Integral applications: mass, centroid, inertia and surface area

Learning goals and model assumptions

Translate volume or surface density into mass and moments, compute a weighted centroid, and distinguish scalar surface area from vector flux. All models are hypothetical mathematical examples. Densities are nonnegative and integrable; a centroid formula additionally requires positive total mass.

For a solid $E$ with volume density $\rho$, mass is $M=\iiint_E\rho\,dV$. Its centroid coordinates are $\overline x=M^{-1}\iiint_E x\rho\,dV$, and similarly for the other coordinates. The moment of inertia about the $z$ axis is $I_z=\iiint_E(x^2+y^2)\rho\,dV$, because $x^2+y^2$ is squared perpendicular distance to that axis. If density has units mass per volume, these quantities have units mass, length and mass times length squared, respectively.

For a planar lamina replace $dV$ by $dA$ and use area density. For a surface use surface density times $dS$. A signed weighting function can define an algebraic moment but is not automatically a physical nonnegative mass density, and a zero total weight cannot be used in the centroid denominator.

1. A weighted triangle is not centered at its uniform centroid

Let $L>0$ and $D=\{x\ge0,y\ge0,x+y\le L\}$ with lamina density $\rho(x,y)=\kappa(x+y)$, where $\kappa>0$. If coordinates have length units, $\kappa$ has units mass per length cubed so that $\rho$ is an area density. By continuous simple-region integration,

$$M=\kappa\int_0^L\int_0^{L-x}(x+y)\,dy\,dx=\frac{\kappa L^3}{3}.$$

The first moment for the $x$ coordinate is

$$\iint_Dx\rho(x,y)\,dA =\kappa\int_0^L\left[x^2(L-x)+\frac{x(L-x)^2}{2}\right]dx=\frac{\kappa L^4}{8}.$$

Symmetry under swapping coordinates gives the same $y$ moment. Thus $\overline x=\overline y=3L/8$. The density is greater farther from the origin, shifting the center away from the uniform-density location $(L/3,L/3)$. Symmetry gives equality of the coordinates, not their numerical value.

The result satisfies $\overline x,\overline y\ge0$ and $\overline x+\overline y=3L/4\le L$. More generally the triangle inequalities survive nonnegative weighting: integrating $x\ge0,y\ge0,x+y\le L$ against $\rho/M$ proves those same centroid inequalities. This is a structural check that the computed center lies in the triangle, independent of the arithmetic.

2. Derive spherical scale before using radial symmetry

Use coordinates

$$T(r,\phi,\theta)=(r\sin\phi\cos\theta,r\sin\phi\sin\theta,r\cos\phi),$$

where $\phi$ is measured from the positive $z$ axis, $0\le\phi\le\pi$, and $\theta$ is azimuth over one full turn. The derivative vectors have lengths $1,r,r\sin\phi$ and are mutually perpendicular with the corresponding positive orientation on the interior. Their scalar triple product is $r^2\sin\phi$, giving

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

To verify the determinant explicitly, the derivative columns in the order $(r,\phi,\theta)$ are

$$T_r=(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi),$$ $$T_\phi=(r\cos\phi\cos\theta,r\cos\phi\sin\theta,-r\sin\phi),\qquad T_\theta=(-r\sin\phi\sin\theta,r\sin\phi\cos\theta,0).$$

Their triple product is $r^2\sin\phi>0$ when $r>0$ and $0<\phi<\pi$. On the open parameter domain $(0,\infty)\times(0,\pi)\times(-\pi,\pi)$ the map is one-to-one onto its slit image, with smooth inverse $r=\sqrt{x^2+y^2+z^2}$, $\phi=\arccos(z/r)$ and the usual smooth slit-plane azimuth. Thus the proved finite-dimensional change-of-variables theorem applies to every compact truncated parameter box inside that domain.

The omitted target regions also need estimates independent of the formula we are proving. The ball of radius $\varepsilon$ is in a cube of volume $8\varepsilon^3$. In a radius-$R$ ball, the two polar caps with angle at most $\delta<\pi/2$ from the axis lie in the box $|x|,|y|\le R\sin\delta$, $|z|\le R$, of volume $8R^3\sin^2\delta$. An azimuth seam of total angular width $2\eta$, with $0<\eta<\pi/2$, lies in $|x|,|z|\le R$, $|y|\le R\sin\eta$, a box of volume $8R^3\sin\eta$. All three bounds vanish with their cutoff. Multiplying by a bound for the continuous target integrand controls its omitted integral, while the parameter estimates below control the other side. This completes the nonsingular-exhaustion argument without using spherical volume to justify its own formula.

The origin, poles and angular seam are coordinate-boundary exceptions. For bounded continuous integrands on a ball, use nonsingular truncated parameter regions and pass to their limit; the omitted parameter integral bounds tend to zero. For a bounded integrand, the radial factor near zero is bounded by a constant times $\int_0^\varepsilon r^2dr=\varepsilon^3/3$, while a polar cap has factor $\int_0^\varepsilon\sin\phi\,d\phi=1-\cos\varepsilon\to0$ and a shrinking azimuth seam has vanishing angle width. The formula is not a license to ignore an arbitrary unbounded singularity at those locations.

For a ball of radius $R>0$ and density $\rho=k r^2$ with $k>0$,

$$M=k\int_0^Rr^4dr\int_0^\pi\sin\phi\,d\phi\int_0^{2\pi}d\theta=\frac{4\pi kR^5}{5}.$$

Reflection symmetry makes each first moment zero, so the center is the origin. For inertia, squared distance to the $z$ axis is $r^2\sin^2\phi$, not the full $r^2$. Therefore

$$I_z=k\int_0^Rr^6dr\int_0^\pi\sin^3\phi\,d\phi\int_0^{2\pi}d\theta =\frac{8\pi kR^7}{21},$$

using $\int_0^\pi\sin^3\phi\,d\phi=4/3$. The ratio $I_z/(MR^2)=10/21$ is dimensionless. Here $k$ has units mass per length to the fifth power, since multiplying it by $r^2$ must give volume density; treating it as a constant ordinary density would produce incorrect dimensions.

3. Surface area includes slope, not just projection

For a graph $z=h(x,y)$ with $h\in C^1$ on a planar region, the parametrization $r(x,y)=(x,y,h(x,y))$ has tangent cross product $(-h_x,-h_y,1)$. Its scalar area element is

$$dS=\sqrt{1+h_x^2+h_y^2}\,dx\,dy.$$

A constant surface density $\sigma$ therefore gives projected-coordinate mass density $\sigma\sqrt{1+h_x^2+h_y^2}$, not generally the constant $\sigma$. Uniform on the actual surface and uniform on its projection are different models.

For the paraboloid graph $h=x^2+y^2$ above the disk of radius $R$, the area is

$$A=2\pi\int_0^Rr\sqrt{1+4r^2}\,dr =\frac\pi6[(1+4R^2)^{3/2}-1].$$

Upward flux of the constant field $(0,0,1)$ over that same graph is instead $\iint_D1\,dA=\pi R^2$, because its dot product with the oriented area vector is one. Surface area accumulates a magnitude; flux accumulates a signed normal component. Their difference is geometric, not a missing constant in integration.

4. A singular vertex can be handled by an actual limit

The cone graph $h(x,y)=a\sqrt{x^2+y^2}$ with $a>0$ is not differentiable at the origin. Away from it, $h_x^2+h_y^2=a^2$. Over the punctured annulus $\varepsilon\le r\le R$, the regular surface area is $\pi(R^2-\varepsilon^2)\sqrt{1+a^2}$. Passing $\varepsilon\downarrow0$ gives lateral area $\pi R^2\sqrt{1+a^2}$.

This conclusion does not come just from saying “one point has zero area.” Every smaller annular contribution between radii $\delta<\varepsilon$ is bounded by $\pi\varepsilon^2\sqrt{1+a^2}$. The Cauchy tail criterion therefore supplies the finite improper surface area and bounds the limiting missing contribution by that same quantity, tending to zero. A different unbounded area factor could fail that limiting argument, just as an unbounded volume integrand can fail at a puncture.

5. Symmetry, bounds and dimensions are independent checks

If $E$ and its density are invariant under $x\mapsto-x$, the integrand $x\rho$ changes sign while the region and volume element stay the same, forcing its integral zero. A symmetric region alone is insufficient when the density breaks that symmetry.

For any nonnegative mass inside a ball of radius $R$, $0\le x^2+y^2\le R^2$ implies $0\le I_z\le MR^2$. A computed inertia outside this interval contradicts the model regardless of any symbolic simplification. A centroid is a weighted average, so multiplying the entire density by a positive constant changes mass and inertia by that factor but leaves the centroid unchanged. Unit analysis, positivity, symmetry and scale laws should be checked separately; agreement with one does not prove all arithmetic correct.

Graded application and modeling tasks

1. Weighted center. Reproduce the triangle mass and first moments, then prove the resulting center must satisfy the triangle's inequalities using only nonnegative weighting.
2. Radial inertia. For the ball density $kr^2$, derive the $r$ and angular factors in $I_z$ and explain why using $r^2$ as distance to the axis would be wrong.
3. Coordinate convention. State which angle is polar in the spherical formula and explain how a different angle convention changes the displayed sine factor and coordinate ranges.
4. Surface versus projection. For uniform surface density on a graph, identify the equivalent density on the projection. Compare the paraboloid's scalar area and constant-vertical-field flux.
5. Singular-point audit. Explain why the cone's vertex does not invalidate the computed limiting area, and state the estimate that would be missing from an unsupported “ignore the point” argument.
6. Integral bound. Prove $0\le I_z\le MR^2$ for a nonnegative density supported in a radius-$R$ ball. Show the radial example obeys it and describe how scaling all lengths affects mass, centroid and inertia when ordinary density is kept fixed.

Full solutions

1. The mass is $\kappa L^3/3$ and each first moment $\kappa L^4/8$, so the center is $(3L/8,3L/8)$. Since every point has $x,y\ge0$ and $x+y\le L$, integrate each inequality against $\rho/M\ge0$, whose total integral is one. The resulting bounds are $\overline x,\overline y\ge0$ and $\overline x+\overline y\le L$, which the calculated pair satisfies. This does not require the density to be constant.
2. The volume element contributes $r^2\sin\phi$, density contributes $kr^2$, and squared axis distance contributes $r^2\sin^2\phi$. Their product is $kr^6\sin^3\phi$. Integrating gives $R^7/7$, $4/3$ and $2\pi$, hence $8\pi kR^7/21$. Full $r^2$ is squared distance to the origin, not to the $z$ axis; it includes the vertical displacement that does not contribute to distance from that axis.
3. Here $\phi\in[0,\pi]$ is measured downward from the positive $z$ axis, and $\theta$ is the full azimuth. If elevation is measured from the equatorial plane instead, its range is $[-\pi/2,\pi/2]$ and the volume factor uses the cosine of elevation. Renaming angles without changing their geometric definitions and ranges can move the trigonometric factor to the wrong variable.
4. The projected density is $\sigma\sqrt{1+h_x^2+h_y^2}$. The paraboloid's area is $\pi[(1+4R^2)^{3/2}-1]/6$, while its upward vertical flux is $\pi R^2$. For $R>0$, the area factor exceeds one except at the center, so scalar area exceeds projected area. Flux uses the oriented cross product dotted with the field, not that cross product's magnitude.
5. The surface is regular on every annulus avoiding the vertex, and the exact area formula there has a finite limit. The omitted area's contribution is bounded by a constant times $\varepsilon^2$, tending to zero. Without such control on neighborhoods of the singular point, zero size of the point alone would not rule out an unbounded accumulated contribution.
6. Multiply $0\le x^2+y^2\le R^2$ by the nonnegative density and integrate, obtaining the bounds. The radial example has ratio $10/21$ strictly between zero and one. If every length scales by $s>0$ and the same ordinary mass-per-volume density pattern is transported without a density multiplier, volume and mass scale by $s^3$, centroid coordinates by $s$, and inertia by $s^5$. If the density law itself changes under scaling, its multiplier must be included separately; one cannot keep a dimensional coefficient fixed and simultaneously assume an unchanged numerical density pattern without checking the model. For the law $\rho=kr^2$ with fixed $k$, scaling radius by $s$ also multiplies density by $s^2$, so mass scales by $s^5$ and inertia by $s^7$, matching the explicit formulas.

Summary and terms

A centroid is a mass-weighted position, not merely a geometric midpoint. Moment of inertia weights mass by squared perpendicular distance to an axis. A surface density multiplies true surface area, while flux uses an oriented normal component. A singular-coordinate limit needs a quantitative neighborhood estimate. Dimensions, symmetry, positivity and scale laws provide essential checks on every integral application.