Why Changing the Parameter Need Not Change the Integral
A parametrization describes a curve by specifying the position of a moving point at each parameter value. The parameter can be interpreted as a clock: one parametrization may move quickly along the curve, while another moves slowly. But if both describe the same traversal, the scalar line integral should depend on the curve and the scalar field, not on the choice of clock.
The reason is that a scalar line integral uses arc length, not elapsed parameter time. A change of parameter changes the speed in the parameter variable, and the arc-length factor changes with it. This tutorial makes that cancellation precise and shows how to use it in computations.
To change the parameter without changing the traversal, take a continuously differentiable, strictly increasing function \(\phi:[\alpha,\beta]\to[a,b]\) with \(\phi(\alpha)=a\) and \(\phi(\beta)=b\). The new parametrization is \(\mathbf{q}(u)=\mathbf{r}(\phi(u))\). It follows the original curve from the same starting point to the same endpoint, in the same order, but may use a different speed.
The Invariance Theorem
The proof uses the chain rule and the one-variable substitution rule. Differentiating the new parametrization gives \(\mathbf{q}'(u)=\mathbf{r}'(\phi(u))\phi'(u)\). Since \(\phi'(u)\geq 0\), taking the norm gives \(\|\mathbf{q}'(u)\|=\|\mathbf{r}'(\phi(u))\|\phi'(u)\). Therefore the new integral is the original integrand evaluated at \(t=\phi(u)\), multiplied by the derivative needed for substitution.
By the chain rule, \(\mathbf{q}'(u)=\mathbf{r}'(\phi(u))\phi'(u)\).
Because \(\phi'(u)\geq0\), \(\|\mathbf{q}'(u)\|=\|\mathbf{r}'(\phi(u))\|\phi'(u)\).
The integral becomes \(\int_\alpha^\beta f(\mathbf{r}(\phi(u)))\|\mathbf{r}'(\phi(u))\|\phi'(u)\,du\).
With \(t=\phi(u)\), the endpoints change from \(u=\alpha,\beta\) to \(t=a,b\), and \(dt=\phi'(u)\,du\). The resulting integral is \(\int_a^b f(\mathbf{r}(t))\|\mathbf{r}'(t)\|\,dt\).
This proves the equality. In particular, the proof does not require \(\phi'(u)>0\) at every point. A zero derivative at a point makes the new parametrization momentarily stationary there, but it does not invalidate the equality.
Worked Example: A Line Segment with a Slower Parameter
Worked Example: A Line Segment with a Slower Parameter
Consider the line segment \(\mathbf{r}(t)=(t,0)\), \(0\leq t\leq1\), and the scalar field \(f(x,y)=x+1\). First calculate the integral using \(t\).
Thus $$ \int_C f\,ds=\int_0^1(t+1)\cdot1\,dt =\left[\frac{t^2}{2}+t\right]_0^1 =\frac{1}{2}+1 =\frac{3}{2}. $$ Now use the increasing change of parameter \(t=u^2\), \(0\leq u\leq1\). The new parametrization is \(\mathbf{q}(u)=(u^2,0)\), and $$ \mathbf{q}'(u)=(2u,0),\qquad \|\mathbf{q}'(u)\|=2u $$ because \(u\geq0\). Also \(f(\mathbf{q}(u))=u^2+1\). Hence $$ \int_C f\,ds=\int_0^1(u^2+1)2u\,du =\int_0^1(2u^3+2u)\,du =\left[\frac{u^4}{2}+u^2\right]_0^1 =\frac12+1 =\frac32. $$ The two calculations agree. Substituting \(t=u^2\) into the original position gives \(\mathbf{r}(u^2)=(u^2,0)=\mathbf{q}(u)\), confirming that both parametrizations describe the same traversal.
Worked Example: A Parabola with a Nonconstant Integrand
Worked Example: A Parabola with a Nonconstant Integrand
Let \(\mathbf{r}(t)=(t,t^2)\), \(0\leq t\leq1\), and \(f(x,y)=x\). Its derivative and speed are $$ \mathbf{r}'(t)=(1,2t),\qquad \|\mathbf{r}'(t)\|=\sqrt{1+4t^2},\qquad f(\mathbf{r}(t))=t. $$ The integral in this parametrization is $$ \int_0^1 t\sqrt{1+4t^2}\,dt. $$ Set \(w=1+4t^2\), so \(dw=8t\,dt\), \(t\,dt=dw/8\), and the endpoints \(t=0,1\) give \(w=1,5\). Therefore $$ \int_0^1 t\sqrt{1+4t^2}\,dt =\frac18\int_1^5w^{1/2}\,dw =\frac18\left[\frac23w^{3/2}\right]_1^5 =\frac{5\sqrt5-1}{12}. $$ Now choose \(t=u^2\), \(0\leq u\leq1\). Then \(\mathbf{q}(u)=(u^2,u^4)\), with $$ \mathbf{q}'(u)=(2u,4u^3),\qquad \|\mathbf{q}'(u)\|=\sqrt{4u^2+16u^6}=2u\sqrt{1+4u^4}, $$ where \(u\geq0\). The field value is \(f(\mathbf{q}(u))=u^2\), so $$ \int_C f\,ds =\int_0^1 2u^3\sqrt{1+4u^4}\,du. $$ Set \(w=1+4u^4\). Then \(dw=16u^3\,du\), \(2u^3\,du=dw/8\), and the endpoints give \(w=1,5\). Thus $$ \int_0^1 2u^3\sqrt{1+4u^4}\,du =\frac18\int_1^5w^{1/2}\,dw =\frac{5\sqrt5-1}{12}. $$ Both parametrizations give the same value. The identity \(\mathbf{r}(u^2)=(u^2,u^4)=\mathbf{q}(u)\) verifies that they describe the same curve traversal.
Worked Example: A Quarter Circle with a Slowed Parameter
Worked Example: A Quarter Circle with a Slowed Parameter
Consider the quarter circle parametrized by \(\mathbf{r}(t)=(\cos t,\sin t)\), \(0\leq t\leq\pi/2\), with \(f(x,y)=x\). Since $$ \mathbf{r}'(t)=(-\sin t,\cos t),\qquad \|\mathbf{r}'(t)\|=\sqrt{\sin^2t+\cos^2t}=1, $$ the integral is $$ \int_C f\,ds=\int_0^{\pi/2}\cos t\,dt =[\sin t]_0^{\pi/2}=1-0=1. $$ For a second parametrization set \(t=\frac{\pi}{2}u^2\), \(0\leq u\leq1\). Then $$ \mathbf{q}(u)=\left(\cos\left(\frac{\pi}{2}u^2\right),\sin\left(\frac{\pi}{2}u^2\right)\right). $$ The chain rule gives $$ \mathbf{q}'(u)=\pi u\left(-\sin\left(\frac{\pi}{2}u^2\right),\cos\left(\frac{\pi}{2}u^2\right)\right), \qquad \|\mathbf{q}'(u)\|=\pi u. $$ The scalar field value is \(\cos(\frac{\pi}{2}u^2)\), so $$ \int_C f\,ds =\int_0^1\cos\left(\frac{\pi}{2}u^2\right)\pi u\,du. $$ Set \(v=\frac{\pi}{2}u^2\), so \(dv=\pi u\,du\). The endpoints become \(v=0\) and \(v=\pi/2\), giving $$ \int_0^{\pi/2}\cos v\,dv =[\sin v]_0^{\pi/2}=1. $$ This agrees with the first calculation. Also, substituting \(t=\frac{\pi}{2}u^2\) into \(\mathbf{r}\) gives exactly \(\mathbf{q}(u)\), so the change only alters the speed along the quarter circle.
What the Conditions Prevent
The increasing change of parameter must cover the original parameter interval exactly once, in order. If the change reverses direction, the speed calculation involves an absolute value, and a separate orientation argument is needed. If it turns around or traverses part of the interval more than once, the integral may count portions of the curve repeatedly. Such a map is not an increasing reparametrization as defined here.
For example, replacing \(t\) by \(u^2\) on \([-1,1]\) is not a one-to-one increasing change from \([0,1]\) onto itself: \(u^2\) runs from \(1\) down to \(0\) and then back to \(1\). Applying the increasing-reparametrization theorem to that interval would be invalid. The issue is not merely algebraic; the proposed parameter does not describe a single traversal in the required order.
Why Reparametrization Is Useful
The invariance theorem allows a calculation to use whichever increasing parametrization makes the integrand or the speed easier to handle. A parameter that simplifies the coordinates can be more useful than one that makes the motion appear uniform. What must remain fixed is the geometric traversal and the scalar field evaluated at the corresponding points.
Whenever two proposed parametrizations are used for the same scalar line integral, check three things: their positions trace the same curve, the parameter change is increasing and covers the interval once, and the speed factor comes from differentiating the full new parametrization. With these conditions in place, the theorem guarantees equal integral values; direct computation in both parameters provides a useful check.
Check Your Understanding
Use the definition and invariance result to answer the following questions.
- For \(\mathbf{q}(u)=\mathbf{r}(\phi(u))\), what chain-rule formula gives the new velocity?
- Why does \(\|\mathbf{q}'(u)\|=\|\mathbf{r}'(\phi(u))\|\phi'(u)\) for an increasing differentiable \(\phi\)?
- Must \(\phi'(u)\) be positive at every point for the invariance theorem to apply? Explain.
- For \(\mathbf{r}(t)=(t,0)\) and \(t=u^2\), what factor would be omitted by using only \(\|\mathbf{r}'(u^2)\|\) as the new speed?
- Why is a parameter change that traverses an interval forward and then backward not an increasing reparametrization?