A wire is shaped like a curve C with linear density f(x, y). You parametrize C two different ways: one traversal takes 2 seconds, another takes 6 seconds over the same path. What is the relationship between the two values of ∫_C f ds?
AThe slower traversal gives a larger value because more time elapses per unit length
BThey are equal — the |r'(t)| factor converts parameter speed to actual arc length, making the result independent of parametrization
CThe faster traversal gives a larger value because the speed term |r'(t)| is larger
DThey differ unless the parametrization is linear in t
The |r'(t)| factor is precisely what makes the scalar line integral parametrization-independent. A faster traversal has a larger |r'(t)|, but the dt interval is correspondingly compressed — the product |r'(t)| dt always equals the actual arc-length element ds. The integral measures mass (or total accumulated quantity) along the curve as a geometric object, not as a function of how you choose to traverse it.
Question 2 Multiple Choice
You compute ∫_C f ds along a curve C from point A to point B. A classmate computes the same integral from B to A (reversing the direction). How do the two results compare?
AThe classmate's result is the negative of yours, because the direction of integration reversed
BThe classmate's result equals yours — scalar line integrals are orientation-independent
CThe classmate's result equals yours only if f is a constant function
DThe classmate's result is twice yours, because the path is traversed in the other direction
Unlike vector-field line integrals (where reversing direction negates the result because the dot product with r'(t) flips sign), scalar line integrals use |r'(t)| — the absolute value of the derivative — which is always positive regardless of direction. Both f and ds are unaffected by orientation, so ∫_C f ds is the same in either direction. This is the key structural difference between scalar and vector line integrals.
Question 3 True / False
The scalar line integral ∫_C f ds gives the same numerical value regardless of which parametrization you choose for the curve C.
TTrue
FFalse
Answer: True
This is a fundamental property of scalar line integrals. The |r'(t)| factor in the formula ∫_a^b f(r(t)) |r'(t)| dt acts as a 'speed correction' that converts changes in the parameter t into actual arc length. No matter how fast or slow you traverse the curve — as long as you traverse the same path — the product f(r(t)) · |r'(t)| dt accumulates the same total quantity along the curve.
Question 4 True / False
The scalar line integral ∫_C f ds changes sign when the direction of traversal along C is reversed, just as a definite integral ∫_a^b f(x) dx changes sign when the limits are swapped.
TTrue
FFalse
Answer: False
This is a common confusion between scalar and vector line integrals. In a single-variable integral, swapping limits introduces a negative sign. But in a scalar line integral, the arc-length element ds = |r'(t)| dt is always positive — reversing direction does NOT flip the sign. The correct analogy for sign-sensitive integrals is the vector line integral ∫_C F · dr, where reversing orientation negates the result because r'(t) (not |r'(t)|) appears in the formula.
Question 5 Short Answer
Why must the |r'(t)| factor be included in the scalar line integral formula ∫_a^b f(r(t)) |r'(t)| dt, rather than just integrating f(r(t)) dt directly?
Think about your answer, then reveal below.
Model answer: The parameter t is not arc length — it is an arbitrary label for positions on the curve. The factor |r'(t)| converts dt (a change in parameter) into ds (an actual infinitesimal length along the curve). Without it, you would be summing f values weighted by parameter increments, which depends on how fast you traverse the curve and gives different answers for different parametrizations. Including |r'(t)| ensures you are integrating f per unit of actual geometric length — giving a result that depends only on the curve as a set of points, not on the arbitrary choice of parameter.
Think of it physically: if f is mass per unit length of a wire, the mass of a tiny piece is f · (length of piece). The length of the piece is |r'(t)| dt, not dt itself. A fast parametrization compresses dt but has large |r'(t)|; a slow one expands dt but has small |r'(t)|. The product is always the same infinitesimal arc length. Omitting |r'(t)| would make the 'mass' calculation depend on traversal speed — a physically meaningless artifact.