Line Integral Preview

Math Calculus • Multivariable Calculus

Written by STEM Calculators Team Published December 30, 2025 Updated February 24, 2026

9. Line Integral Preview

Compute \(\displaystyle \int_C P\,dx + Q\,dy\) using a parametric path \(x=x(t),\,y=y(t)\), \(t\in[a,b]\). Includes a field + path graph (pan/zoom) and a conservative-field (path-independence) check.

Vector field

\(P(x,y)\): \(Q(x,y)\):

Input supports pi, e, sqrt(2), trig (sin, cos), etc. Use * for multiplication.

Path \(C\) (parametric)

\(x(t)\): \(y(t)\): \(a\) (t min): \(b\) (t max): Probe \(t_0\): Samples:

The calculator evaluates \[ \int_a^b \big(P(x(t),y(t))\,x'(t)+Q(x(t),y(t))\,y'(t)\big)\,dt. \]

Graph settings (optional)

x min: x max: y min: y max: Field arrows: Arrow density: Normalize arrows: Arrow scale: Show grid: Engine preference:

Ready

Enter \(P,Q\) and a path, then click “Compute”.

Graph

Field + path (pan/zoom)

Axes units are in your \(x,y\) input units. The curve is the path \(C\). Arrows show \(\langle P,Q\rangle\). Hover to probe values. Drag to pan; mouse wheel to zoom; double-click resets.

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Theory

Line integrals of a vector field

For a 2D vector field \(\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle\), the line integral along a curve \(C\) is \[ \int_C \mathbf{F}\cdot d\mathbf{r} =\int_C P\,dx + Q\,dy. \] Physically, this often represents work done by the field as you move along the path.

Parametrization formula

If \(C\) is parametrized by \(x=x(t)\), \(y=y(t)\) for \(t\in[a,b]\), then \[ dx=x'(t)\,dt,\qquad dy=y'(t)\,dt, \] so the integral becomes \[ \int_C P\,dx + Q\,dy =\int_a^b \Big(P(x(t),y(t))\,x'(t)+Q(x(t),y(t))\,y'(t)\Big)\,dt. \]

Conservative fields and path independence

A field \(\mathbf{F}=\langle P,Q\rangle\) is conservative on a simply connected region if there exists a potential function \(\phi\) such that \(\nabla \phi=\mathbf{F}\). In that case, \[ \int_C \mathbf{F}\cdot d\mathbf{r}=\phi(B)-\phi(A), \] meaning the value depends only on the endpoints (path independent).

A common test in 2D is the curl condition: \[ \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0 \] (together with a simply connected domain) \(\Rightarrow\) conservative.

Worked example

Compute \(\int_C y\,dx + x\,dy\) along the line from \((0,0)\) to \((1,1)\). Parametrize with \(x=t\), \(y=t\), \(t\in[0,1]\). Then \(dx=dt\), \(dy=dt\), so \[ \int_0^1 \big(t\,dt+t\,dt\big)=\int_0^1 2t\,dt=t^2\Big|_0^1=1. \] Here \(\langle P,Q\rangle=\langle y,x\rangle\) is conservative (curl \(=1-1=0\)), and a potential is \(\phi=xy\).

Green’s theorem preview (closed curves)

If \(C\) is a positively oriented, simple closed curve bounding region \(R\), Green’s theorem says \[ \oint_C P\,dx+Q\,dy=\iint_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA. \] The calculator can provide a rough numerical preview when the path is (approximately) closed.

Frequently Asked Questions

How do you compute integral_C P dx + Q dy using a parametric curve?

If C is given by x=x(t), y=y(t) for t in [a,b], then dx = x'(t) dt and dy = y'(t) dt. The line integral becomes integral_a^b (P(x(t),y(t)) x'(t) + Q(x(t),y(t)) y'(t)) dt.

What does it mean if the vector field is conservative in this line integral tool?

A conservative field has a potential function phi with grad phi = <P,Q>, so the line integral depends only on the endpoints, not the path. A common 2D test is dQ/dx - dP/dy = 0 on a simply connected region.

Why does the calculator ask for a probe t0?

The probe t0 lets you inspect the curve point (x(t0), y(t0)) and related field values at a specific parameter value. It helps connect the integral computation to the behavior of the field along the path.

How do arrow normalization and arrow scale change the vector field plot?

Normalization makes arrows similar in length so direction is easier to compare across the grid. Arrow scale changes the displayed arrow length, which can improve readability when the field magnitude varies widely.