University Calculus III — Vector Fields and Line Integrals Worksheet
The block where integration leaves regions behind and runs along curves, and where most marks are lost on orientation and on choosing the tool rather than on the integral itself. Reading a vector field from its formula and from a plot; integrating a density along a wire with respect to arc length; computing work as a line integral of a force along a parametrized path, one piece at a time; testing whether a field is conservative and building its potential function; using the fundamental theorem for line integrals so that only the endpoints matter; turning a closed line integral into a double integral with Green's theorem, with the orientation checked; and computing curl and divergence and reading what they say. Have a look on this page, then print the free PDF when you want to write on it.
Practice worksheet — free PDF
No email, no account, no watermark. Teachers: photocopy it for your classes freely. Worked solutions and 9 harder problems come with the University Calculus III bundle.
10 of the 13 questions
Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one. 10 of the 13 questions are printed below. The other 3 are built on a diagram or a table of values that does not translate to the page, so they are in the free PDF — marked below where they would have come.
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Q1Vector Fields
Let .
- Compute at the six points , , , , and , and draw each vector on the grid with its tail at the point where it was computed. Each square of the grid is one unit, and the origin is where the axes cross.
- Let be the position vector of a point. Show that at every point, and express in terms of .
- Use (b) to describe the whole field in one sentence.
A blank Cartesian grid for this question is on the printable PDF.
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Q2Vector Fields
This question is built around a diagram or a table of values. Open it in the PDF.
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Q3Line Integrals of Scalar Functions
A thin wire lies along the curve , , with lengths in centimetres. Its linear density at the point is grams per centimetre.
- Show that .
- Find the mass of the wire, .
- Find the length of the wire and hence its average density.
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Q4Line Integrals of Vector Fields and Work
A force field , in newtons, acts on a particle that moves along the parabola from to , with distances in metres. Find the work done by , .
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Q5Line Integrals of Vector Fields and Work
This question is built around a diagram or a table of values. Open it in the PDF.
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Q6Conservative Fields and Potential Functions
A plane vector field whose components have continuous partial derivatives on a region with no holes (a simply connected region, such as the whole plane) is conservative on that region exactly when throughout it.
Let on .
- Writing , check that , and say why this shows that is conservative, checking each hypothesis of the test above.
- Find a potential function , that is, a function with .
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Q7Conservative Fields and Potential Functions
A vector field whose components have continuous partial derivatives on all of is conservative exactly when Let , where and are constants.
- Find the values of and for which is conservative.
- For those values, find a potential function .
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Q8The Fundamental Theorem for Line Integrals
Let and .
- Write out .
- Evaluate , where is the curve , .
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Q9The Fundamental Theorem for Line Integrals
The force field , in newtons, has the potential function ; distances are in metres. Let , and .
- Verify that .
- Find the work done by on a particle that moves from to along any smooth curve.
- The particle then continues from to . Find the work done on this second leg, and the total work from to .
- How much work does do on a particle that leaves and returns to ?
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Q10Green's Theorem
Let be the boundary of the triangle with vertices , and , traversed counterclockwise. Use Green's theorem to evaluate
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Q11Green's Theorem
This question is built around a diagram or a table of values. Open it in the PDF.
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Q12Curl and Divergence
Let .
- Compute the divergence and the curl .
- Evaluate both at the point .
- Is conservative? Give the reason.
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Q13Synthesis — drawing on several topics in this unit
A force field , in newtons, acts in the plane; distances are in metres.
- Show that is not conservative.
- Use Green's theorem to find the work done by on a particle that goes once counterclockwise round the circle .
- Find the work done by along the diameter from to .
- Deduce, using Green's theorem on the upper half-disk, the work done along the upper semicircle from counterclockwise to .
The 9 challenge problems for this topic are a separate, paid sheet and are not reproduced here.
What does this set assume? This set stands on CEGEP Calculus I, CEGEP Calculus II and CEGEP Linear Algebra, and on the earlier Calculus III sets, and re-teaches none of them. Parametrizing a segment, a circle and the graph of a function, the dot product and the length of a vector, partial derivatives and equality of mixed partials, and double integrals over triangles and in polar coordinates are all used without comment; every single integral that remains is a CEGEP Calculus II integral chosen to stay light. Scope is the core that university multivariable courses share: vector fields, line integrals of scalar functions and of vector fields, work, conservative fields and potential functions, the fundamental theorem for line integrals, Green's theorem, and curl and divergence, placed here so that the curl test and Green's theorem sit side by side. Surface integrals, flux through a surface, Stokes' theorem and the divergence theorem open the next set. No differential equation is solved, so flow lines are not here, and exact equations are read as a differential-equations topic, not this one. Every question is settled by hand, with no calculator.
Which course is this for? In the public course calendars of Montreal universities, this material is part of the courses numbered MAST 219, ENGR 233, MAT1410 and MAT165. Each course orders and weights the topics its own way, so check your own outline for what your exam covers. Which sets match your course.
The rest of this unit
The worksheet is the practice, and it is free. Two more printable documents cover the same unit and come with the University Calculus III bundle: read the notes first, work this sheet, then sit the test closed-book. See what each one covers.
What the Vector Fields and Line Integrals notes cover About the Vector Fields and Line Integrals unit test
How to do every concept on this sheet
This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.
A vector field is a vector attached to every point
Q1 gives a plane field by its formula and asks you to compute it at six points and draw each vector with its tail at the point where it was computed. That last instruction is the whole convention: the arrow at (4, 4) starts at (4, 4), not at the origin. The grid is one unit per square, so each arrow is drawn by moving its two components from its own point.
at the point (x, y), draw F(x, y) from (x, y) to (x, y) + F(x, y)Parts (b) and (c) turn six arrows into one sentence. A dot product of zero with the position vector says the arrow is perpendicular to the line from the origin — which is the same as tangent to the circle through that point centred at the origin. The length in terms of the distance from the origin says how the arrows grow. Part (c) asks you to put those two facts together, and the direction of turning is read off any one arrow you drew in part (a).
Q2 runs the other way: a plot, shown in the PDF, and four candidate formulas. The question lists three features you can read from the plot, and asks you to pick the one formula that matches and, for each of the other three, to name one feature it fails and a point where the failure shows.
Matching a formula to a plot
Test each feature against the formula, not the formula against your memory of the plot.
- 1Constant along vertical lines
A field is identical along every vertical line exactly when neither component involves y.
- 2Evaluate on the y-axis
Put x = 0 in each candidate and see whether the arrow is horizontal and points right.
- 3Check signs on each side
Pick one point to the right of the y-axis and one to the left, and read the signs of the two components.
- 4Give a witness
For a rejected formula, one feature it fails and one specific point is the whole answer; the question says one feature is enough.
Line integrals of scalar functions
Q3 is a wire in space with a density that varies along it, and the mass is the integral of the density with respect to arc length. Everything is converted to the parameter t: the density is evaluated on the curve, and ds becomes the speed times dt.
∫C f ds = ∫ab f(r(t)) ‖r′(t)‖ dtPart (a) is set up to hand you a clean speed. The square root of the sum of the squared components is a perfect square here — look for it rather than expanding and stopping. Once the speed is a polynomial, part (b) is a polynomial integral. Part (c) divides the mass by the length, and the average density has a built-in check: it must lie between the smallest and largest density the wire actually has.
Substitute the curve into the density. The density is given as a function of x, y and z; on the wire each of those is a function of t. Integrating the density with x, y or z still in it is the most common slip in this topic.
Line integrals of vector fields and work
Work adds up the component of the force along the direction of travel, so the integrand is a dot product with r′(t), not a length. Q4 is the direct case: a force along a parabola from one point to another. Parametrize the parabola with x as the parameter, evaluate the field on the curve, dot it with the derivative, and integrate between the parameter values of the two endpoints.
W = ∫C F · dr = ∫ab F(r(t)) · r′(t) dtThe direction of travel matters here, unlike in Q3. Reversing a curve changes the sign of r′(t) and therefore the sign of the work. Check that your parametrization starts at the stated start point before integrating.
Q5 is built on a figure, shown in the PDF: a curve in two pieces, a quarter circle followed by a straight segment. A line integral over a piecewise curve is the sum of the integrals over the pieces, each with a parametrization of its own.
A curve in pieces
One parametrization per piece, each checked at both ends.
- 1The arc
Use the radius times (cos t, sin t), with the t-interval chosen so that the arc starts and ends where the figure says, in the direction of its arrow.
- 2The segment
Start point + t × (end − start), for 0 ≤ t ≤ 1. That form cannot run the wrong way.
- 3Integrate each
Evaluate the field on each piece, dot with that piece's r′(t), integrate over that piece's interval.
- 4Add, then sanity-check the sign
Compare the direction of the field with the direction of travel along each piece; the sign of each contribution should agree with that picture.
Conservative fields and potential functions
A field is conservative when it is the gradient of some function f, its potential. Q6 states the test for a plane field and asks you to use it with its hypotheses: compare ∂P/∂y with ∂Q/∂x, and say why equality settles the matter here — the components have continuous partial derivatives, and the region is the whole plane, which has no holes.
The test has a hypothesis about the region, and part (a) asks you to check it. Equal cross partials on a region with a hole do not guarantee a potential. On the whole plane they do; say so, rather than writing only the two derivatives.
Building a potential function
Integrate one component, then let the others fix what is left.
- 1Integrate P in x
The constant of integration is a function of the other variables, not a number: write it as h(y), or g(y, z) in space.
- 2Differentiate in y and compare with Q
Everything but h′(y) should match. What does not match is h′(y), and it must not contain x.
- 3In space, repeat with z and R
Each comparison narrows the unknown function by one variable.
- 4Check
Take the gradient of your f and compare it with the field, component by component.
Q7 is the same machinery in three dimensions, with two constants left open. There are three cross-partial conditions in space, and each one is an equation that must hold at every point: compare coefficients, and note which conditions hold whatever the constants are. Only once the constants are fixed is there a field to find a potential for in part (b).
The fundamental theorem for line integrals
When the field is a gradient, the line integral is the change in the potential between the endpoints — the path in between does not enter. Q8 hands you the potential and a curve whose parametrization would be unpleasant to integrate along. Part (a) is the gradient; for part (b), the only thing you need from the curve is where it starts and where it ends.
∫C ∇f · dr = f(end point) − f(start point)Read the endpoints off the parametrization. Put the first and last values of t into r(t). A cosine of a multiple of π is the usual place for a sign to go missing.
Q9 is the physical reading: a force field with a stated potential, three points in space, and work in joules. Part (a) verifies the potential; after that every part is a subtraction of two values of f. Part (c) asks for two legs and the total, which is a check in itself — the total must equal the difference between the last point and the first. Part (d) is the same subtraction with a special pair of endpoints.
Path independence is a property of the field, not the path. The shortcut in Q8 and Q9 is licensed only because each field is stated to be a gradient. For a field you have not shown to be conservative, the endpoints are not enough.
Green's theorem
Green's theorem trades a line integral round a closed curve for a double integral over the region inside it. It needs a simple closed curve, traversed counterclockwise, and components with continuous partial derivatives on the region.
∮C P dx + Q dy = ∬D (∂Q/∂x − ∂P/∂y) dAQ10 is the direct case: the boundary of a triangle, counterclockwise, with the integral written in P dx + Q dy form. Name P and Q, form ∂Q/∂x − ∂P/∂y, and describe the triangle by vertical or horizontal strips, as in the double-integrals set. The integrand that survives is usually simpler than either P or Q, which is the point of the theorem.
Check the hypotheses in a line before you use it. Closed, simple, counterclockwise, and smooth components on the region. When a question says "use Green's theorem", that line is part of the answer.
Q11 is built on a figure, shown in the PDF: a half-disk whose boundary is traversed clockwise. Two decisions carry it. First, why Green's theorem at all — look at which terms of P and Q a direct evaluation would have to antidifferentiate, and at what happens to each of them when you form ∂Q/∂x − ∂P/∂y. Second, the orientation: the theorem is stated for the counterclockwise boundary, and the curve here goes the other way.
Green's theorem on a region the wrong way round
Compute as if counterclockwise, then correct once.
- 1Form the integrand
∂Q/∂x − ∂P/∂y. Terms of P that do not contain y, and terms of Q that do not contain x, drop out here.
- 2Describe the region
A half-disk centred at the origin is a polar rectangle; see whether the integrand is simpler in polar form too.
- 3Evaluate the double integral
Remember the r in dA = r dr dθ.
- 4Fix the orientation
The double integral is the counterclockwise value. Traversing the boundary clockwise reverses the sign of the line integral.
Curl and divergence
Q12 asks for both of the derivatives a space field has. Divergence is a scalar: the sum of each component differentiated in its own variable. Curl is a vector, most reliably computed as the formal determinant of a 3 × 3 array with the unit vectors in the first row, the partial derivative operators in the second and the components in the third.
∇ · F = Px + Qy + Rz, ∇ × F = (Ry − Qz, Pz − Rx, Qx − Py)The middle component of the curl is where signs go wrong. Expanding the determinant, the second component carries a minus sign; written out as above, it is Pz − Rx, not Rx − Pz. Check it against the cyclic pattern x → y → z before you evaluate at the point in part (b).
Part (c) asks whether the field is conservative, with the reason. The curl of a gradient field is the zero vector, because each of its components is a difference of mixed partial derivatives that are equal. So the curl you computed in part (a) is the test: decide whether it is the zero vector everywhere, and say which way that settles the question.
The synthesis question
Q13 follows one force field through four tools. Part (a) is Q6's test, used in the other direction: showing that ∂P/∂y and ∂Q/∂x are not equal is enough to rule out a potential, with no hypothesis about the region to check. Part (b) is Q10's method on a disk; look at what ∂Q/∂x − ∂P/∂y turns out to be, and the double integral may need nothing more than an area. Part (c) is Q4's method on a segment of the x-axis: parametrize it and notice what happens to the dy term there. Part (d) puts (b) and (c) together — the semicircle and the diameter close up into the boundary of a half-disk, so Green's theorem on that half-disk gives their sum, and subtracting the diameter's contribution leaves the semicircle's.
Closing a curve is a technique, not just a theorem. An open curve can be turned into a closed one by adding a piece that is easy to integrate over. Green's theorem then handles the closed curve, and the easy piece is subtracted off. Check that the added piece is traversed so that the whole boundary is counterclockwise.
Preview all 10 pages
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Getting the most out of it
Decide the tool before you parametrize
Before any computation, ask three questions in order: is the field a gradient, is the curve closed, and is what is left simple to parametrize? A gradient means only endpoints; a closed curve means Green's theorem may be shorter; only when neither helps is a direct parametrization the right move.
Check every parametrization at both ends
Put the first and last parameter values into r(t) and compare them with the curve's stated start and end points. Half the sign errors in this block come from a parametrization that runs the right curve the wrong way.
Write ds and dr differently, on purpose
ds is the speed ‖r′(t)‖ dt, a positive length; dr is the vector r′(t) dt. Writing which one you are using at the top of each integral keeps a scalar integral from picking up an orientation, and a work integral from losing one.
Draw the orientation on the region for Green's theorem
Sketch the region, put an arrow on its boundary, and check that the region is on your left as you walk along it. If the curve in the question goes the other way, write the sign change down before you start the double integral, not after.
Verify a potential by taking its gradient
A potential function is checked in one line: differentiate it in each variable and compare with the components of the field. It costs less than any other check in this set and catches a forgotten h(y) every time.
Want the solutions, or something more challenging?
The worksheet, the questions above and every explanation on this page stay free permanently. Three more PDFs exist for this topic — the worked answer key, a harder problem set, and the answer key to that. They come with the University Calculus III Solutions Bundle, beside the unit notes and the unit test, which is what keeps the rest of the series free.
What else exists for Vector Fields and Line Integrals
Three PDFs · 17 pages · all three are in the bundle below.
- Answer key — 3 pages. All 13 questions worked step by step, including the restrictions and the justifications. Not a list of final answers.
- Challenge problems — 9 pages, 9 problems. A separate sheet at exam-plus difficulty covering the same 7 concepts. Harder than anything on the free sheet.
- Challenge answer key — 5 pages. Every challenge problem worked to the same standard, with the checks shown.
- PDF, letter size, print-ready.
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Common questions
Is this worksheet really free?
Yes — the questions are on this page to read, and the PDF downloads directly, no email and no account. The one paid item is optional: the complete University Calculus III Solutions Bundle, which covers every set at this level.
Which university courses is this for?
The course codes listed on this page are taken from the public course calendars of universities that teach multivariable calculus after CEGEP. Each course orders and weights the topics its own way — some meet line integrals before multiple integrals, some place curl and divergence before line integrals and others after Green's theorem, and some stop at Green's theorem altogether — so check the outline for your own section to see where this set falls in your term.
What do I need to know before starting this set?
CEGEP Calculus I and II and CEGEP Linear Algebra, and the earlier Calculus III sets. You should be able to parametrize segments, circles and graphs, take dot products and lengths of vectors, compute partial derivatives, and evaluate double integrals over triangles and in polar coordinates.
What is the difference between integrating with respect to ds and along dr?
An integral with respect to arc length, ds, adds up a scalar quantity along the curve — such as the mass of a wire — and does not depend on the direction of travel. An integral along dr adds up the component of a vector field in the direction of travel, such as work, and changes sign when the curve is reversed.
How do I know whether a vector field is conservative?
In the plane, compare ∂P/∂y with ∂Q/∂x; in space, check all three cross-partial conditions, which together say the curl is the zero vector. Equality proves the field conservative when the components have continuous partial derivatives on a region with no holes, such as the whole plane or all of space. Inequality anywhere proves it is not.
When should I use Green's theorem instead of parametrizing?
When the curve is closed and ∂Q/∂x − ∂P/∂y is simpler than P and Q themselves — especially when a term of P or Q has no elementary antiderivative, or the curve is made of several pieces. Check that the curve is traversed counterclockwise, or account for the sign if it is not.
Why does the fundamental theorem for line integrals ignore the path?
Because along any curve, the dot product of a gradient with r′(t) is the derivative of the potential along that curve, by the chain rule. Integrating a derivative gives the change in the function between the endpoints, so what the curve does in between drops out — provided the field really is a gradient.
Are surface integrals, Stokes' theorem or the divergence theorem in this set?
No. They open the next set, which starts from parametric surfaces. This set stops at Green's theorem and at curl and divergence as derivatives of a field, which the next set then puts to work.
Can teachers use this in class?
Yes. Print and photocopy it for your own classes freely — I just ask that the tutorinmontreal.ca footer stays on the page.
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