Work against a constant force along a curve

∫C F·dr where F = (0, -9.8, 0) along r(t) = (t, t^2, 0) equals -196/5 ≈ -39.2. A worked vector field line integral with an editable calculator.

∫C F·dr where F = (0, -9.8, 0) = -196/5 ≈ -39.2. A constant downward force does work that depends only on the height change, not the path taken.

Vector field F = (P, Q, R)

Each component in terms of x, y and z

Curve C: r(t) = (x(t), y(t), z(t))

Leave z(t) as 0 for a plane curve

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Frequently asked questions

What is the value of this line integral?

∫C F·dr where F = (0, -9.8, 0) = -196/5 ≈ -39.2. A constant downward force does work that depends only on the height change, not the path taken.

What curve is being integrated along?

The curve is parametrised as r(t) = (t, t^2, 0) for t from 0 to 2, and it has arc length 4.646784.

Does the direction of travel matter?

Yes. Reversing the orientation of the curve flips the sign of a vector line integral, because the direction of dr reverses.

About this Calculator

∫C F·dr where F = (0, -9.8, 0) along r(t) = (t, t^2, 0) equals -196/5 ≈ -39.2. A worked vector field line integral with an editable calculator.

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