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line integrals practice problems & Exercises with Solutions

Type
Practice problems
Problems
13
Levels
Easy · Medium · Hard

Line integrals are a fundamental tool in multivariable calculus, extending ordinary integration to functions evaluated along curves in two- or three-dimensional space. Mastering line integrals requires fluency in curve parameterization, integration with respect to arc length, and computation of line integrals of vector fields. This page offers a structured progression of problems covering scalar line integrals, vector field integrals, the Fundamental Theorem of Line Integrals, conservative vector fields, path independence, and Green’s Theorem, all key LSI concepts that appear consistently across calculus III curricula.

Contents5 sections

Scalar Line Integrals with Respect to Arc Length

These problems focus on integrating a scalar function f(x,y) f(x, y) or f(x,y,z) f(x, y, z) along a curve C C with respect to arc length ds ds . The key formula is ∫Cf ds=∫abf(r(t)) ∥r′(t)∥ dt\int_C f\,ds = \int_a^b f(\mathbf{r}(t))\,\|\mathbf{r}'(t)\|\,dt. Begin by parameterizing the curve, computing ∥r′(t)∥ \|\mathbf{r}'(t)\| , and substituting into the integrand.

Problem 1: Line Integral Along a Line Segment in 2D

Easy

Let C C be the line segment from (0,0) (0, 0) to (3,4) (3, 4) .

  1. Parameterize the curve C C using a parameter t∈[0,1] t \in [0, 1] .
  2. Evaluate the line integral ∫C(x+y) ds\displaystyle\int_C (x + y)\,ds.
Hint
Use the linear parameterization r(t)=(1−t)(0,0)+t(3,4) \mathbf{r}(t) = (1-t)(0,0) + t(3,4) . Compute ∥r′(t)∥ \|\mathbf{r}'(t)\| from the constant velocity vector, then substitute x=3t x = 3t and y=4t y = 4t into the integrand.
View Solution
Solution to question 1:

Parameterize: r(t)=(3t, 4t) \mathbf{r}(t) = (3t,\, 4t) , t∈[0,1] t \in [0,1] . Then r′(t)=(3,4) \mathbf{r}'(t) = (3, 4) and ∥r′(t)∥=9+16=5 \|\mathbf{r}'(t)\| = \sqrt{9 + 16} = 5 .

Solution to question 2: ∫C(x+y) ds=∫01(3t+4t)⋅5 dt=5∫017t dt=35∫01t dt=35⋅12=352 \int_C (x+y)\,ds = \int_0^1 (3t + 4t)\cdot 5\,dt = 5\int_0^1 7t\,dt = 35\int_0^1 t\,dt = 35\cdot\frac{1}{2} = \frac{35}{2}

Problem 2: Line Integral Along a Circular Arc

Easy

Let C C be the upper semicircle of radius 2 2 centered at the origin, traversed counterclockwise from (2,0) (2, 0) to (−2,0) (-2, 0) .

  1. Write a parameterization of C C and compute ∥r′(t)∥ \|\mathbf{r}'(t)\| .
  2. Evaluate ∫Cy ds\displaystyle\int_C y\,ds.
Hint
Use the standard trigonometric parameterization x=2cos⁡t x = 2\cos t , y=2sin⁡t y = 2\sin t for t∈[0,π] t \in [0, \pi] . On this arc, y=2sin⁡t≥0 y = 2\sin t \geq 0 .
View Solution
Solution to question 1:

r(t)=(2cos⁡t, 2sin⁡t) \mathbf{r}(t) = (2\cos t,\, 2\sin t) , t∈[0,π] t \in [0, \pi] . Then r′(t)=(−2sin⁡t, 2cos⁡t) \mathbf{r}'(t) = (-2\sin t,\, 2\cos t) and ∥r′(t)∥=4sin⁡2t+4cos⁡2t=2 \|\mathbf{r}'(t)\| = \sqrt{4\sin^2 t + 4\cos^2 t} = 2 .

Solution to question 2: ∫Cy ds=∫0π2sin⁡t⋅2 dt=4∫0πsin⁡t dt=4[−cos⁡t]0π=4(1+1)=8 \int_C y\,ds = \int_0^{\pi} 2\sin t \cdot 2\,dt = 4\int_0^{\pi}\sin t\,dt = 4\Big[-\cos t\Big]_0^{\pi} = 4(1+1) = 8

Problem 3: Scalar Integral Along a Parabola

Medium

Consider the parabolic arc y=x2 y = x^2 from (0,0) (0, 0) to (1,1) (1, 1) .

  1. Parameterize C C using x=t x = t and find ds ds in terms of dt dt .
  2. Evaluate ∫Cxy ds\displaystyle\int_C xy\,ds.
Hint
With r(t)=(t,t2) \mathbf{r}(t) = (t, t^2) , compute ∥r′(t)∥=1+(2t)2=1+4t2 \|\mathbf{r}'(t)\| = \sqrt{1 + (2t)^2} = \sqrt{1 + 4t^2} . On C C , x=t x = t and y=t2 y = t^2 , so xy=t⋅t2=t3 xy = t \cdot t^2 = t^3 .
View Solution
Solution to question 1:

r(t)=(t,t2) \mathbf{r}(t) = (t, t^2) , t∈[0,1] t \in [0,1] . Then ds=1+4t2 dt ds = \sqrt{1 + 4t^2}\,dt .

Solution to question 2: ∫Cxy ds=∫01t31+4t2 dt \int_C xy\,ds = \int_0^1 t^3\sqrt{1+4t^2}\,dt

Let u=1+4t2 u = 1 + 4t^2 , so du=8t dt du = 8t\,dt and t2=u−14 t^2 = \dfrac{u-1}{4} . Then:

=∫15u−14⋅u⋅du8=132∫15(u3/2−u1/2) du=132[2u5/25−2u3/23]15 = \int_1^5 \frac{u-1}{4}\cdot\sqrt{u}\cdot\frac{du}{8} = \frac{1}{32}\int_1^5 (u^{3/2} - u^{1/2})\,du = \frac{1}{32}\left[\frac{2u^{5/2}}{5} - \frac{2u^{3/2}}{3}\right]_1^5 =132(2⋅2555−2⋅553−25+23)=132(105−1053+415)=132⋅2053+4480≈0.4696 = \frac{1}{32}\left(\frac{2\cdot25\sqrt{5}}{5} - \frac{2\cdot5\sqrt{5}}{3} - \frac{2}{5} + \frac{2}{3}\right) = \frac{1}{32}\left(10\sqrt{5} - \frac{10\sqrt{5}}{3} + \frac{4}{15}\right) = \frac{1}{32}\cdot\frac{20\sqrt{5}}{3} + \frac{4}{480} \approx 0.4696

Line Integrals with Respect to Coordinate Variables

When the integrand is expressed as P dx+Q dy P\,dx + Q\,dy (or P dx+Q dy+R dz P\,dx + Q\,dy + R\,dz in 3D), we substitute the parameterization directly into x x , y y , and their differentials. Unlike arc-length integrals, these are sensitive to the orientation of C C . This section develops the skill of setting up and evaluating such integrals over line segments, circular arcs, and piecewise paths.

Problem 4: Integral of the Form P dx+Q dy P\,dx + Q\,dy Over a Line Segment

Easy

Let C C be the directed line segment from (1,0) (1, 0) to (0,1) (0, 1) .

  1. Parameterize C C and express dx dx and dy dy in terms of dt dt .
  2. Evaluate ∫Cy2 dx+x2 dy\displaystyle\int_C y^2\,dx + x^2\,dy.
Hint
Use r(t)=(1−t, t) \mathbf{r}(t) = (1-t,\, t) for t∈[0,1] t \in [0,1] . Then dx=−dt dx = -dt and dy=dt dy = dt . Substitute x=1−t x = 1-t and y=t y = t .
View Solution
Solution to question 1:

x=1−t x = 1-t , y=t y = t , t∈[0,1] t \in [0,1] ; dx=−dt dx = -dt , dy=dt dy = dt .

Solution to question 2: ∫Cy2 dx+x2 dy=∫01[t2(−1)+(1−t)2(1)]dt=∫01[−t2+1−2t+t2]dt=∫01(1−2t) dt \int_C y^2\,dx + x^2\,dy = \int_0^1 \left[t^2(-1) + (1-t)^2(1)\right]dt = \int_0^1\left[-t^2 + 1 - 2t + t^2\right]dt = \int_0^1(1-2t)\,dt =[t−t2]01=0 = \Big[t - t^2\Big]_0^1 = 0

Problem 5: Piecewise Path (Triangle Boundary)

Medium

Let C C be the boundary of the triangle with vertices A=(0,0) A=(0,0) , B=(2,0) B=(2,0) , D=(0,2) D=(0,2) , traversed counterclockwise.

  1. Decompose C C into three directed line segments C1 C_1 , C2 C_2 , C3 C_3 .
  2. Evaluate ∫Cxy dx+(x+y) dy\displaystyle\int_C xy\,dx + (x+y)\,dy by summing the contributions of each piece.
Hint
Counterclockwise means A→B→D→A A \to B \to D \to A . Parameterize each edge separately: C1 C_1 : from (0,0)(0,0) to (2,0)(2,0); C2 C_2 : from (2,0)(2,0) to (0,2)(0,2); C3 C_3 : from (0,2)(0,2) to (0,0)(0,0).
View Solution
Solution to question 1:

C1 C_1 : (t,0) (t, 0) , t∈[0,2] t\in[0,2] . C2 C_2 : (2−t,t) (2-t, t) , t∈[0,2] t\in[0,2] . C3 C_3 : (0,2−t) (0, 2-t) , t∈[0,2] t\in[0,2] .

Solution to question 2:

On C1 C_1 : y=0 y=0 , dy=0 dy=0 , dx=dt dx=dt .

I1=∫02t⋅0⋅dt+0=0 I_1 = \int_0^2 t\cdot 0\cdot dt + 0 = 0

On C2 C_2 : x=2−t x=2-t , y=t y=t , dx=−dt dx=-dt , dy=dt dy=dt .

I2=∫02[(2−t)t(−1)+(2−t+t)(1)]dt=∫02[−2t+t2+2]dt=[−t2+t33+2t]02=−4+83+4=83 I_2 = \int_0^2\left[(2-t)t(-1) + (2-t+t)(1)\right]dt = \int_0^2\left[-2t+t^2+2\right]dt = \left[-t^2+\frac{t^3}{3}+2t\right]_0^2 = -4+\frac{8}{3}+4 = \frac{8}{3}

On C3 C_3 : x=0 x=0 , dx=0 dx=0 , y=2−t y=2-t , dy=−dt dy=-dt .

I3=∫020+(0+2−t)(−1) dt=∫02(t−2) dt=[t22−2t]02=2−4=−2 I_3 = \int_0^2 0 + (0+2-t)(-1)\,dt = \int_0^2(t-2)\,dt = \left[\frac{t^2}{2}-2t\right]_0^2 = 2-4 = -2 ∫C=I1+I2+I3=0+83−2=23 \int_C = I_1 + I_2 + I_3 = 0 + \frac{8}{3} - 2 = \frac{2}{3}

Line Integrals of Vector Fields and Work

The line integral of a vector field F=⟨P,Q,R⟩\mathbf{F} = \langle P, Q, R \rangle along a curve C C is defined as ∫CF⋅dr=∫CP dx+Q dy+R dz\int_C \mathbf{F}\cdot d\mathbf{r} = \int_C P\,dx + Q\,dy + R\,dz. Physically, this computes the work done by the force field F\mathbf{F} along the path C C . Problems in this section require careful attention to the direction of C C and the structure of F\mathbf{F}.

Problem 6: Work Integral Over a Circular Path

Easy

A particle moves counterclockwise along the full unit circle x2+y2=1 x^2 + y^2 = 1 under the force F(x,y)=⟨−y, x⟩ \mathbf{F}(x,y) = \langle -y,\, x \rangle .

  1. Set up ∫CF⋅dr\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r} using the parameterization r(t)=(cos⁡t,sin⁡t) \mathbf{r}(t) = (\cos t, \sin t) .
  2. Evaluate the integral and interpret the result as work.
Hint
Compute F(r(t))⋅r′(t) \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) and integrate over t∈[0,2π] t \in [0, 2\pi] . Note that sin⁡2t+cos⁡2t=1 \sin^2 t + \cos^2 t = 1 .
View Solution
Solution to question 1:

r′(t)=(−sin⁡t,cos⁡t) \mathbf{r}'(t) = (-\sin t, \cos t) . Then F(r(t))=(−sin⁡t,cos⁡t) \mathbf{F}(\mathbf{r}(t)) = (-\sin t, \cos t) .

F⋅r′=(−sin⁡t)(−sin⁡t)+(cos⁡t)(cos⁡t)=sin⁡2t+cos⁡2t=1 \mathbf{F}\cdot\mathbf{r}' = (-\sin t)(-\sin t) + (\cos t)(\cos t) = \sin^2 t + \cos^2 t = 1 Solution to question 2: W=∫02π1 dt=2π W = \int_0^{2\pi} 1\,dt = 2\pi

The force field F=⟨−y,x⟩\mathbf{F} = \langle -y, x \rangle is always tangent to the unit circle and aligned with the direction of motion, so it does positive work of 2π 2\pi joules (in appropriate units).

Problem 7: Vector Field Integral Along a Space Curve (Helix)

Medium

Let F(x,y,z)=⟨y, −x, z⟩ \mathbf{F}(x,y,z) = \langle y,\, -x,\, z \rangle and let C C be the helix r(t)=(cos⁡t, sin⁡t, t) \mathbf{r}(t) = (\cos t,\, \sin t,\, t) for t∈[0,2π] t \in [0, 2\pi] .

  1. Compute F(r(t)) \mathbf{F}(\mathbf{r}(t)) and r′(t) \mathbf{r}'(t) .
  2. Evaluate ∫CF⋅dr\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}.
Hint
Along the helix, x=cos⁡t x = \cos t , y=sin⁡t y = \sin t , z=t z = t . Substitute into F\mathbf{F} and compute the dot product with r′=(−sin⁡t,cos⁡t,1)\mathbf{r}' = (-\sin t, \cos t, 1). Split the resulting integral into recognizable parts.
View Solution
Solution to question 1:

F(r(t))=(sin⁡t, −cos⁡t, t) \mathbf{F}(\mathbf{r}(t)) = (\sin t,\, -\cos t,\, t) and r′(t)=(−sin⁡t, cos⁡t, 1) \mathbf{r}'(t) = (-\sin t,\, \cos t,\, 1) .

Solution to question 2: F⋅r′=(sin⁡t)(−sin⁡t)+(−cos⁡t)(cos⁡t)+t⋅1=−sin⁡2t−cos⁡2t+t=−1+t \mathbf{F}\cdot\mathbf{r}' = (\sin t)(-\sin t) + (-\cos t)(\cos t) + t\cdot 1 = -\sin^2 t - \cos^2 t + t = -1 + t ∫CF⋅dr=∫02π(t−1) dt=[t22−t]02π=4π22−2π=2π2−2π \int_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}(t - 1)\,dt = \left[\frac{t^2}{2} - t\right]_0^{2\pi} = \frac{4\pi^2}{2} - 2\pi = 2\pi^2 - 2\pi

Problem 8: Work Along a Piecewise Path in 3D

Hard

Let F(x,y,z)=⟨yz, xz, xy⟩\mathbf{F}(x,y,z) = \langle yz,\, xz,\, xy \rangle. Consider the piecewise path C=C1∪C2 C = C_1 \cup C_2 where C1 C_1 goes from (0,0,0) (0,0,0) to (1,0,0) (1,0,0) along the x x -axis, and C2 C_2 goes from (1,0,0) (1,0,0) to (1,1,1) (1,1,1) along the straight line.

  1. Evaluate ∫C1F⋅dr\displaystyle\int_{C_1} \mathbf{F}\cdot d\mathbf{r}.
  2. Evaluate ∫C2F⋅dr\displaystyle\int_{C_2} \mathbf{F}\cdot d\mathbf{r}.
  3. Find the total work done along C C .
Hint
On C1 C_1 : y=z=0 y = z = 0 , so F=0\mathbf{F} = \mathbf{0}. On C2 C_2 : parameterize as (1,t,t) (1, t, t) for t∈[0,1] t \in [0,1] . Notice that F(1,t,t)=(t2,t,t)\mathbf{F}(1,t,t) = (t^2, t, t) and dr=(0,1,1) dtd\mathbf{r} = (0, 1, 1)\,dt.
View Solution
Solution to question 1:

C1 C_1 : r(t)=(t,0,0) \mathbf{r}(t) = (t,0,0) , t∈[0,1] t\in[0,1] . Then F=(0⋅0, t⋅0, t⋅0)=(0,0,0)\mathbf{F} = (0\cdot 0,\, t\cdot 0,\, t\cdot 0) = (0,0,0).

∫C1F⋅dr=0 \int_{C_1}\mathbf{F}\cdot d\mathbf{r} = 0 Solution to question 2:

C2 C_2 : r(t)=(1,t,t) \mathbf{r}(t) = (1,t,t) , t∈[0,1] t\in[0,1] , so dr=(0,1,1) dt d\mathbf{r} = (0,1,1)\,dt .

F(1,t,t)=(t⋅t,  1⋅t,  1⋅t)=(t2,t,t) \mathbf{F}(1,t,t) = (t\cdot t,\; 1\cdot t,\; 1\cdot t) = (t^2, t, t) ∫C2F⋅dr=∫01(t2⋅0+t⋅1+t⋅1) dt=∫012t dt=1 \int_{C_2}\mathbf{F}\cdot d\mathbf{r} = \int_0^1(t^2\cdot 0 + t\cdot 1 + t\cdot 1)\,dt = \int_0^1 2t\,dt = 1 Solution to question 3: Wtotal=0+1=1 W_{\text{total}} = 0 + 1 = 1

Conservative Vector Fields and the Fundamental Theorem of Line Integrals

A vector field F\mathbf{F} is conservative if it equals the gradient of a scalar potential function f f , i.e., F=∇f\mathbf{F} = \nabla f. When F\mathbf{F} is conservative, the Fundamental Theorem of Line Integrals states that ∫CF⋅dr=f(r(b))−f(r(a))\int_C \mathbf{F}\cdot d\mathbf{r} = f(\mathbf{r}(b)) - f(\mathbf{r}(a)), making computation independent of the path. The test for conservatism in 2D is ∂P/∂y=∂Q/∂x\partial P/\partial y = \partial Q/\partial x.

Problem 9: Identifying a Conservative Field and Finding a Potential

Medium

Let F(x,y)=⟨2xy+1,  x2−3y2⟩ \mathbf{F}(x,y) = \langle 2xy + 1,\; x^2 - 3y^2 \rangle .

  1. Verify that F\mathbf{F} is conservative by checking the appropriate partial derivatives.
  2. Find a potential function f(x,y) f(x, y) such that ∇f=F\nabla f = \mathbf{F}.
  3. Use the Fundamental Theorem to evaluate ∫CF⋅dr\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r} from (0,0) (0,0) to (2,1) (2,1) .
Hint
Check ∂P/∂y=2x\partial P/\partial y = 2x and ∂Q/∂x=2x\partial Q/\partial x = 2x. To find f f , integrate fx=2xy+1 f_x = 2xy+1 with respect to x x , then differentiate the result with respect to y y and match with Q Q .
View Solution
Solution to question 1: ∂P∂y=2x,∂Q∂x=2x  ⟹  F is conservative. \frac{\partial P}{\partial y} = 2x, \qquad \frac{\partial Q}{\partial x} = 2x \implies \mathbf{F} \text{ is conservative.} Solution to question 2:

Integrate fx=2xy+1 f_x = 2xy+1 : f=x2y+x+g(y) f = x^2y + x + g(y) .

Then fy=x2+g′(y)=x2−3y2  ⟹  g′(y)=−3y2  ⟹  g(y)=−y3 f_y = x^2 + g'(y) = x^2 - 3y^2 \implies g'(y) = -3y^2 \implies g(y) = -y^3 .

f(x,y)=x2y+x−y3 f(x,y) = x^2 y + x - y^3 Solution to question 3: ∫CF⋅dr=f(2,1)−f(0,0)=(4⋅1+2−1)−0=5 \int_C \mathbf{F}\cdot d\mathbf{r} = f(2,1) - f(0,0) = (4\cdot1 + 2 - 1) - 0 = 5

Problem 10: Path Independence (Non-Conservative vs. Conservative)

Medium

Consider F(x,y)=⟨y2,  2xy⟩\mathbf{F}(x,y) = \langle y^2,\; 2xy \rangle and two paths from A=(0,0) A=(0,0) to B=(1,2) B=(1,2) : the straight line C1 C_1 and the parabola C2 C_2 given by y=2x2 y = 2x^2 .

  1. Verify that F\mathbf{F} is conservative and find its potential function.
  2. Evaluate ∫C1F⋅dr\int_{C_1}\mathbf{F}\cdot d\mathbf{r} directly using the straight-line parameterization.
  3. Confirm path independence by computing ∫C2F⋅dr\int_{C_2}\mathbf{F}\cdot d\mathbf{r} via the potential function.
Hint
Check ∂y(y2)=2y\partial_y(y^2) = 2y and ∂x(2xy)=2y\partial_x(2xy) = 2y. For the potential, integrate fx=y2f_x = y^2 to get f=xy2+g(y)f = xy^2 + g(y), then use fy=2xy+g′(y)=2xyf_y = 2xy + g'(y) = 2xy to find gg.
View Solution
Solution to question 1:

∂y(y2)=2y=∂x(2xy)\partial_y(y^2) = 2y = \partial_x(2xy), so F\mathbf{F} is conservative.

From fx=y2⇒f=xy2+g(y)f_x = y^2 \Rightarrow f = xy^2 + g(y). Then fy=2xy+g′(y)=2xy⇒g′(y)=0⇒g=Cf_y = 2xy + g'(y) = 2xy \Rightarrow g'(y) = 0 \Rightarrow g = C.

f(x,y)=xy2 f(x,y) = xy^2 Solution to question 2:

C1 C_1 : (t,2t) (t, 2t) , t∈[0,1] t\in[0,1] ; dr=(1,2) dt d\mathbf{r} = (1, 2)\,dt .

∫C1=∫01[(2t)2+2t(2t)⋅2]dt=∫01(4t2+8t2) dt=12∫01t2 dt=4 \int_{C_1} = \int_0^1 \left[(2t)^2 + 2t(2t)\cdot 2\right]dt = \int_0^1(4t^2 + 8t^2)\,dt = 12\int_0^1 t^2\,dt = 4 Solution to question 3: ∫C2F⋅dr=f(1,2)−f(0,0)=1⋅4−0=4✓ \int_{C_2}\mathbf{F}\cdot d\mathbf{r} = f(1,2) - f(0,0) = 1\cdot 4 - 0 = 4 \checkmark

Problem 11: Conservative Field in 3D (Finding a Potential Function)

Hard

Let F(x,y,z)=⟨2xy+z2,  x2,  2xz⟩\mathbf{F}(x,y,z) = \langle 2xy + z^2,\; x^2,\; 2xz \rangle.

  1. Verify that F\mathbf{F} is conservative by checking that ∇×F=0\nabla \times \mathbf{F} = \mathbf{0}.
  2. Find a potential function ff such that ∇f=F\nabla f = \mathbf{F}.
  3. Evaluate ∫CF⋅dr\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r} along any path from (1,0,1)(1,0,1) to (2,1,3)(2,1,3).
Hint
For the curl, check three conditions: F3,y=F2,zF_{3,y} = F_{2,z}, F1,z=F3,xF_{1,z} = F_{3,x}, and F2,x=F1,yF_{2,x} = F_{1,y}. To find ff, start by integrating fx=2xy+z2f_x = 2xy + z^2 with respect to xx, then determine g(y,z)g(y,z) by matching fyf_y and fzf_z.
View Solution
Solution to question 1: ∇×F=∣ijk∂x∂y∂z2xy+z2x22xz∣ \nabla\times\mathbf{F} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ 2xy+z^2 & x^2 & 2xz\end{vmatrix}

Compute: F3,y−F2,z=0−0=0F_{3,y} - F_{2,z} = 0 - 0 = 0; F1,z−F3,x=2z−2z=0F_{1,z} - F_{3,x} = 2z - 2z = 0; F2,x−F1,y=2x−2x=0F_{2,x} - F_{1,y} = 2x - 2x = 0. Hence ∇×F=0\nabla\times\mathbf{F} = \mathbf{0}. ✓\checkmark

Solution to question 2:

fx=2xy+z2⇒f=x2y+xz2+g(y,z)f_x = 2xy + z^2 \Rightarrow f = x^2y + xz^2 + g(y,z).

fy=x2+gy=x2⇒gy=0⇒g=h(z)f_y = x^2 + g_y = x^2 \Rightarrow g_y = 0 \Rightarrow g = h(z).

fz=2xz+h′(z)=2xz⇒h′(z)=0⇒h=Cf_z = 2xz + h'(z) = 2xz \Rightarrow h'(z) = 0 \Rightarrow h = C.

f(x,y,z)=x2y+xz2 f(x,y,z) = x^2y + xz^2 Solution to question 3: ∫CF⋅dr=f(2,1,3)−f(1,0,1)=(4⋅1+2⋅9)−(1⋅0+1⋅1)=(4+18)−1=21 \int_C\mathbf{F}\cdot d\mathbf{r} = f(2,1,3) - f(1,0,1) = (4\cdot1 + 2\cdot9) - (1\cdot0 + 1\cdot1) = (4+18) - 1 = 21

Green’s Theorem and Its Applications

Green’s Theorem relates a line integral around a simple closed curve C C to a double integral over the enclosed region D D : ∮CP dx+Q dy=∬D(∂Q∂x−∂P∂y)dA\oint_C P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA. It is particularly powerful for evaluating line integrals over closed curves and for computing areas. The curve must be positively oriented (counterclockwise). These problems develop fluency in recognizing when Green’s Theorem applies and how to exploit it efficiently.

Problem 12: Green’s Theorem Over a Rectangle

Medium

Let C C be the positively oriented boundary of the rectangle with vertices (0,0) (0,0) , (3,0) (3,0) , (3,2) (3,2) , (0,2) (0,2) .

  1. State the conditions under which Green’s Theorem applies here.
  2. Use Green’s Theorem to evaluate ∮C(xy) dx+(x2+y) dy\displaystyle\oint_C (xy)\,dx + (x^2 + y)\,dy.
Hint
Identify P=xy P = xy and Q=x2+y Q = x^2 + y . Compute ∂Q/∂x−∂P/∂y=2x−x=x\partial Q/\partial x - \partial P/\partial y = 2x - x = x. The double integral becomes ∬Dx dA\iint_D x\,dA over the rectangle [0,3]×[0,2][0,3]\times[0,2].
View Solution
Solution to question 1:

P P and Q Q have continuous partial derivatives on and inside C C , and C C is a simple, closed, positively oriented curve. Green’s Theorem applies.

Solution to question 2: ∮C=∬D(∂Q∂x−∂P∂y)dA=∬Dx dA=∫03∫02x dy dx=∫032x dx=[x2]03=9 \oint_C = \iint_D\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA = \iint_D x\,dA = \int_0^3\int_0^2 x\,dy\,dx = \int_0^3 2x\,dx = \Big[x^2\Big]_0^3 = 9

Problem 13: Green’s Theorem with a Non-Simply-Connected Region

Hard

Let F(x,y)=⟨−yx2+y2,  xx2+y2⟩\mathbf{F}(x,y) = \left\langle \dfrac{-y}{x^2+y^2},\; \dfrac{x}{x^2+y^2} \right\rangle, which is defined everywhere except the origin.

  1. Compute ∂Q/∂x−∂P/∂y\partial Q/\partial x - \partial P/\partial y on the domain R2∖{(0,0)}\mathbb{R}^2 \setminus \{(0,0)\}. What do you observe?
  2. Evaluate ∮C1F⋅dr\displaystyle\oint_{C_1}\mathbf{F}\cdot d\mathbf{r} where C1 C_1 is the unit circle traversed counterclockwise. Is Green’s Theorem directly applicable? Explain.
  3. Evaluate ∮C2F⋅dr\displaystyle\oint_{C_2}\mathbf{F}\cdot d\mathbf{r} where C2 C_2 is the circle of radius 2 2 traversed counterclockwise, and explain what the result reveals about the topology of the domain.
Hint
Compute the partial derivatives carefully: you will find ∂Q/∂x−∂P/∂y=0\partial Q/\partial x - \partial P/\partial y = 0 away from the origin. For C1 C_1 , use x=cos⁡t x = \cos t , y=sin⁡t y = \sin t and compute the dot product directly. Green’s Theorem requires the enclosed region to be simply connected; it fails here because the origin is excluded.
View Solution
Solution to question 1:

Let r2=x2+y2 r^2 = x^2 + y^2 . Then:

∂Q∂x=r2−x⋅2xr4=y2−x2r4,∂P∂y=−r2+y⋅2yr4=y2−x2r4 \frac{\partial Q}{\partial x} = \frac{r^2 - x\cdot 2x}{r^4} = \frac{y^2 - x^2}{r^4}, \qquad \frac{\partial P}{\partial y} = \frac{-r^2 + y\cdot 2y}{r^4} = \frac{y^2 - x^2}{r^4} ∂Q∂x−∂P∂y=0for all (x,y)≠(0,0). \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 \quad \text{for all } (x,y) \neq (0,0).

This is zero everywhere in the domain, yet F\mathbf{F} is not conservative because its domain is not simply connected.

Solution to question 2:

Parameterize C1 C_1 : x=cos⁡t x=\cos t , y=sin⁡t y=\sin t , dr=(−sin⁡t,cos⁡t) dt d\mathbf{r} = (-\sin t, \cos t)\,dt .

F(r(t))=(−sin⁡t,cos⁡t) \mathbf{F}(\mathbf{r}(t)) = (-\sin t, \cos t) ∮C1F⋅dr=∫02π[sin⁡2t+cos⁡2t]dt=2π \oint_{C_1}\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}\left[\sin^2 t + \cos^2 t\right]dt = 2\pi

Green’s Theorem is not directly applicable: the enclosed disk contains the singularity (0,0)(0,0) where F\mathbf{F} is undefined, so the region is not simply connected in the required sense.

Solution to question 3:

The same calculation with C2 C_2 : x=2cos⁡t x=2\cos t , y=2sin⁡t y=2\sin t .

F(r(t))=(−2sin⁡t4,2cos⁡t4)=(−sin⁡t2,cos⁡t2),dr=(−2sin⁡t,2cos⁡t) dt \mathbf{F}(\mathbf{r}(t)) = \left(\frac{-2\sin t}{4}, \frac{2\cos t}{4}\right) = \left(\frac{-\sin t}{2}, \frac{\cos t}{2}\right), \qquad d\mathbf{r} = (-2\sin t, 2\cos t)\,dt ∮C2=∫02π(sin⁡2t+cos⁡2t)dt=2π \oint_{C_2} = \int_0^{2\pi}\left(\sin^2 t + \cos^2 t\right)dt = 2\pi

Both circles give 2π2\pi, regardless of radius. This illustrates that the integral is a topological invariant: any simple closed curve encircling the singularity once counterclockwise yields the same value 2π2\pi. This reflects the fact that F\mathbf{F} has a non-trivial circulation around the origin.