read Cartesian Bounds

For a vertically simple region, the notation

$$ \iint_D f(x,y)\,dA = \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx $$

means: fix \(x\) between \(a\) and \(b\), let \(y\) run from the lower edge \(g_1(x)\) to the upper edge \(g_2(x)\), and then let \(x\) range across the region. For a horizontally simple region, the roles are reversed:

$$ \iint_D f(x,y)\,dA = \int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy. $$

In either order, the inner limits are the two endpoints of a slice through the region. The outer limits give the full range of the variable held fixed while taking those slices. If the endpoints change formula, split the outer interval and add the resulting integrals. This partitions the area integral; it does not add extra boundary segments to \(C\).

Worked Example: A Disk

Worked Example: A Disk

Set up the circulation and flux area integrals for the disk centered at \((1,-1)\) with radius \(2\).

Step 1: Write the disk condition. A point \((x,y)\) is in the disk when its squared distance from the center is at most the squared radius:

$$ (x-1)^2+(y+1)^2\le 4. $$

Step 2: Find vertical limits. The leftmost and rightmost \(x\)-coordinates are \(1-2=-1\) and \(1+2=3\). For each fixed \(x\), solve the circle equation for \(y\):

$$ (y+1)^2=4-(x-1)^2, \qquad y=-1\pm\sqrt{4-(x-1)^2}. $$

The minus sign gives the lower half and the plus sign gives the upper half. Thus the disk is described by \(-1\le x\le 3\) and \(-1-\sqrt{4-(x-1)^2}\le y\le -1+\sqrt{4-(x-1)^2}\).

Step 3: Insert the Green’s theorem integrands. The resulting setups are

$$ \oint_C P\,dx+Q\,dy = \int_{-1}^{3} \int_{-1-\sqrt{4-(x-1)^2}}^{-1+\sqrt{4-(x-1)^2}} (Q_x-P_y)\,dy\,dx, $$
$$ \oint_C P\,dy-Q\,dx = \int_{-1}^{3} \int_{-1-\sqrt{4-(x-1)^2}}^{-1+\sqrt{4-(x-1)^2}} (P_x+Q_y)\,dy\,dx. $$

Check: At the extreme \(x\)-values \(-1\) and \(3\), the square root is \(0\), so the vertical slice collapses to the points \((-1,-1)\) and \((3,-1)\). At the center line \(x=1\), the limits are \(-1-2=-3\) and \(-1+2=1\), the disk’s bottom and top. The limits cover the disk from bottom to top without extending beyond its circle.

Worked Example: A Triangle

Worked Example: A Triangle

Set up Green’s theorem area integrals over the triangle with vertices \((0,0)\), \((4,0)\), and \((1,3)\).

Step 1: Identify the boundary lines. The edge from \((0,0)\) to \((1,3)\) has slope \((3-0)/(1-0)=3\), so its equation is \(y=3x\). The edge from \((1,3)\) to \((4,0)\) has slope \((0-3)/(4-1)=-1\). Using the point \((1,3)\), its equation is \(y-3=-(x-1)\), or \(y=4-x\). The third edge is \(y=0\).

Step 2: Use vertical slices. The top boundary changes at the vertex with \(x=1\). For \(0\le x\le 1\), a slice runs from \(y=0\) to \(y=3x\). For \(1\le x\le 4\), it runs from \(y=0\) to \(y=4-x\). Therefore:

$$ \iint_D f(x,y)\,dA = \int_0^1\int_0^{3x} f(x,y)\,dy\,dx + \int_1^4\int_0^{4-x} f(x,y)\,dy\,dx. $$

Step 3: Set up both Green’s theorem forms. Substituting the appropriate density for \(f\) gives:

$$ \oint_C P\,dx+Q\,dy = \int_0^1\int_0^{3x}(Q_x-P_y)\,dy\,dx + \int_1^4\int_0^{4-x}(Q_x-P_y)\,dy\,dx, $$
$$ \oint_C P\,dy-Q\,dx = \int_0^1\int_0^{3x}(P_x+Q_y)\,dy\,dx + \int_1^4\int_0^{4-x}(P_x+Q_y)\,dy\,dx. $$

Check: At \(x=0\), the lower and upper limits are both \(0\); at \(x=1\), both formulas for the upper edge give \(3\); at \(x=4\), the limits again meet at \(0\). The slices rise from the bottom edge to the sloping top edge across the full range \(0\le x\le4\).

The same triangle can be described with horizontal slices. Solving the left edge \(y=3x\) for \(x\) gives \(x=y/3\). Solving \(y=4-x\) gives \(x=4-y\). For \(0\le y\le3\), a horizontal slice runs from the left edge to the right edge:

$$ \iint_D f(x,y)\,dA = \int_0^3\int_{y/3}^{4-y} f(x,y)\,dx\,dy. $$

At \(y=0\), these limits give \(0\le x\le4\); at \(y=3\), both give \(x=1\). The endpoints meet at the triangle’s top vertex, as they should.

Worked Example: A Piecewise-Bounded Trapezoid

Worked Example: A Piecewise-Bounded Trapezoid

Set up Green’s theorem area integrals over the trapezoid with vertices \((0,0)\), \((3,0)\), \((2,2)\), and \((1,2)\), in counterclockwise order.

Step 1: Find the edges that control vertical slices. The left sloping edge from \((0,0)\) to \((1,2)\) has equation \(y=2x\). The right sloping edge from \((3,0)\) to \((2,2)\) has slope \((2-0)/(2-3)=-2\); using \((3,0)\), its equation is \(y=-2(x-3)=6-2x\). The bottom and top edges are \(y=0\) and \(y=2\).

Step 2: Split where the slice endpoints change. For \(0\le x\le1\), the slice runs from \(y=0\) to \(y=2x\). For \(1\le x\le2\), it runs from \(y=0\) to \(y=2\). For \(2\le x\le3\), it runs from \(y=0\) to \(y=6-2x\). Hence:

$$ \begin{aligned} \iint_D f(x,y)\,dA ={}& \int_0^1\int_0^{2x} f(x,y)\,dy\,dx\\ &+\int_1^2\int_0^2 f(x,y)\,dy\,dx\\ &+\int_2^3\int_0^{6-2x} f(x,y)\,dy\,dx. \end{aligned} $$

Step 3: Use those bounds in Green’s theorem. The circulation setup is

$$ \begin{aligned} \oint_C P\,dx+Q\,dy ={}& \int_0^1\int_0^{2x}(Q_x-P_y)\,dy\,dx\\ &+\int_1^2\int_0^2(Q_x-P_y)\,dy\,dx\\ &+\int_2^3\int_0^{6-2x}(Q_x-P_y)\,dy\,dx. \end{aligned} $$

The flux setup uses the same three regions of integration, replacing \(Q_x-P_y\) by \(P_x+Q_y\).

Check: At \(x=0\) and \(x=3\), the slice has zero height. At \(x=1\), both adjoining upper-bound formulas give \(2\); at \(x=2\), the middle and right-hand formulas also give \(2\). The three intervals meet without a gap or overlap of positive area.

Horizontal slices give a shorter description. The left edge is \(x=y/2\). The right edge \(y=6-2x\) gives \(x=3-y/2\). For \(0\le y\le2\), each slice runs between these two edges:

$$ \iint_D f(x,y)\,dA = \int_0^2\int_{y/2}^{3-y/2} f(x,y)\,dx\,dy. $$

At \(y=0\), the limits are \(x=0\) and \(x=3\); at \(y=2\), they are \(x=1\) and \(x=2\). These are exactly the endpoints of the bottom and top edges. Either order is valid; horizontal slices avoid the three-piece split.

Why the Slicing Method Works

The bounds are not guesses: they encode every point in the region exactly once, apart from shared edges, which have zero area. For example, in the trapezoid, a vertical slice with \(x\) between \(0\) and \(1\) intersects the region from the bottom edge to \(y=2x\). Between \(1\) and \(2\), it reaches the top edge \(y=2\). Between \(2\) and \(3\), it reaches \(y=6-2x\). These intervals partition the full horizontal range \([0,3]\); within each interval, the lower and upper functions describe precisely the points of the slice that belong to \(D\).

For an integrable density \(f\), the iterated integrals over these slices therefore add to \(\iint_D f\,dA\), by the standard change in order and evaluation of double integrals on simple regions. The same reasoning applies to \(Q_x-P_y\) and \(P_x+Q_y\) when the hypotheses of Green’s theorem hold. Splitting limits is simply a way to represent one region as the union of simpler vertically or horizontally simple pieces.

Practical rule: Sketch or identify the boundary, choose vertical or horizontal slices, solve the relevant boundary equations for the inner variable, and split the outer interval wherever the slice endpoints change. Then use the resulting bounds with the Green’s theorem density.

A Common Bounds Error

For the triangle with vertices \((0,0)\), \((4,0)\), and \((1,3)\), a tempting but incorrect setup is

$$ \int_0^4\int_0^{4-x} f(x,y)\,dy\,dx. $$

This uses the edge \(y=4-x\) across the entire interval \(0\le x\le4\), even though that edge is the triangle’s upper boundary only when \(1\le x\le4\). For instance, at \(x=0\), the incorrect bounds allow \(0\le y\le4\), while the triangle narrows to the single point \((0,0)\). The error both includes points outside the triangle and fails to follow its actual top edge over \(0\le x\le1\).

The correction is to split at the vertex \(x=1\) and use the correct edge on each interval:

$$ \int_0^1\int_0^{3x} f(x,y)\,dy\,dx + \int_1^4\int_0^{4-x} f(x,y)\,dy\,dx. $$

A related mistake is to treat the split at \(x=1\), or at another change in slice limits, as an additional boundary edge. It is only a seam in the area-integral description. Green’s theorem’s curve \(C\) consists of the actual exterior boundary segments, each traversed once in the positive direction.

Putting the Setup Together

1
Identify the region and its exterior edges.
Record the extreme \(x\)- and \(y\)-values and note the vertices where an edge or slice endpoint changes.
2
Choose a slicing direction.
Use vertical slices for \(dy\,dx\), or horizontal slices for \(dx\,dy\). Prefer the order with fewer pieces when convenient.
3
Find the slice endpoints.
Solve the relevant boundary equations for the inner variable, and check which endpoint is lower or left and which is upper or right.
4
Split wherever the formulas change.
Add the integrals for the pieces so the full region is covered once, with no gaps or extra area.
5
Insert the Green’s theorem density.
Use \(Q_x-P_y\) for circulation or \(P_x+Q_y\) for outward flux, with the same region bounds.

Good bounds provide a transparent description of the area enclosed by the curve. Checking the extreme slices and every point where the formulas change is a reliable way to catch a missing boundary piece before applying Green’s theorem.

Check Your Understanding

For each question, focus on the region described by the limits before considering any evaluation.

  1. For the disk \((x-2)^2+(y+1)^2\le9\), what are its vertical \(x\)-range and its lower and upper \(y\)-limits?
  2. For a triangle with vertices \((0,0)\), \((3,0)\), and \((0,2)\), write bounds using vertical slices.
  3. What geometric event indicates that the outer interval may need to be split when using vertical slices?
  4. For a vertically simple region, which Green’s theorem density belongs in the circulation double integral?
  5. Why is a vertical line used to split a piecewise iterated integral not an additional edge of the Green’s theorem boundary?

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