Choose the variable and bounds
One curve, two ways to slice it
The curve is \(y = \sqrt[3]{x}\) from \((0,\,0)\) to \((8,\,2)\), with four orange segments laid along it. The corners sit at equal steps of \(x\), at \(x = 0,\,2,\,4,\,6,\,8\), so the curve is sliced by \(x\). The label under the x-axis shows that \(x\) runs from 0 to 8.
Press Slice by y. The corners slide to equal steps of \(y\), at \(y = 0,\,0.5,\,1,\,1.5,\,2\), and the label moves to the y-axis, where \(y\) runs from 0 to 2.
The arc is the same in both pictures, so its length is the same. Only the variable changes, and its bounds, 0 to 8 or 0 to 2, change with it.
The same arc, either variable
The thick segment is the hypotenuse of a right triangle whose green legs are the small changes \(dx\) and \(dy\), so \(ds = \sqrt{dx^2 + dy^2}\).
If you slice by \(x\), you take \(dx\) out of the root, so \(ds = \sqrt{1 + (dy/dx)^2}\,dx\). If you slice by \(y\), you take \(dy\) out, so \(ds = \sqrt{1 + (dx/dy)^2}\,dy\).
Drag the far end. Sliced by \(x\), the bounds are \(a\) and \(b\), the x-values of the ends, in orange on the x-axis. Sliced by \(y\), they are \(c\) and \(d\), the y-values of the same two ends.
Both ways measure the same arc, so the derivative in the root and the bounds must belong to the one variable you chose.
Notation
If the curve is \(y = f(x)\), slice by \(x\). The bounds \(a\) and \(b\) are the x-values of the two ends.
\[\cy{L} = \int_a^b \sqrt{1 + \left(\dfrac{dy}{dx}\right)^2}\,dx\]If the curve is \(x = g(y)\), slice by \(y\). The bounds \(c\) and \(d\) are the y-values of the two ends.
\[\cy{L} = \int_c^d \sqrt{1 + \left(\dfrac{dx}{dy}\right)^2}\,dy\]The derivative must match the variable of integration. The form with \(dx\) uses \(dy/dx\), and the form with \(dy\) uses \(dx/dy\).
Slice by a variable in which the curve is a function, so that each value of it gives one point of the arc. The derivative must also exist and be continuous on the interval.
A curve is a function of \(x\) when each vertical line meets it once. It is a function of \(y\) when each horizontal line meets it once.
When it fails
The curve \(y = \sqrt[3]{x}\) has a vertical tangent at the origin. A vertical line has no slope, so \(dy/dx\) does not exist at \(x = 0\) and the \(x\) form has no integrand there.
Write the same curve as \(x = y^3\). Then \(dx/dy = 3y^2\), which is 0 at \(y = 0\). This derivative exists and is continuous, so the \(y\) form works.
Example 1: x given in terms of y
Write the integral for the length of \(x = \dfrac{y^2}{2} + 1\) from \((3,\,2)\) to \((9,\,4)\).
The equation gives \(x\) in terms of \(y\), so slice by \(y\) and find \(dx/dy\).
\[\dfrac{dx}{dy} = y\]Square the derivative, add 1, and take the root. The thick segment is about \(\sqrt{1 + y^2}\,\Delta y\) long.
\[\sqrt{1 + y^2}\]The bounds are the y-values of the ends, 2 and 4, not the faint x-values 3 and 9.
\[L = \int_2^4 \sqrt{1 + y^2}\,dy\]
Example 2: a curve with a square root
Write the length of \(y = 3 + \sqrt{x}\) from \((1,\,4)\) to \((9,\,6)\) as an integral over \(y\).
The integral is over \(y\), so solve for \(x\). Because \(y = 3 + \sqrt{x}\), \(\sqrt{x} = y - 3\), and squaring both sides gives \(x\). Then find \(dx/dy\).
\[x = (y - 3)^2, \qquad \dfrac{dx}{dy} = 2(y - 3)\]Square the derivative, add 1, and take the root.
\[\sqrt{1 + 4(y - 3)^2}\]A tempting wrong move: Use the bounds 1 and 9 from the original problem.
\[\int_1^9 \sqrt{1 + 4(y - 3)^2}\,dy\]The integral runs along \(y\), but 1 and 9 are x-values. The red stretch continues to \(y = 9\), where \(x = 36\), far past the end \((9,\,6)\), so the integral measures a longer arc.
The bounds are the y-values of the ends, 4 and 6.
\[L = \int_4^6 \sqrt{1 + 4(y - 3)^2}\,dy\]
Practice
The curve \(y = \ln x\) runs from \((1,\,0)\) to \((3,\,\ln 3)\). Write its length as an integral over \(x\), and as an integral over \(y\).
For the integral over \(y\), solve for \(x\) first. The bounds are the values of \(y\) at the two ends.
Solution
Slice by \(x\). The derivative is \(1/x\), and the x-values of the ends are 1 and 3.
\[L = \int_1^3 \sqrt{1 + \dfrac{1}{x^2}}\,dx\]To slice by \(y\), solve for \(x\) and find \(dx/dy\).
\[x = e^{y}, \qquad \dfrac{dx}{dy} = e^{y}\]The bounds are the y-values of the ends, 0 and \(\ln 3\).
\[L = \int_0^{\ln 3} \sqrt{1 + e^{2y}}\,dy\]
Questions
The graph shows \(y = \sqrt[3]{x}\) from \((0,\,0)\) to \((8,\,2)\). The tangent line at the origin is vertical.
The length is written as an integral over \(y\). What are the bounds?
- 0 to 2
- 0 to 8
- 2 to 8
Answer
0 to 2
Which statement about the vertical tangent at the origin is correct?
- \(dy/dx\) does not exist at \(x = 0\), so the \(x\) form has no integrand at the left end
- \(dx/dy\) does not exist at \(y = 0\), so the \(y\) form fails
- Neither form can be used, because the curve is vertical
- The curve is not a function of \(y\)
Answer
\(dy/dx\) does not exist at \(x = 0\), so the \(x\) form has no integrand at the left end
Which integral gives the length of \(x = 4 - y^2\) from \((3,\,1)\) to \((0,\,2)\)?
- \(\displaystyle\int_1^2 \sqrt{1 + 4y^2}\,dy\)
- \(\displaystyle\int_0^3 \sqrt{1 + 4y^2}\,dy\)
- \(\displaystyle\int_1^2 \sqrt{1 + \dfrac{1}{4y^2}}\,dy\)
- \(\displaystyle\int_1^2 \sqrt{1 + 2y}\,dy\)
Answer
\(\displaystyle\int_1^2 \sqrt{1 + 4y^2}\,dy\)
Summary
You slice by a variable in which the curve is a function with a continuous derivative, and that derivative goes in the root.
The integral runs along that variable, so the bounds are its values at the two ends of the arc.
Slice by \(y\) when the curve is given as \(x = g(y)\), or when \(dy/dx\) does not exist at an end, as for \(y = \sqrt[3]{x}\) at the origin.
This is one of four skills in topic 8.13 of Which & Why: AP Calculus.