Unit 5: Multivariable integration and applications

MTH165 — Mathematics For Engineers 9 min read

I. Foundations of Multivariable Integration

Multivariable integration extends the definite integral from intervals to regions in two or three dimensions. It accumulates quantities distributed over an area or volume by partitioning the domain into small elements, forming Riemann sums, and taking a limit.

A. Fundamental framework

The central principle is that a multiple integral sums infinitesimal contributions throughout a multidimensional region.

Integrand: A function represents the density or magnitude being accumulated. For example, (f(x,y)) may describe mass per unit area, while (f(x,y,z)) may describe mass per unit volume.
Domain: A double integral uses a planar region (R), whereas a triple integral uses a solid region (V).
Elements of integration: The symbols (dA) and (dV) represent infinitesimal area and volume:
TEXT
```
  dA = dx dy, dy dx, r dr dθ, or another transformed area element
  dV = dx dy dz, r dr dθ dz, ρ² sinφ dρ dφ dθ, or another volume element
```
Here (x,y,z) are Cartesian coordinates; (r,\theta) are polar or cylindrical coordinates; and (\rho,\phi,\theta) are spherical coordinates.
Existence: A continuous function on a closed, bounded region is integrable. Many bounded functions with only limited discontinuities are also integrable.
Geometric interpretation: If (f(x,y)\geq0), its double integral gives the volume below the surface (z=f(x,y)) and above (R).
Fubini’s principle: Under suitable continuity or integrability conditions, a multiple integral can be evaluated as successive single-variable integrals.

II. Double Integrals — Accumulation Over a Plane Region

A. double integrals

A double integral accumulates the values of a two-variable function over a planar region.

Definition by a Riemann sum: Divide (R) into small subregions of area (\Delta A_i), choose a sample point in each, and take the limit:
TEXT
```
  ∬_R f(x,y) dA = lim Σ f(x_i,y_i) ΔA_i
```
Here (R) is the planar domain, (f) is the integrand, ((x_i,y_i)) is a sample point, and (\Delta A_i) is the area of the (i)-th subregion.
Rectangular region: For (R=[a,b]\times[c,d]), continuity permits either order:
TEXT
```
  ∬_R f(x,y) dA
  = ∫_a^b ∫_c^d f(x,y) dy dx
  = ∫_c^d ∫_a^b f(x,y) dx dy
```
Here (a,b) are the (x)-limits and (c,d) are the (y)-limits.
Type I region: Vertical slices describe a region bounded below and above by functions of (x):
TEXT
```
  ∬_R f(x,y) dA = ∫_a^b ∫_{g₁(x)}^{g₂(x)} f(x,y) dy dx
```
Here (a\leq x\leq b), while (g₁(x)\leq y\leq g₂(x)).
Type II region: Horizontal slices describe a region bounded left and right by functions of (y):
TEXT
```
  ∬_R f(x,y) dA = ∫_c^d ∫_{h₁(y)}^{h₂(y)} f(x,y) dx dy
```
Here (c\leq y\leq d), while (h₁(y)\leq x\leq h₂(y)).
Evaluation rule: Evaluate the inner integral first, treating the outer variable as constant, and then evaluate the resulting outer integral.
Worked example: Over (0\leq x\leq1), (0\leq y\leq2),
TEXT
```
  ∫_0^1 ∫_0^2 (x+y) dy dx
  = ∫_0^1 (2x+2) dx
  = 3
```
The value (3) is the accumulated quantity over the rectangle.

III. Reversing Iterated Integrals — Adapting Slices to the Region

A. change of order of integration

Changing the order rewrites an iterated integral with the opposite slicing direction without changing its value.

Purpose: Reversal may simplify limits, remove a difficult inner antiderivative, or make the geometry easier to represent.
Geometric requirement: The original bounds must first be translated into inequalities describing the region; bounds cannot simply be interchanged.
Procedure:
1. Read the original region: Extract the outer interval and the inner variable’s bounds.
2. Sketch or identify boundaries: Find intersection points of the bounding curves.
3. Choose the new slice: Use horizontal slices instead of vertical slices, or conversely.
4. Write new limits: The new outer limits must be constants; the new inner limits may depend on the outer variable.
Region splitting: If one pair of functions cannot describe all slices in the new direction, divide the region and write a sum of integrals.
Worked example: Consider
TEXT
```
  I = ∫_0^1 ∫_x^1 f(x,y) dy dx
```
Here (I) is the integral and (f) is an integrable function. The original inequalities are
TEXT
```
  0 ≤ x ≤ 1,   x ≤ y ≤ 1.
```
The same triangular region satisfies
TEXT
```
  0 ≤ y ≤ 1,   0 ≤ x ≤ y.
```
Therefore,
TEXT
```
  I = ∫_0^1 ∫_0^y f(x,y) dx dy.
```
Verification: Both descriptions must cover every point of the original region exactly once and introduce no additional points.

IV. Triple Integrals — Accumulation Through a Solid

A. triple integrals

A triple integral accumulates a function throughout a three-dimensional solid.

Definition: Subdivide a solid (V) into small cells and form the limiting sum:
TEXT
```
  ∭_V f(x,y,z) dV = lim Σ f(x_i,y_i,z_i) ΔV_i
```
Here (f) is the integrand, ((x_i,y_i,z_i)) is a sample point, and (\Delta V_i) is a small cell volume.
Rectangular box: For constant coordinate limits,
TEXT
```
  ∭_V f dV = ∫_a^b ∫_c^d ∫_p^q f(x,y,z) dz dy dx
```
Here (a,b), (c,d), and (p,q) are the respective limits for (x,y,z).
General solid: If (D) is the projection of (V) onto the (xy)-plane and the solid lies between two surfaces,
TEXT
```
  ∭_V f dV = ∬_D ∫_{u(x,y)}^{v(x,y)} f(x,y,z) dz dA
```
Here (u(x,y)) and (v(x,y)) are the lower and upper surfaces.
Order of evaluation: Any of the six Cartesian orders may be used when valid bounds can be constructed. The innermost limits describe entry into and exit from the solid.
Physical uses: If (\delta(x,y,z)) is volume density, total mass is
TEXT
```
  M = ∭_V δ(x,y,z) dV
```
Here (M) is mass and (\delta) is mass per unit volume.
Worked example: For the unit box and (f=x+y+z),
TEXT
```
  ∫_0^1 ∫_0^1 ∫_0^1 (x+y+z) dz dy dx = 3/2.
```
Symmetry gives an average contribution of (1/2) from each coordinate.

V. Coordinate Transformations — Matching Coordinates to Geometry

A. change of variables

A change of variables replaces inconvenient coordinates with variables that simplify the region, integrand, or both.

General transformation: Let (x=x(u,v)) and (y=y(u,v)). Then
TEXT
```
  ∬_R f(x,y) dx dy
  = ∬_S f(x(u,v),y(u,v)) |J| du dv
```
Here (S) is the transformed region and (J) is the Jacobian determinant.
Jacobian: The Jacobian measures local area scaling:
TEXT
```
  J = ∂(x,y)/∂(u,v)
    = (∂x/∂u)(∂y/∂v) − (∂x/∂v)(∂y/∂u).
```
Absolute value is required because area and volume elements are nonnegative.
Polar coordinates: Circular regions are often simplified by
TEXT
```
  x = r cosθ,   y = r sinθ,   dA = r dr dθ.
```
Here (r\geq0) is radial distance and (\theta) is polar angle. The factor (r) is the Jacobian.
Cylindrical coordinates:
TEXT
```
  x = r cosθ,   y = r sinθ,   z = z,   dV = r dr dθ dz.
```
These coordinates suit cylinders, cones, and solids symmetric about the (z)-axis.
Spherical coordinates:
TEXT
```
  x = ρ sinφ cosθ
  y = ρ sinφ sinθ
  z = ρ cosφ
  dV = ρ² sinφ dρ dφ dθ
```
Here (\rho\geq0) is distance from the origin, (\phi) is measured from the positive (z)-axis, and (\theta) is the azimuthal angle.
Selection principle: Coordinate surfaces should align with boundaries; circles suggest polar coordinates, cylinders suggest cylindrical coordinates, and spheres suggest spherical coordinates.

VI. Area and Volume by Double Integration

A. application of double integrals to calculate area and volume

Integrating the constant function (1) gives area, while integrating a nonnegative height function gives volume.

Area of a plane region:
- Formula:
  TEXT
```
     A = ∬_R 1 dA.
```
  Here (A) is area and (R) is the planar region.
- Cartesian form:
  TEXT
```
     A = ∫_a^b [g₂(x) − g₁(x)] dx.
```
  Here (g₂-g₁) is the vertical slice height.
- Polar form:
  TEXT
```
     A = ∫_α^β ∫_{r₁(θ)}^{r₂(θ)} r dr dθ.
```
  Here (\alpha,\beta) are angular limits and (r₁,r₂) are radial boundaries.
Volume under a surface:
- Formula: If (z=f(x,y)\geq0),
  TEXT
```
     V = ∬_R f(x,y) dA.
```
  Here (V) is volume and (f) is height above (R).
- Between surfaces:
  TEXT
```
     V = ∬_R [z_top(x,y) − z_bottom(x,y)] dA.
```
  Here (z{\text{top}}) and (z{\text{bottom}}) are the upper and lower surfaces.
- Units: Area has squared units; height times area produces cubic units.
- Worked example: Under (z=4-x-y) above (0\leq x\leq1), (0\leq y\leq1),
  TEXT
```
  V = ∫_0^1 ∫_0^1 (4-x-y) dy dx = 3.
```
  Thus the volume is (3) cubic units.

VII. Volume by Triple Integration

A. application of triple integrals to calculate volume

The volume of a solid is obtained by integrating unit volume density throughout the solid.

Fundamental formula:
TEXT
```
  V = ∭_E 1 dV.
```
Here (V) denotes volume and (E) denotes the solid region.
Cartesian description:
TEXT
```
  V = ∬_D ∫_{u(x,y)}^{v(x,y)} 1 dz dA
    = ∬_D [v(x,y) − u(x,y)] dA.
```
Here (D) is the projection onto the (xy)-plane, and (u,v) are bounding surfaces. This shows the equivalence between double- and triple-integral volume formulas.
Cylindrical description:
TEXT
```
  V = ∭_E r dr dθ dz.
```
The Jacobian factor (r) must be retained.
Spherical description:
TEXT
```
  V = ∭_E ρ² sinφ dρ dφ dθ.
```
The factor (\rho^2\sin\phi) accounts for radial and angular scaling.
Worked example: For a sphere of radius (a>0),
TEXT
```
  V = ∫_0^{2π} ∫_0^π ∫_0^a ρ² sinφ dρ dφ dθ
    = (2π)(2)(a³/3)
    = 4πa³/3.
```
Here (a) is the sphere’s radius, confirming the standard sphere-volume formula.
Method choice: Use the coordinate system producing the simplest boundaries and fewest subdivisions; a correct Jacobian and complete, non-overlapping limits are essential.