Module 39: Double integrals, changing order, and polar area elements

A. Prerequisites and observable goals#

B. The central idea: accumulate over an input region#

A double integral $\iint_Df(x,y)\dd A$ sums height times tiny area over a planar region. For nonnegative $f$, it is the volume under a graph; for density, it is mass; for a general signed function, it is signed accumulation. The area element belongs to the input plane. It is not the curved surface area of the graph, which is a later topic.

C. Iterated integrals and coordinates#

For a continuous function on a rectangle, Fubini’s theorem permits either iterated order. On a vertically simple region $D=\{(x,y):a\le x\le b,\ g_1(x)\le y\le g_2(x)\}$,

The inner variable changes first; the other is held constant. Horizontally simple regions use $x$ bounds depending on $y$. A region may need splitting when changing order. For more general improper integrals, absolute integrability or other applicable theorems must be checked; swapping an order across singularities is not universally valid. [O3Double]

In polar coordinates, $x=r\cos\theta$, $y=r\sin\theta$ and

Thus $\iint_Df\dd A=\iint_E f(r\cos\theta,r\sin\theta)r\dd r\dd\theta$ on a suitable once-covered region. The factor $r$ changes area, not the function value. A nonnegative density $\rho$ gives mass $m=\iint_D\rho\dd A$, with centroid coordinates $\bar x=\iint x\rho/m$, $\bar y=\iint y\rho/m$. The average of $f$ over positive finite area $A(D)$ is $(1/A(D))\iint_Df\dd A$.

D. Why the polar factor is necessary#

A small polar cell has radial thickness $\Delta r$ and arc-length width approximately $r\Delta\theta$, so its area is approximately $r\Delta r\Delta\theta$. Equal changes in angle sweep larger cells at larger radii. Ignoring $r$ would weight the disk’s center and edge incorrectly. The Jacobian module derives the factor algebraically.

Changing integration order does not mean mechanically swapping differentials and leaving the bounds unchanged. It means describing the same set of input points by a different family of slices. The region is the invariant object; the inequalities must be rebuilt from its geometry.

E. A region-first workflow#

Draw or describe the region with inequalities. Decide which slice direction makes the integrand and bounds easier. Integrate the inner variable while treating the outer one as constant, then integrate the resulting expression over its full outer interval. For order reversal, project the region onto the new outer axis and solve its boundary equations for the new inner variable. For a circular or radial region, test polar coordinates and explicitly include the area factor before integrating.

F. Worked examples#

Reverse order to make an integral elementary. Consider $\int_0^1\int_x^1e^{y^2}\dd y\dd x$. The region is $0\le x\le y\le1$. Reversing gives

The original inner antiderivative is not elementary, but integration across horizontal widths introduces the factor needed for substitution.

A disk integral. Over $x^2+y^2\le4$, the integrand $x^2+y^2$ becomes $r^2$. Therefore $\iint_D(x^2+y^2)\dd A=\int_0^{2\pi}\int_0^2r^3\dd r\dd\theta=8\pi$. The $r^3$ combines the transformed height $r^2$ with the area factor $r$.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

H. Complete solutions: Module 39#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

Open every solution

I. Retention and readiness#

Retrieve M17.05 and M32.08. Explain why the inner bounds may depend on the outer variable but not on the variable currently being integrated after its endpoint evaluation. Advance when you can draw the region implied by someone else’s bounds and detect an omitted polar factor.