Setting Up Length Integrals in Either Graph Orientation

Choosing a Graph Orientation

A curve can sometimes be described either by giving \(y\) as a function of \(x\), or by giving \(x\) as a function of \(y\). The choice matters when setting up an arc-length integral: it determines both the derivative in the integrand and the coordinate used for the bounds.

The key question is not simply which equation looks more familiar. It is whether the chosen description represents the entire curve segment as a single-valued graph, and whether its bounds cover exactly the segment whose length is wanted.

Definition: A curve segment is written as \(y=f(x)\) on \([a,b]\) when each \(x\) in that interval has exactly one corresponding point \((x,f(x))\) on the segment. It is written as \(x=g(y)\) on \([c,d]\) when each \(y\) in that interval has exactly one corresponding point \((g(y),y)\) on the segment.

The Two Graph-Length Formulas

For a continuously differentiable function \(y=f(x)\) on \([a,b]\), the arc length of its graph is given by integrating with respect to \(x\). The derivative in the integrand is \(dy/dx=f'(x)\).

L=\int_a^b \sqrt{1+\bigl(f'(x)\bigr)^2}\,dx.

If the curve is instead written as \(x=g(y)\) for \(c\le y\le d\), integrate with respect to \(y\). The derivative is now \(dx/dy=g'(y)\), and the bounds are \(y\)-values.

L=\int_c^d \sqrt{1+\bigl(g'(y)\bigr)^2}\,dy.

These formulas require a suitable graph description and a derivative that exists and is continuous on the interval. If a curve has corners or the derivative fails to be defined at an endpoint, the interval may need to be split or an improper integral may be needed. The examples here use intervals where the formulas apply directly.

Rule: For \(y=f(x)\), use \(f'(x)\) and \(x\)-bounds. For \(x=g(y)\), use \(g'(y)\) and \(y\)-bounds. The variable of integration, derivative, and limits must all match the chosen orientation.

Why the \(x=g(y)\) Formula Works

The formula for \(x=g(y)\) follows from the ordinary graph formula by exchanging the roles of the horizontal and vertical coordinates. To see this precisely, describe the same points using new coordinates \(X=y\) and \(Y=x\). In these coordinates, the curve is the graph \(Y=g(X)\).

The graph formula applied to \(Y=g(X)\), over \(c\le X\le d\), gives

L=\int_c^d \sqrt{1+\left(\frac{dY}{dX}\right)^2}\,dX.

Because \(Y=x\) and \(X=y\), we have \(dY/dX=dx/dy=g'(y)\) and \(dX=dy\). Substituting these equalities yields

L=\int_c^d \sqrt{1+\bigl(g'(y)\bigr)^2}\,dy.

Thus the second formula is not a separate rule to memorize: it is the graph formula with the coordinate roles exchanged. The same reasoning also explains why using \(dy\) requires the derivative \(dx/dy\), rather than \(dy/dx\).

Worked Examples: Setting Up Both Orientations for a Line

Worked Example: A Line Segment in Either Orientation

Find a length integral for the segment of \(y=2x+1\) from \(x=0\) to \(x=3\), and write the same segment using \(x\) as a function of \(y\).

First use \(y=f(x)=2x+1\). The derivative is \(f'(x)=2\), and the stated \(x\)-bounds are 0 and 3. Therefore the setup is

L=\int_0^3\sqrt{1+2^2}\,dx =\int_0^3\sqrt{5}\,dx.

To express the line in the other orientation, solve for \(x\):

y=2x+1 \quad\Longrightarrow\quad x=\frac{y-1}{2}.

At \(x=0\), \(y=2(0)+1=1\). At \(x=3\), \(y=2(3)+1=7\). Thus the matching \(y\)-bounds are 1 and 7. Since \(g(y)=(y-1)/2\), its derivative is \(g'(y)=1/2\), so the alternative setup is

L=\int_1^7\sqrt{1+\left(\frac12\right)^2}\,dy =\int_1^7\sqrt{\frac54}\,dy.

Check the endpoint correspondence: \(g(1)=(1-1)/2=0\) and \(g(7)=(7-1)/2=3\). Both integrals cover precisely the segment from \((0,1)\) to \((3,7)\). They are two setups for the same length, not two different pieces of the line.

When One Orientation Avoids Multiple Branches

A curve may be a single-valued graph in one orientation but not the other. For instance, a sideways parabola \(x=g(y)\) can assign one \(x\)-value to every \(y\) in an interval, even though some \(x\)-values correspond to two different \(y\)-values. In that situation, the \(x=g(y)\) description can cover the whole chosen segment without combining separate branches.

Worked Example: A Sideways Parabola with a Fold

Set up the length integral for the portion of \(x=y^2+1\) with \(-1\le y\le2\). Determine why this portion cannot be written as one graph \(y=f(x)\) over its full \(x\)-range.

The chosen orientation is \(x=g(y)\), where \(g(y)=y^2+1\). Differentiating with respect to \(y\) gives \(g'(y)=2y\). The \(y\)-bounds are already given, so the length integral is

L=\int_{-1}^2\sqrt{1+(2y)^2}\,dy =\int_{-1}^2\sqrt{1+4y^2}\,dy.

To check the corresponding \(x\)-values, evaluate \(g(-1)=(-1)^2+1=2\), \(g(0)=0^2+1=1\), and \(g(2)=2^2+1=5\). The curve runs from \((2,-1)\) to \((1,0)\) to \((5,2)\). The \(x\)-coordinate decreases and then increases, but that does not cause a problem for the \(x=g(y)\) formula: \(y\) itself still runs through the interval from \(-1\) to 2 exactly once.

This segment is not a single graph \(y=f(x)\) over its full \(x\)-range. For example, \(x=2\) occurs at both \(y=-1\) and \(y=1\), since \((-1)^2+1=2\) and \(1^2+1=2\). One input \(x=2\) would therefore need two output values. The \(x=g(y)\) description avoids that branch problem.

Bounds Must Match the Chosen Variable

Bounds are the coordinate values at the ends of the chosen interval, not automatically the coordinates that appear most prominently in the original equation. If integrating with respect to \(x\), identify the leftmost and rightmost \(x\)-values of the intended graph segment. If integrating with respect to \(y\), identify its lowest and highest \(y\)-values. Then check that the interval traces exactly the intended portion.

When changing orientation, solve the equation for the other variable and translate the endpoints. Do not merely copy the old limits. If the curve turns back in the coordinate being used as the independent variable, a single-valued graph may not cover the full segment; split the curve into suitable pieces or use the other orientation.

Worked Example: Converting Bounds to the Inverse Orientation

Set up the length of the curve \(y=x^{1/3}\) from \(x=1\) to \(x=8\), first as \(y=f(x)\) and then as \(x=g(y)\).

For the \(y=f(x)\) orientation, \(f'(x)=\frac{1}{3}x^{-2/3}\). The stated \(x\)-bounds give

L=\int_1^8 \sqrt{1+\left(\frac{1}{3}x^{-2/3}\right)^2}\,dx =\int_1^8\sqrt{1+\frac{1}{9x^{4/3}}}\,dx.

For the other orientation, cube both sides to obtain \(x=y^3\), so \(g'(y)=3y^2\). Translate the endpoints: when \(x=1\), \(y=1^{1/3}=1\); when \(x=8\), \(y=8^{1/3}=2\). Therefore

L=\int_1^2\sqrt{1+(3y^2)^2}\,dy =\int_1^2\sqrt{1+9y^4}\,dy.

The endpoint check in the inverse description is \(g(1)=1^3=1\) and \(g(2)=2^3=8\). These are the original \(x\)-coordinates, so the two setups describe the same segment. In particular, the inverse-orientation limits are 1 and 2, not 1 and 8: those are the \(y\)-values, while 1 and 8 are the \(x\)-values.

A Common Orientation Error

A frequent mistake is to switch the differential but keep the derivative from the other orientation. Consider \(x=y^2+1\), with \(-1\le y\le2\). The derivative needed when integrating with respect to \(y\) is \(dx/dy=2y\).

A wrong setup would be to use \(dy/dx\) in the \(dy\)-formula, perhaps writing \(\int_{-1}^2\sqrt{1+(dy/dx)^2}\,dy\). That derivative is not even a single-valued function of \(x\) on this whole curve: at \(x=2\), the curve has both \(y=-1\) and \(y=1\). Moreover, the formula for \(x=g(y)\) specifically requires \(dx/dy\).

The correction is to retain the chosen orientation consistently: \(x=g(y)=y^2+1\), \(g'(y)=2y\), and \(y\)-bounds \(-1\) and 2. The correct setup is

L=\int_{-1}^2\sqrt{1+(2y)^2}\,dy.

Checking the endpoint values, \(g(-1)=2\) and \(g(2)=5\), confirms that the bounds refer to the intended \(y\)-interval. A correct derivative with incorrect limits—or correct limits with a derivative from the other orientation—does not give a valid setup.

A Practical Setup Check

Before proceeding with a length integral, verify the description and the limits as a set. A useful check is to state the curve in the chosen orientation, identify its independent-variable interval, and evaluate the dependent variable at the interval endpoints. If the dependent coordinate reverses direction within the interval, ask whether the curve is still single-valued in the chosen orientation. For \(x=g(y)\), changes in the direction of \(x\) as \(y\) increases are allowed; repeating a \(y\)-value for two different points would be the problem.

Choose the graph orientation.
Use \(y=f(x)\) if each chosen \(x\) identifies one point on the segment; use \(x=g(y)\) if each chosen \(y\) does.

Differentiate the dependent coordinate.
For \(y=f(x)\), calculate \(dy/dx\). For \(x=g(y)\), calculate \(dx/dy\).

Set bounds in the independent coordinate.
Use \(x\)-values with \(dx\), and \(y\)-values with \(dy\). Translate endpoints when changing orientation.

Check coverage and endpoints.
Confirm that the chosen interval traces exactly the desired segment and that its endpoint values match the intended curve points.

The purpose of choosing an orientation carefully is to describe the segment exactly once with a single-valued graph. Once that description and its bounds are right, the formula follows directly: the square root contains one plus the square of the derivative for that orientation, and the differential matches its independent variable.

Check Your Understanding

For each question, identify the graph orientation, derivative, and bounds before writing the length integral.

For \(y=3x-2\) from \(x=1\) to \(x=4\), what is the arc-length setup in the \(y=f(x)\) orientation?
For \(x=y^2+4\) on \(-2\le y\le1\), write the length integral and state whether the whole segment is a single graph \(y=f(x)\).
If a segment is written as \(y=f(x)\) for \(2\le x\le5\), which derivative and differential belong in its graph-length formula?
For \(x=y^3\) with \(1\le y\le3\), what are the corresponding \(x\)-bounds, and what are the bounds when integrating with respect to \(y\)?
In an \(x=g(y)\) setup, what endpoint and single-valuedness checks help confirm that the bounds cover the intended segment?