Multiple Integrals & Change of Variables

A multiple integral accumulates density over a region. The hard part is usually not antiderivatives; it is describing each physical point exactly once, with the correct local area or volume scale.

From Riemann sums to integrals #

For a bounded region $D\subset\mathbb R^2$, a double integral is the limit

$$\iint_D f\,dA =\lim_{\max\mathop{\mathrm{diam}}D_k\to0} \sum_k f(\xi_k)\,\mathop{\mathrm{area}}(D_k),$$

when the limit exists independently of the partition and sample points. Triple integrals use volume elements in the same way.

This definition explains the core applications:

$$\mathop{\mathrm{area}}(D)=\iint_D1\,dA,\qquad m=\iint_D \rho\,dA,$$

and, for $m>0$,

$$\bar x=\frac1m\iint_D x\rho\,dA,\qquad \bar y=\frac1m\iint_D y\rho\,dA.$$

Units are a useful audit. If $\rho$ is mass per area and $dA$ is area, $\rho\,dA$ must have units of mass.

A physical Riemann sum: a nonuniform plate #

Consider a thin rectangular plate occupying $0\le x\le2\ \mathrm m$ and $0\le y\le1\ \mathrm m$. With $x$ expressed numerically in metres, suppose its surface density is

$$\rho(x,y)=(3+x)\ \mathrm{kg/m^2}.$$

A small patch has approximate mass $\rho(x,y)\Delta A$; adding all patches and shrinking them produces

$$m=\int_0^2\int_0^1(3+x)\,dy\,dx =\left[3x+\frac{x^2}{2}\right]_0^2 =8\ \mathrm{kg}.$$

Its horizontal center of mass is

$$\bar x =\frac1{8}\int_0^2\int_0^1x(3+x)\,dy\,dx =\frac1{8}\left(6+\frac83\right) =\frac{13}{12}\ \mathrm m \approx1.083\ \mathrm m.$$

The geometric midpoint is $1\ \mathrm m$, so the shift toward the denser right side is a physical check on the calculation. Integrating a density means weighting each local value by the amount of area it occupies.

Iterated integrals and region descriptions #

For a Type I region

$$D=\{(x,y):a\le x\le b,\ g_1(x)\le y\le g_2(x)\},$$

the integral becomes

$$\iint_D f(x,y)\,dA =\int_a^b\int_{g_1(x)}^{g_2(x)}f(x,y)\,dy\,dx.$$

For a Type II region

$$D=\{(x,y):c\le y\le d,\ h_1(y)\le x\le h_2(y)\},$$

it becomes

$$\int_c^d\int_{h_1(y)}^{h_2(y)}f(x,y)\,dx\,dy.$$

The inner bounds may depend only on variables integrated later. A bound containing an already-integrated variable is not meaningful.

Changing order safely #

Do not manipulate the symbols alone. Reconstruct the region:

1. translate every bound into inequalities;
2. sketch or otherwise characterize the intersection;
3. project onto the new outer axis;
4. find which boundary is entered and exited;
5. split the region where those boundaries change.

A single $x$-simple region may require two or more $y$-simple pieces.

For nonnegative measurable functions, Tonelli’s theorem permits order exchange even when the value is infinite. For absolutely integrable functions, Fubini’s theorem permits order exchange and gives finite iterated integrals. In an elementary continuous-on-compact setting these conditions are automatic, but improper or sign-changing examples require care.

Change of variables #

Let $T:G\to D$ be

$$T(u,v)=(x(u,v),y(u,v)).$$

Under the standard change-of-variables hypotheses—$T$ is one-to-one on the relevant pieces, continuously differentiable, and has nonzero Jacobian except on harmless boundary/null sets—

$$\iint_D f(x,y)\,dx\,dy = \iint_G f(T(u,v)) \left|\det\frac{\partial(x,y)}{\partial(u,v)}\right|du\,dv.$$

The absolute value is essential. The determinant records signed orientation, whereas area is nonnegative.

Why the determinant appears #

Near $(u_0,v_0)$,

$$T(u_0+\Delta u,v_0+\Delta v) \approx T(u_0,v_0) +T_u\Delta u+T_v\Delta v.$$

A small parameter rectangle therefore maps approximately to the parallelogram spanned by $T_u\Delta u$ and $T_v\Delta v$. Its area is

$$|\det[T_u\ T_v]|\,\Delta u\,\Delta v.$$

A zero Jacobian means the linearized map collapses area. It is a warning: the coordinate map may be singular there, and the theorem cannot simply be invoked pointwise.

Polar coordinates #

With

$$x=r\cos\theta,\qquad y=r\sin\theta,$$

the Jacobian is

$$\det \begin{bmatrix} \cos\theta&-r\sin\theta\\ \sin\theta&r\cos\theta \end{bmatrix} =r,$$

so

$$dA=r\,dr\,d\theta.$$

The factor $r$ is geometric: a small angular change $d\theta$ sweeps arc length $r\,d\theta$.

Worked example: radial integrand on an annulus #

Evaluate

$$I=\iint_D(x^2+y^2)\,dA,\qquad D=\{(x,y):1\le x^2+y^2\le4\}.$$

In polar coordinates, $1\le r\le2$, $0\le\theta\le2\pi$, and $x^2+y^2=r^2$:

$$I=\int_0^{2\pi}\int_1^2 r^2(r\,dr\,d\theta) =2\pi\left[\frac{r^4}{4}\right]_1^2 =\frac{15\pi}{2}.$$

Three checks agree: the answer is positive, its scale lies between area times minimum and maximum integrand, and the integrand’s radial symmetry eliminated all angular dependence.

Polar coordinates are not automatically simpler. They help when both the region and integrand become simpler enough to offset the extra factor $r$.

Triple integrals and coordinate systems #

A $z$-simple solid has the form

$$E=\{(x,y,z):(x,y)\in D,\ u(x,y)\le z\le v(x,y)\},$$

so

$$\iiint_E f\,dV =\iint_D\int_u^v f(x,y,z)\,dz\,dA.$$

Cylindrical coordinates use

$$x=r\cos\theta,\quad y=r\sin\theta,\quad z=z, \qquad dV=r\,dr\,d\theta\,dz.$$

Using the convention in which $\phi$ is the polar angle down from the positive $z$-axis, spherical coordinates are

$$x=\rho\sin\phi\cos\theta,\quad y=\rho\sin\phi\sin\theta,\quad z=\rho\cos\phi,$$

with

$$dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.$$

Other texts swap the meanings of $\theta$ and $\phi$. Never import bounds without checking the diagram and formulas.

Geometry-first bound construction #

For a spherical or cylindrical region, fix an angular direction first and ask which radial intervals actually belong to the solid. This prevents a common complement error.

Suppose two surfaces give radial boundaries $\rho=a(\phi)$ and $\rho=b(\phi)$. “Inside the first and outside the second” means

$$b(\phi)\le \rho\le a(\phi)$$

only where $b(\phi)\le a(\phi)$. Reversing the radial bounds computes the opposite crescent, even if every subsequent antiderivative is correct.

Before integrating:

• test one interior point against every inequality;
• test one point from the supposed excluded region;
• verify that lower bound $\le$ upper bound over the angular range;
• compare the result with a containing volume or area.

These checks catch the common error of integrating the complementary spherical crescent while every subsequent antiderivative remains internally consistent.

Applications #

For a solid with density $\rho(x,y,z)$,

$$m=\iiint_E \rho\,dV,\qquad \bar x=\frac1m\iiint_E x\rho\,dV,$$

with analogous formulas for $\bar y,\bar z$.

Symmetry can prove a moment is zero, but only if both the region and density have the required symmetry. A symmetric region with an asymmetric density generally has an off-center centroid.

Moments of inertia are weighted by squared distance from an axis:

$$I_z=\iiint_E(x^2+y^2)\rho\,dV.$$

This differs from the polar moment of the geometric region only when density is constant.

Failure modes #

• Integrating before identifying the region.
• Reversing order without splitting where the active boundary changes.
• Omitting $|J|$, or using $J$ with its sign as an area factor.
• Using a non-one-to-one coordinate map without decomposing the parameter domain.
• Forgetting the polar/cylindrical factor $r$ or spherical factor $\rho^2\sin\phi$.
• Mixing spherical-angle conventions.
• Choosing $0\le\theta\le2\pi$ when a smaller interval already covers the region—or covering it twice with signed radii.
• Inferring a zero centroid coordinate from geometric symmetry while ignoring density.
• Accepting a positive-looking numerical answer without checking that the bounds represent the requested side of each surface.