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.
Scalar Line Integrals with Respect to Arc Length
These problems focus on integrating a scalar function \( f(x, y) \) or \( f(x, y, z) \) along a curve \( C \) with respect to arc length \( ds \). The key formula is \(\int_C f\,ds = \int_a^b f(\mathbf{r}(t))\,\|\mathbf{r}'(t)\|\,dt\). Begin by parameterizing the curve, computing \( \|\mathbf{r}'(t)\| \), and substituting into the integrand.
Problem 1: Line Integral Along a Line Segment in 2D
Easy
Let \( C \) be the line segment from \( (0, 0) \) to \( (3, 4) \).
- Parameterize the curve \( C \) using a parameter \( t \in [0, 1] \).
- Evaluate the line integral \(\displaystyle\int_C (x + y)\,ds\).
Hint
View Solution
Parameterize: \( \mathbf{r}(t) = (3t,\, 4t) \), \( t \in [0,1] \). Then \( \mathbf{r}'(t) = (3, 4) \) and \( \|\mathbf{r}'(t)\| = \sqrt{9 + 16} = 5 \).
Solution to question 2:
\[
\int_C (x+y)\,ds = \int_0^1 (3t + 4t)\cdot 5\,dt = 5\int_0^1 7t\,dt = 35\int_0^1 t\,dt = 35\cdot\frac{1}{2} = \frac{35}{2}
\]
Problem 2: Line Integral Along a Circular Arc
Easy
Let \( C \) be the upper semicircle of radius \( 2 \) centered at the origin, traversed counterclockwise from \( (2, 0) \) to \( (-2, 0) \).
- Write a parameterization of \( C \) and compute \( \|\mathbf{r}'(t)\| \).
- Evaluate \(\displaystyle\int_C y\,ds\).
Hint
View Solution
\( \mathbf{r}(t) = (2\cos t,\, 2\sin t) \), \( t \in [0, \pi] \). Then \( \mathbf{r}'(t) = (-2\sin t,\, 2\cos t) \) and \( \|\mathbf{r}'(t)\| = \sqrt{4\sin^2 t + 4\cos^2 t} = 2 \).
Solution to question 2:
\[
\int_C y\,ds = \int_0^{\pi} 2\sin t \cdot 2\,dt = 4\int_0^{\pi}\sin t\,dt = 4\Big[-\cos t\Big]_0^{\pi} = 4(1+1) = 8
\]
Problem 3: Scalar Integral Along a Parabola
Medium
Consider the parabolic arc \( y = x^2 \) from \( (0, 0) \) to \( (1, 1) \).
- Parameterize \( C \) using \( x = t \) and find \( ds \) in terms of \( dt \).
- Evaluate \(\displaystyle\int_C xy\,ds\).
Hint
View Solution
\( \mathbf{r}(t) = (t, t^2) \), \( t \in [0,1] \). Then \( ds = \sqrt{1 + 4t^2}\,dt \).
Solution to question 2:
\[
\int_C xy\,ds = \int_0^1 t^3\sqrt{1+4t^2}\,dt
\]
Let \( u = 1 + 4t^2 \), so \( du = 8t\,dt \) and \( t^2 = \dfrac{u-1}{4} \). Then:
\[
= \int_1^5 \frac{u-1}{4}\cdot\sqrt{u}\cdot\frac{du}{8} = \frac{1}{32}\int_1^5 (u^{3/2} – u^{1/2})\,du = \frac{1}{32}\left[\frac{2u^{5/2}}{5} – \frac{2u^{3/2}}{3}\right]_1^5
\]
\[
= \frac{1}{32}\left(\frac{2\cdot25\sqrt{5}}{5} – \frac{2\cdot5\sqrt{5}}{3} – \frac{2}{5} + \frac{2}{3}\right) = \frac{1}{32}\left(10\sqrt{5} – \frac{10\sqrt{5}}{3} + \frac{4}{15}\right) = \frac{1}{32}\cdot\frac{20\sqrt{5}}{3} + \frac{4}{480} \approx 0.4696
\]
Line Integrals with Respect to Coordinate Variables
When the integrand is expressed as \( P\,dx + Q\,dy \) (or \( P\,dx + Q\,dy + R\,dz \) in 3D), we substitute the parameterization directly into \( x \), \( y \), and their differentials. Unlike arc-length integrals, these are sensitive to the orientation of \( C \). 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 \( P\,dx + Q\,dy \) Over a Line Segment
Easy
Let \( C \) be the directed line segment from \( (1, 0) \) to \( (0, 1) \).
- Parameterize \( C \) and express \( dx \) and \( dy \) in terms of \( dt \).
- Evaluate \(\displaystyle\int_C y^2\,dx + x^2\,dy\).
Hint
View Solution
\( x = 1-t \), \( y = t \), \( t \in [0,1] \); \( dx = -dt \), \( dy = dt \).
Solution to question 2:
\[
\int_C y^2\,dx + x^2\,dy = \int_0^1 \left[t^2(-1) + (1-t)^2(1)\right]dt = \int_0^1\left[-t^2 + 1 – 2t + t^2\right]dt = \int_0^1(1-2t)\,dt
\]
\[
= \Big[t – t^2\Big]_0^1 = 0
\]
Problem 5: Piecewise Path — Triangle Boundary
Medium
Let \( C \) be the boundary of the triangle with vertices \( A=(0,0) \), \( B=(2,0) \), \( D=(0,2) \), traversed counterclockwise.
- Decompose \( C \) into three directed line segments \( C_1 \), \( C_2 \), \( C_3 \).
- Evaluate \(\displaystyle\int_C xy\,dx + (x+y)\,dy\) by summing the contributions of each piece.
Hint
View Solution
\( C_1 \): \( (t, 0) \), \( t\in[0,2] \). \( C_2 \): \( (2-t, t) \), \( t\in[0,2] \). \( C_3 \): \( (0, 2-t) \), \( t\in[0,2] \).
Solution to question 2:
On \( C_1 \): \( y=0 \), \( dy=0 \), \( dx=dt \).
\[
I_1 = \int_0^2 t\cdot 0\cdot dt + 0 = 0
\]
On \( C_2 \): \( x=2-t \), \( y=t \), \( dx=-dt \), \( dy=dt \).
\[
I_2 = \int_0^2\left[(2-t)t(-1) + (2-t+t)(1)\right]dt = \int_0^2\left[-2t+t^2+2\right]dt = \left[-t^2+\frac{t^3}{3}+2t\right]_0^2 = -4+\frac{8}{3}+4 = \frac{8}{3}
\]
On \( C_3 \): \( x=0 \), \( dx=0 \), \( y=2-t \), \( dy=-dt \).
\[
I_3 = \int_0^2 0 + (0+2-t)(-1)\,dt = \int_0^2(t-2)\,dt = \left[\frac{t^2}{2}-2t\right]_0^2 = 2-4 = -2
\]
\[
\int_C = I_1 + I_2 + I_3 = 0 + \frac{8}{3} – 2 = \frac{2}{3}
\]
Line Integrals of Vector Fields and Work
The line integral of a vector field \(\mathbf{F} = \langle P, Q, R \rangle\) along a curve \( C \) is defined as \(\int_C \mathbf{F}\cdot d\mathbf{r} = \int_C P\,dx + Q\,dy + R\,dz\). Physically, this computes the work done by the force field \(\mathbf{F}\) along the path \( C \). Problems in this section require careful attention to the direction of \( C \) and the structure of \(\mathbf{F}\).
Problem 6: Work Integral Over a Circular Path
Easy
A particle moves counterclockwise along the full unit circle \( x^2 + y^2 = 1 \) under the force \( \mathbf{F}(x,y) = \langle -y,\, x \rangle \).
- Set up \(\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}\) using the parameterization \( \mathbf{r}(t) = (\cos t, \sin t) \).
- Evaluate the integral and interpret the result as work.
Hint
View Solution
\( \mathbf{r}'(t) = (-\sin t, \cos t) \). Then \( \mathbf{F}(\mathbf{r}(t)) = (-\sin t, \cos t) \).
\[
\mathbf{F}\cdot\mathbf{r}’ = (-\sin t)(-\sin t) + (\cos t)(\cos t) = \sin^2 t + \cos^2 t = 1
\]
Solution to question 2:
\[
W = \int_0^{2\pi} 1\,dt = 2\pi
\]
The force field \(\mathbf{F} = \langle -y, x \rangle\) is always tangent to the unit circle and aligned with the direction of motion, so it does positive work of \( 2\pi \) joules (in appropriate units).
Problem 7: Vector Field Integral Along a Space Curve (Helix)
Medium
Let \( \mathbf{F}(x,y,z) = \langle y,\, -x,\, z \rangle \) and let \( C \) be the helix \( \mathbf{r}(t) = (\cos t,\, \sin t,\, t) \) for \( t \in [0, 2\pi] \).
- Compute \( \mathbf{F}(\mathbf{r}(t)) \) and \( \mathbf{r}'(t) \).
- Evaluate \(\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}\).
Hint
View Solution
\( \mathbf{F}(\mathbf{r}(t)) = (\sin t,\, -\cos t,\, t) \) and \( \mathbf{r}'(t) = (-\sin t,\, \cos t,\, 1) \).
Solution to question 2:
\[
\mathbf{F}\cdot\mathbf{r}’ = (\sin t)(-\sin t) + (-\cos t)(\cos t) + t\cdot 1 = -\sin^2 t – \cos^2 t + t = -1 + t
\]
\[
\int_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}(t – 1)\,dt = \left[\frac{t^2}{2} – t\right]_0^{2\pi} = \frac{4\pi^2}{2} – 2\pi = 2\pi^2 – 2\pi
\]
Problem 8: Work Along a Piecewise Path in 3D
Hard
Let \(\mathbf{F}(x,y,z) = \langle yz,\, xz,\, xy \rangle\). Consider the piecewise path \( C = C_1 \cup C_2 \) where \( C_1 \) goes from \( (0,0,0) \) to \( (1,0,0) \) along the \( x \)-axis, and \( C_2 \) goes from \( (1,0,0) \) to \( (1,1,1) \) along the straight line.
- Evaluate \(\displaystyle\int_{C_1} \mathbf{F}\cdot d\mathbf{r}\).
- Evaluate \(\displaystyle\int_{C_2} \mathbf{F}\cdot d\mathbf{r}\).
- Find the total work done along \( C \).
Hint
View Solution
\( C_1 \): \( \mathbf{r}(t) = (t,0,0) \), \( t\in[0,1] \). Then \(\mathbf{F} = (0\cdot 0,\, t\cdot 0,\, t\cdot 0) = (0,0,0)\).
\[
\int_{C_1}\mathbf{F}\cdot d\mathbf{r} = 0
\]
Solution to question 2:
\( C_2 \): \( \mathbf{r}(t) = (1,t,t) \), \( t\in[0,1] \), so \( d\mathbf{r} = (0,1,1)\,dt \).
\[
\mathbf{F}(1,t,t) = (t\cdot t,\; 1\cdot t,\; 1\cdot t) = (t^2, t, t)
\]
\[
\int_{C_2}\mathbf{F}\cdot d\mathbf{r} = \int_0^1(t^2\cdot 0 + t\cdot 1 + t\cdot 1)\,dt = \int_0^1 2t\,dt = 1
\]
Solution to question 3:
\[
W_{\text{total}} = 0 + 1 = 1
\]
Conservative Vector Fields and the Fundamental Theorem of Line Integrals
A vector field \(\mathbf{F}\) is conservative if it equals the gradient of a scalar potential function \( f \), i.e., \(\mathbf{F} = \nabla f\). When \(\mathbf{F}\) is conservative, the Fundamental Theorem of Line Integrals states that \(\int_C \mathbf{F}\cdot d\mathbf{r} = f(\mathbf{r}(b)) – f(\mathbf{r}(a))\), making computation independent of the path. The test for conservatism in 2D is \(\partial P/\partial y = \partial Q/\partial x\).
Problem 9: Identifying a Conservative Field and Finding a Potential
Medium
Let \( \mathbf{F}(x,y) = \langle 2xy + 1,\; x^2 – 3y^2 \rangle \).
- Verify that \(\mathbf{F}\) is conservative by checking the appropriate partial derivatives.
- Find a potential function \( f(x, y) \) such that \(\nabla f = \mathbf{F}\).
- Use the Fundamental Theorem to evaluate \(\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}\) from \( (0,0) \) to \( (2,1) \).
Hint
View Solution
\[
\frac{\partial P}{\partial y} = 2x, \qquad \frac{\partial Q}{\partial x} = 2x \implies \mathbf{F} \text{ is conservative.}
\]
Solution to question 2:
Integrate \( f_x = 2xy+1 \): \( f = x^2y + x + g(y) \).
Then \( f_y = x^2 + g'(y) = x^2 – 3y^2 \implies g'(y) = -3y^2 \implies g(y) = -y^3 \).
\[
f(x,y) = x^2 y + x – y^3
\]
Solution to question 3:
\[
\int_C \mathbf{F}\cdot d\mathbf{r} = f(2,1) – f(0,0) = (4\cdot1 + 2 – 1) – 0 = 5
\]
Problem 10: Path Independence — Non-Conservative vs. Conservative
Medium
Consider \(\mathbf{F}(x,y) = \langle y^2,\; 2xy \rangle\) and two paths from \( A=(0,0) \) to \( B=(1,2) \): the straight line \( C_1 \) and the parabola \( C_2 \) given by \( y = 2x^2 \).
- Verify that \(\mathbf{F}\) is conservative and find its potential function.
- Evaluate \(\int_{C_1}\mathbf{F}\cdot d\mathbf{r}\) directly using the straight-line parameterization.
- Confirm path independence by computing \(\int_{C_2}\mathbf{F}\cdot d\mathbf{r}\) via the potential function.
Hint
View Solution
\(\partial_y(y^2) = 2y = \partial_x(2xy)\), so \(\mathbf{F}\) is conservative.
From \(f_x = y^2 \Rightarrow f = xy^2 + g(y)\). Then \(f_y = 2xy + g'(y) = 2xy \Rightarrow g'(y) = 0 \Rightarrow g = C\).
\[
f(x,y) = xy^2
\]
Solution to question 2:
\( C_1 \): \( (t, 2t) \), \( t\in[0,1] \); \( d\mathbf{r} = (1, 2)\,dt \).
\[
\int_{C_1} = \int_0^1 \left[(2t)^2 + 2t(2t)\cdot 2\right]dt = \int_0^1(4t^2 + 8t^2)\,dt = 12\int_0^1 t^2\,dt = 4
\]
Solution to question 3:
\[
\int_{C_2}\mathbf{F}\cdot d\mathbf{r} = f(1,2) – f(0,0) = 1\cdot 4 – 0 = 4 \checkmark
\]
Problem 11: Conservative Field in 3D — Finding a Potential Function
Hard
Let \(\mathbf{F}(x,y,z) = \langle 2xy + z^2,\; x^2,\; 2xz \rangle\).
- Verify that \(\mathbf{F}\) is conservative by checking that \(\nabla \times \mathbf{F} = \mathbf{0}\).
- Find a potential function \(f\) such that \(\nabla f = \mathbf{F}\).
- Evaluate \(\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}\) along any path from \((1,0,1)\) to \((2,1,3)\).
Hint
View Solution
\[
\nabla\times\mathbf{F} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ 2xy+z^2 & x^2 & 2xz\end{vmatrix}
\]
Compute: \(F_{3,y} – F_{2,z} = 0 – 0 = 0\); \(F_{1,z} – F_{3,x} = 2z – 2z = 0\); \(F_{2,x} – F_{1,y} = 2x – 2x = 0\). Hence \(\nabla\times\mathbf{F} = \mathbf{0}\). \(\checkmark\)
Solution to question 2:
\(f_x = 2xy + z^2 \Rightarrow f = x^2y + xz^2 + g(y,z)\).
\(f_y = x^2 + g_y = x^2 \Rightarrow g_y = 0 \Rightarrow g = h(z)\).
\(f_z = 2xz + h'(z) = 2xz \Rightarrow h'(z) = 0 \Rightarrow h = C\).
\[
f(x,y,z) = x^2y + xz^2
\]
Solution to question 3:
\[
\int_C\mathbf{F}\cdot d\mathbf{r} = f(2,1,3) – f(1,0,1) = (4\cdot1 + 2\cdot9) – (1\cdot0 + 1\cdot1) = (4+18) – 1 = 21
\]
Green’s Theorem and Its Applications
Green’s Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the enclosed region \( D \):
\(\oint_C P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} – \frac{\partial P}{\partial y}\right)dA\).
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 \( C \) be the positively oriented boundary of the rectangle with vertices \( (0,0) \), \( (3,0) \), \( (3,2) \), \( (0,2) \).
- State the conditions under which Green’s Theorem applies here.
- Use Green’s Theorem to evaluate \(\displaystyle\oint_C (xy)\,dx + (x^2 + y)\,dy\).
Hint
View Solution
\( P \) and \( Q \) have continuous partial derivatives on and inside \( C \), and \( C \) is a simple, closed, positively oriented curve. Green’s Theorem applies.
Solution to question 2:
\[
\oint_C = \iint_D\left(\frac{\partial Q}{\partial x} – \frac{\partial P}{\partial y}\right)dA = \iint_D x\,dA = \int_0^3\int_0^2 x\,dy\,dx = \int_0^3 2x\,dx = \Big[x^2\Big]_0^3 = 9
\]
Problem 13: Green’s Theorem with a Non-Simply-Connected Region
Hard
Let \(\mathbf{F}(x,y) = \left\langle \dfrac{-y}{x^2+y^2},\; \dfrac{x}{x^2+y^2} \right\rangle\), which is defined everywhere except the origin.
- Compute \(\partial Q/\partial x – \partial P/\partial y\) on the domain \(\mathbb{R}^2 \setminus \{(0,0)\}\). What do you observe?
- Evaluate \(\displaystyle\oint_{C_1}\mathbf{F}\cdot d\mathbf{r}\) where \( C_1 \) is the unit circle traversed counterclockwise. Is Green’s Theorem directly applicable? Explain.
- Evaluate \(\displaystyle\oint_{C_2}\mathbf{F}\cdot d\mathbf{r}\) where \( C_2 \) is the circle of radius \( 2 \) traversed counterclockwise, and explain what the result reveals about the topology of the domain.
Hint
View Solution
Let \( r^2 = x^2 + y^2 \). Then:
\[
\frac{\partial Q}{\partial x} = \frac{r^2 – x\cdot 2x}{r^4} = \frac{y^2 – x^2}{r^4}, \qquad \frac{\partial P}{\partial y} = \frac{-r^2 + y\cdot 2y}{r^4} = \frac{y^2 – x^2}{r^4}
\]
\[
\frac{\partial Q}{\partial x} – \frac{\partial P}{\partial y} = 0 \quad \text{for all } (x,y) \neq (0,0).
\]
This is zero everywhere in the domain, yet \(\mathbf{F}\) is not conservative because its domain is not simply connected.
Solution to question 2:
Parameterize \( C_1 \): \( x=\cos t \), \( y=\sin t \), \( d\mathbf{r} = (-\sin t, \cos t)\,dt \).
\[
\mathbf{F}(\mathbf{r}(t)) = (-\sin t, \cos t)
\]
\[
\oint_{C_1}\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}\left[\sin^2 t + \cos^2 t\right]dt = 2\pi
\]
Green’s Theorem is not directly applicable: the enclosed disk contains the singularity \((0,0)\) where \(\mathbf{F}\) is undefined, so the region is not simply connected in the required sense.
Solution to question 3:
The same calculation with \( C_2 \): \( x=2\cos t \), \( y=2\sin t \).
\[
\mathbf{F}(\mathbf{r}(t)) = \left(\frac{-2\sin t}{4}, \frac{2\cos t}{4}\right) = \left(\frac{-\sin t}{2}, \frac{\cos t}{2}\right), \qquad d\mathbf{r} = (-2\sin t, 2\cos t)\,dt
\]
\[
\oint_{C_2} = \int_0^{2\pi}\left(\sin^2 t + \cos^2 t\right)dt = 2\pi
\]
Both circles give \(2\pi\), 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 \(2\pi\). This reflects the fact that \(\mathbf{F}\) has a non-trivial circulation around the origin.