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Multiple, Line & Surface Integrals

Useful preparation: Hessians & Multivariable Extrema

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Meaning & notation

Integration needs a quantity and a geometric measure: area for a sheet, arc length for a wire, and surface area for a surface. Vector fields additionally introduce directional work and flux.

Multiple integral: Accumulation using area or volume. Iterated integration accumulates one coordinate at a time, with bounds describing the region.
Scalar and vector line integrals: Scalar line integrals weight arc length; work integrals project vector fields onto directed displacement.
Surface integral and flux: Scalar surface integrals weight area; flux first projects a vector field onto an oriented unit normal.

Conditions before calculation

Continuity on the bounded elementary regions here permits iterated integration. Changing order requires new bounds describing the same region.
Coordinate changes require the absolute Jacobian determinant and a valid one-to-one description apart from boundary overlaps. Here polar coordinates use $\displaystyle r \ge 0$ , so the area factor is r; a signed-radius parameterization requires $\displaystyle \left|r\right|$ and a check for repeated coverage.
Use piecewise smooth curves and regular surfaces. Reversing orientation negates work and flux, but preserves scalar integrals.

Why the method works

1. Describe the domain before integrating

Changing slice widths require variable inner bounds. Polar angle increments produce longer arcs farther from the origin, giving the radial area factor.

\iint_D f\,dA=\int_a^b\int_{\ell(x)}^{u(x)}f(x,y)\,dy\,dx,\qquad dA=r\,dr\,d\theta,\quad r\ge0

2. Convert a curve to a parameter interval

Speed converts parameter width into length. Work instead uses the velocity vector itself, preserving direction.

\int_C\rho\,ds=\int_a^b\rho(\mathbf r(t))\|\mathbf r'(t)\|\,dt,\qquad \int_C\mathbf F\cdot d\mathbf r=\int_a^b\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt

3. Convert a surface patch to parameter area

Parameter tangents span a small parallelogram. Their cross-product magnitude scales area, while its direction chooses the flux normal.

dS=\|\mathbf R_u\times\mathbf R_v\|\,du\,dv,\qquad \iint_S\mathbf F\cdot\mathbf n\,dS=\iint_U\mathbf F(\mathbf R)\cdot(\mathbf R_u\times\mathbf R_v)\,du\,dv

Worked examples

EXAMPLE 1

Find the mass of a triangular sheet with the displayed density.

Vertical slices run from the horizontal axis to the sloping edge. The density is nonnegative throughout this triangle.
$D=\{(x,y):x\ge0,\ y\ge0,\ x+y\le2\},\qquad \rho(x,y)=x+y$
Integrate with respect to y while holding x fixed. The upper bound depends on x, reflecting the tapering region.
$M=\int_0^2\int_0^{2-x}(x+y)\,dy\,dx=\int_0^2\left[x(2-x)+\frac{(2-x)^2}{2}\right]dx$
Simplify and perform the outer integral. The triangle's area is 2, so the resulting average density provides a consistency check.
$M=\int_0^2\left(2-\frac{x^2}{2}\right)dx=\frac83,\qquad \rho_{\mathrm{avg}}=\frac{M}{2}=\frac43$

Conclusion: Mass is eight thirds; average density lies within the density range from zero to two.

EXAMPLE 2

Compare accumulation on a sloping wire and tilted surface, explaining why parameter size alone is insufficient.

The wire parameter runs from zero to one, but the spatial displacement is longer. Compute its speed before integrating.
$\mathbf r(t)=\langle t,2t,0\rangle,\quad 0\le t\le1;\qquad \|\mathbf r'(t)\|=\sqrt5,\quad \int_C1\,ds=\sqrt5$
The surface is a tilted unit parameter square. Its two tangent vectors determine the actual area multiplier.
$\mathbf R(u,v)=\langle u,v,u\rangle,\quad 0\le u,v\le1;\qquad \mathbf R_u=\langle1,0,1\rangle,\quad \mathbf R_v=\langle0,1,0\rangle$
Use the cross-product magnitude for scalar area. For upward flux of a unit vertical field, use the oriented cross product instead.
$\mathbf R_u\times\mathbf R_v=\langle-1,0,1\rangle,\quad \iint_S1\,dS=\sqrt2,\qquad \iint_S\langle0,0,1\rangle\cdot\mathbf n\,dS=1$

Conclusion: Wire length, surface area, and vertical flux respectively equal square root of 5, square root of 2, and 1.

Pitfalls & counterexamples

Tempting claim: The parameter differential alone is always the geometric measure.

Correction: Parameter intervals and patches must be scaled into actual length or area; the needed factor depends on the geometry.

Test the claim: The wire parameter interval has length one, but the physical wire has length square root of 5.

Check your understanding

What changes when the chosen surface normal is reversed?

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Compare your reasoning

Flux changes sign; a scalar surface-area integral does not.

Flux contains the signed normal direction, while scalar accumulation uses a nonnegative area magnitude.

Connect this idea

These measures support mass, work, and transport models. Green's, Stokes', and the divergence theorems later relate certain interior accumulations to boundary accumulations under their respective hypotheses.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.