Integrating Functions Independent of One Coordinate

Why One Coordinate Can Disappear

A double integral over a rectangle accumulates values across both coordinates. But if the integrand does not change as one coordinate changes, then integrating across that coordinate simply repeats the same value over a fixed length. The double integral can then be reduced to a single integral multiplied by that length.

The key is to match the multiplier to the coordinate the function does not depend on. If the integrand depends on \(x\), the multiplier is the rectangle’s side length in the \(y\)-direction. If it depends on \(y\), the multiplier is the side length in the \(x\)-direction.

Rectangle and Coordinate-Independent Integrands

Let \(R=[a,b]\times[c,d]\) be a rectangle, where \(a\le b\) and \(c\le d\). Its horizontal side length is \(b-a\), and its vertical side length is \(d-c\). Suppose \(\phi\) is continuous on the relevant interval. An integrand that depends only on \(x\) has the form \(f(x,y)=\phi(x)\); one that depends only on \(y\) has the form \(f(x,y)=\psi(y)\).

Definition: An integrand is independent of \(y\) if its value is determined by \(x\) alone, so it can be written \(f(x,y)=\phi(x)\) throughout the rectangle. It is independent of \(x\) if it can be written \(f(x,y)=\psi(y)\).

For continuous integrands on a rectangle, Fubini’s theorem allows the double integral to be evaluated as an iterated integral in either order. Applying it gives the reduction formulas:

\int_c^d\int_a^b \phi(x)\,dx\,dy =(d-c)\int_a^b \phi(x)\,dx

\int_a^b\int_c^d \psi(y)\,dy\,dx =(b-a)\int_c^d \psi(y)\,dy.

In the first formula, \(\phi(x)\) does not vary with \(y\), so the inner \(x\)-integral is repeated for every \(y\) from \(c\) to \(d\). In the second, \(\psi(y)\) does not vary with \(x\), so the inner \(y\)-integral is repeated across the \(x\)-interval.

Why the Side-Length Factor Appears

Consider first an integrand \(f(x,y)=\phi(x)\) and integrate with respect to \(y\) first. For each fixed \(x\), the value \(\phi(x)\) is constant as \(y\) runs from \(c\) to \(d\). Thus its integral over that interval is the constant value multiplied by the interval’s length. The remaining integral accumulates those values across \(x\).

\int_a^b\int_c^d \phi(x)\,dy\,dx =\int_a^b\bigl(\phi(x)(d-c)\bigr)\,dx =(d-c)\int_a^b\phi(x)\,dx.

This also explains the result through rectangular Riemann sums. Partition the \(x\)-interval into subintervals of widths \(\Delta x_i\), and the \(y\)-interval into subintervals of widths \(\Delta y_j\). Since the sampled value depends only on \(x\), call it \(\phi(x_i^*)\). The sum over all rectangles is

\sum_i\sum_j \phi(x_i^*)\,\Delta x_i\,\Delta y_j =\sum_i \phi(x_i^*)\,\Delta x_i\left(\sum_j\Delta y_j\right) =(d-c)\sum_i\phi(x_i^*)\,\Delta x_i.

The \(y\)-subinterval widths add to the full length \(d-c\). As the partitions are refined, the \(x\)-sum approaches \(\int_a^b\phi(x)\,dx\). Therefore the double integral is \((d-c)\int_a^b\phi(x)\,dx\). The argument for an integrand depending only on \(y\) is the same with the roles of the coordinates exchanged.

Theorem: If \(\phi\) is continuous on \([a,b]\), then for \(R=[a,b]\times[c,d]\), \(\iint_R\phi(x)\,dA=(d-c)\int_a^b\phi(x)\,dx\). If \(\psi\) is continuous on \([c,d]\), then \(\iint_R\psi(y)\,dA=(b-a)\int_c^d\psi(y)\,dy\). Continuity ensures the integrands are integrable on the rectangle, so Fubini’s theorem applies.

The formulas do not require the integrand to be positive. If it takes both positive and negative values, the double integral is still signed, and the same reduction applies. If either side length is zero, the rectangle has zero area and the corresponding double integral is zero.

Worked Examples

Worked Example: An Integrand Depending Only on \(x\)

Evaluate \(\iint_R \frac{1}{1+x}\,dA\) on \(R=[0,3]\times[-1,2]\).

The integrand depends on \(x\), not \(y\). The \(y\)-side length is \(2-(-1)=3\). Therefore, reduce the double integral to this length times an integral in \(x\):

\iint_R \frac{1}{1+x}\,dA =3\int_0^3\frac{1}{1+x}\,dx.

An antiderivative of \(1/(1+x)\) is \(\ln(1+x)\) on this interval. Hence

3\int_0^3\frac{1}{1+x}\,dx =3\left[\ln(1+x)\right]_0^3 =3\bigl(\ln 4-\ln 1\bigr) =3\ln 4.

Check by integrating in the other order: for each fixed \(x\), integrating \(1/(1+x)\) over \(y\in[-1,2]\) gives \(3/(1+x)\). Integrating that over \(x\in[0,3]\) again gives \(3\ln 4\), as required.

Worked Example: An Integrand Depending Only on \(y\)

Evaluate \(\iint_R \sqrt{y+1}\,dA\) on \(R=[2,5]\times[0,3]\).

This time the integrand depends only on \(y\), so use the \(x\)-side length, \(5-2=3\):

\iint_R\sqrt{y+1}\,dA =3\int_0^3\sqrt{y+1}\,dy.

Since \(\sqrt{y+1}=(y+1)^{1/2}\), an antiderivative is \(\frac{2}{3}(y+1)^{3/2}\). Evaluating at the bounds gives

3\int_0^3\sqrt{y+1}\,dy =3\left[\frac{2}{3}(y+1)^{3/2}\right]_0^3 =3\cdot\frac{2}{3}\left(4^{3/2}-1^{3/2}\right) =3\cdot\frac{14}{3} =14.

Check in the opposite order: for each \(y\), integrating \(\sqrt{y+1}\) over \(x\in[2,5]\) multiplies it by \(3\). The remaining \(y\)-integral is exactly \(3\int_0^3\sqrt{y+1}\,dy=14\).

Worked Example: A Rational Integrand Depending Only on \(x\)

Evaluate \(\iint_R \frac{x}{x+1}\,dA\) on \(R=[0,1]\times[-2,2]\).

The integrand is independent of \(y\). The \(y\)-side length is \(2-(-2)=4\), so

\iint_R\frac{x}{x+1}\,dA =4\int_0^1\frac{x}{x+1}\,dx.

Rewrite the fraction as \(x/(x+1)=1-1/(x+1)\). Then

4\int_0^1\left(1-\frac{1}{x+1}\right)\,dx =4\left[x-\ln(x+1)\right]_0^1 =4\bigl((1-\ln 2)-(0-\ln 1)\bigr) =4(1-\ln 2).

As a check, integrating the integrand over \(y\in[-2,2]\) first gives \(4x/(x+1)\). Its integral from \(x=0\) to \(x=1\) is \(4(1-\ln 2)\), agreeing with the result.

A Common Mistake: Multiplying by the Wrong Length

The multiplier is the length of the side in the coordinate that the integrand does not depend on. It is not automatically the \(x\)-length or the \(y\)-length, and it is not the side length corresponding to the remaining single integral.

For example, in the first worked example the integrand depends on \(x\), while the rectangle has \(x\)-length \(3\) and \(y\)-length \(3\); that particular rectangle cannot expose a mistaken choice because the lengths happen to match. Consider instead \([0,2]\times[1,5]\) with integrand \(1/(1+x)\). Its \(x\)-length is \(2\), and its \(y\)-length is \(4\). A wrong calculation would use the \(x\)-length as the multiplier:

\text{Wrong:}\qquad 2\int_0^2\frac{1}{1+x}\,dx=2\ln 3.

But the integrand is independent of \(y\), so the multiplier must be the \(y\)-length, \(5-1=4\). The correct calculation is

\iint_R\frac{1}{1+x}\,dA =4\int_0^2\frac{1}{1+x}\,dx =4\left[\ln(1+x)\right]_0^2 =4\ln 3.

The check is direct: for every fixed \(x\), the value \(1/(1+x)\) is repeated over the \(y\)-interval of length \(4\). Its contribution at that \(x\) must therefore be \(4/(1+x)\), not \(2/(1+x)\).

Choosing the Single Integral

The quickest way to apply the rule is to identify the coordinate that does not appear in the integrand, find the length of the corresponding side, and keep the bounds of the coordinate that does appear. The result should have one single integral and one length factor.

Integrand	Coordinate absent	Multiply by	Single integral
\(\phi(x)\)	\(y\)	\(d-c\)	\(\int_a^b\phi(x)\,dx\)
\(\psi(y)\)	\(x\)	\(b-a\)	\(\int_c^d\psi(y)\,dy\)

This reduction is useful because it removes an integration step without changing the value of the double integral. It also offers a quick check on setup: if the integrand is \(\phi(x)\), the remaining integration bounds must be the \(x\)-bounds, while the multiplier must come from the \(y\)-bounds. Thinking in terms of repeated values across the rectangle makes the rule easier to remember and helps distinguish it from simply multiplying by the rectangle’s area.

Check Your Understanding

Use the coordinate-independent rule and state which side length is used in each case.

For \(R=[1,4]\times[0,2]\), write \(\iint_R \ln(x+1)\,dA\) as a length times a single integral.
For \(R=[-1,1]\times[2,6]\), write \(\iint_R\sqrt{y}\,dA\) as a length times a single integral.
Why does integrating a function \(\phi(x)\) with respect to \(y\) produce a factor of \(d-c\)?
In the formula for \(\psi(y)\), which interval supplies the multiplier, and why?
What is the double integral of a continuous function \(\phi(x)\) over \([a,b]\times[c,c]\), and what geometric feature explains the result?