Unit 6: Integral Calculus

MTH174 — Engineering Mathematics 8 min read

I. Orientation

Integral calculus extends one-dimensional accumulation to regions and solids. A double integral accumulates a function over a two-dimensional region, while a triple integral accumulates it through a three-dimensional volume. These ideas depend on partitioning a domain into small elements, multiplying function values by element measures, and taking a limiting sum.

Governing principle: A definite integral represents accumulated quantity; in two dimensions the elemental measure is (dA), and in three dimensions it is (dV).
Geometric interpretation: (\iint_R 1\,dA) gives the area of a plane region (R), while (\iiint_V 1\,dV) gives the volume of a solid (V).
Iterated integration: Fubini’s theorem permits a multiple integral to be evaluated as successive single integrals when the function is continuous or suitably integrable over the region.
Coordinate convention: Cartesian coordinates use (dA=dx\,dy) or (dy\,dx), and (dV=dx\,dy\,dz) in an appropriate order.
Change of variables: A transformation replaces difficult coordinates by convenient ones and requires a Jacobian determinant to convert area or volume elements.
Boundary discipline: Limits must describe exactly the region or solid; reversing an order or changing coordinates is valid only when the corresponding limits and differential element are changed consistently.

II. Double Integrals — Accumulation over Plane Regions

A double integral of (f(x,y)) over a region (R) is defined as the limit of sums over small area elements and is written (\iint_R f(x,y)\,dA).

A. Double integrals

This subsection establishes the notation, interpretation, and evaluation of double integrals over rectangular and non-rectangular regions.

Definition: If (R=[a,b]\times[c,d]), then
TEXT
```
  ∬_R f(x,y) dA
  = ∫_a^b ∫_c^d f(x,y) dy dx
```
Here, (x) and (y) are Cartesian coordinates, (a,b,c,d) are boundary constants, and (dA) is an infinitesimal area.
Iterated evaluation: Integrate first with respect to the innermost variable. For example, in (dy\,dx), treat (x) as constant during the (y)-integration.
Non-rectangular region: If (R) lies between (y=g_1(x)) and (y=g_2(x)), with (a\le x\le b), then
TEXT
```
  ∬_R f(x,y) dA
  = ∫_a^b ∫_{g_1(x)}^{g_2(x)} f(x,y) dy dx
```
The lower curve must lie below the upper curve throughout the stated interval.
Geometric meaning: If (f(x,y)\ge0), the integral gives the volume under (z=f(x,y)) and above (R). If (f) changes sign, positive and negative contributions are algebraically combined.
Basic properties: Linearity gives (\iint_R(\alpha f+\beta g)dA=\alpha\iint_Rf\,dA+\beta\iint_Rg\,dA), where (\alpha,\beta) are constants. Integrals over adjacent regions can be added.

B. Change of order of integration

Changing the order of integration rewrites the same region so that the other variable is integrated first, often simplifying the calculation.

Purpose: Convert an integral such as (\inta^b\int{g_1(x)}^{g_2(x)}f(x,y)\,dy\,dx) into an equivalent (dx\,dy) form when the inner antiderivative is difficult.
Geometric procedure:
- Sketch the curves and identify the complete region.
- Find the overall minimum and maximum values of the new inner variable.
- Express the new inner limits using the opposite boundary curves.
Simple reversal: For the triangular region (0\le x\le1), (x\le y\le1), the equivalent description is (0\le y\le1), (0\le x\le y):
TEXT
```
  ∫_0^1 ∫_x^1 f(x,y) dy dx
  = ∫_0^1 ∫_0^y f(x,y) dx dy
```
Splitting requirement: If a horizontal or vertical line intersects the region in separate patterns, split the integral. A single pair of limits is valid only when the boundary description remains unchanged.
Worked example: Since (y=x^2) and (y=1) enclose (0\le x\le1), the order (dy\,dx) is
TEXT
```
  ∫_0^1 ∫_{x^2}^1 f(x,y) dy dx.
```
Solving (y=x^2) for (x) gives (x=\sqrt y), so the reversed form is
TEXT
```
  ∫_0^1 ∫_0^{√y} f(x,y) dx dy.
```

C. Change of variables

A change of variables replaces ((x,y)) by new coordinates ((u,v)) so that the region or integrand becomes simpler.

Transformation: Let (x=x(u,v)) and (y=y(u,v)). The original area element is converted using the Jacobian:
TEXT
```
  dA = |∂(x,y)/∂(u,v)| du dv
```
The determinant is
TEXT
```
  ∂(x,y)/∂(u,v)
  = | x_u  x_v |
    | y_u  y_v |
  = x_u y_v - x_v y_u.
```
Conditions: The transformation should be one-to-one, differentiable, and have a nonzero Jacobian in the relevant region, except possibly on boundary or isolated points.

Integral formula:

TEXT

  ∬_R f(x,y) dA
  = ∬_S f(x(u,v),y(u,v))
      |∂(x,y)/∂(u,v)| du dv.

Here, (S) is the transformed region.

Polar coordinates: For circular regions, (x=r\cos\theta), (y=r\sin\theta), and (dA=r\,dr\,d\theta). The factor (r) is the absolute Jacobian.
Worked example: For the disk (x^2+y^2\le a^2), polar limits are (0\le r\le a), (0\le\theta\le2\pi). Its area is
TEXT
```
  ∬_R 1 dA
  = ∫_0^{2π} ∫_0^a r dr dθ
  = πa².
```

D. Application of double integrals to calculate area and volume

Double integrals calculate planar area directly and volume under a surface when the integrand represents height.

Area of a plane region: Set the integrand equal to (1):
TEXT
```
  Area(R) = ∬_R 1 dA.
```
For (R) between (y=g_1(x)) and (y=g_2(x)), this becomes (\int_a^b[g_2(x)-g_1(x)]dx).
Volume under a surface: If (z=f(x,y)\ge0) over (R), then
TEXT
```
  Volume = ∬_R f(x,y) dA.
```
The surface is (z=f(x,y)), and the base is (R) in the (xy)-plane.
Volume between surfaces: If (z=f(x,y)) is above (z=g(x,y)), then
TEXT
```
  Volume = ∬_R [f(x,y)-g(x,y)] dA.
```
Mass and weighted quantities: For surface density (\rho(x,y)), mass is (M=\iint_R\rho(x,y)dA). The density must have units such as kilograms per square metre, making (M) measured in kilograms.
Centroid: For uniform lamina, (\bar x=\frac{1}{A}\iint_Rx\,dA) and (\bar y=\frac{1}{A}\iint_Ry\,dA), where (A) is the area. Symmetry can make one coordinate zero without direct integration.

III. Triple Integrals — Accumulation through Solids

A triple integral extends double integration to a solid region (V\subset\mathbb R^3), accumulating a function over infinitesimal volume elements.

A. Triple integrals

This subsection describes evaluation in Cartesian coordinates and the role of three-dimensional limits.

Definition: Over a rectangular box (V=[a,b]\times[c,d]\times[e,f]),
TEXT
```
  ∭_V F(x,y,z) dV
  = ∫_a^b ∫_c^d ∫_e^f F(x,y,z) dz dy dx.
```
Here, (F) is a scalar function and (dV=dx\,dy\,dz) is an infinitesimal volume.
Iterated integration: The order may be (dz\,dy\,dx), (dx\,dz\,dy), or any other valid order, provided the limits describe the same solid.
General solid: If (V) is vertically bounded by (z=h_1(x,y)) and (z=h_2(x,y)) above a base (R), then
TEXT
```
  ∭_V F dV
  = ∬_R ∫_{h_1(x,y)}^{h_2(x,y)}
      F(x,y,z) dz dA.
```
Constant integrand: For (F=1), the triple integral measures volume. For density (\rho(x,y,z)), it measures mass:
TEXT
```
  M = ∭_V ρ(x,y,z) dV.
```
Coordinate choices: Cartesian coordinates suit boxes and planes; cylindrical coordinates suit circular cylinders; spherical coordinates suit spheres and radial boundaries.

B. Application of triple integrals to calculate volume

Triple integrals calculate the volume of a three-dimensional solid by integrating the constant function (1) over its entire domain.

Volume formula:
TEXT
```
  Volume(V) = ∭_V 1 dV.
```
The result has cubic units, such as (\text{m}^3), when all coordinates are measured in metres.
Cartesian description: For a solid (0\le x\le a), (0\le y\le b), (0\le z\le c),
TEXT
```
  Volume = ∫_0^a ∫_0^b ∫_0^c 1 dz dy dx
         = abc.
```
This recovers the rectangular-box formula.
Cylindrical coordinates: Use (x=r\cos\theta), (y=r\sin\theta), (z=z), with
TEXT
```
  dV = r dz dr dθ.
```
The factor (r) is required because planar polar area contributes (r\,dr\,d\theta).
Spherical coordinates: Use (x=\rho\sin\phi\cos\theta), (y=\rho\sin\phi\sin\theta), (z=\rho\cos\phi), where (\rho) is radial distance, (\phi) is measured from the positive (z)-axis, and (\theta) is the azimuthal angle. Then
TEXT
```
  dV = ρ² sinφ dρ dφ dθ.
```
Worked example: For a sphere of radius (a), limits are (0\le\rho\le a), (0\le\phi\le\pi), (0\le\theta\le2\pi):
TEXT
```
  Volume
  = ∫_0^{2π} ∫_0^π ∫_0^a ρ² sinφ dρ dφ dθ
  = 4πa³/3.
```
Symmetry and decomposition: A solid may be divided into simpler solids, or symmetry may reduce limits. For example, integrating over one octant and multiplying by (8) is valid for a sphere because all octants have equal volume.
Physical applications: Triple integrals also give mass, moments, centre of mass, and average value. The average of (F) over (V) is
TEXT
```
  F_average = (1/Volume(V)) ∭_V F(x,y,z) dV.
```
The denominator ensures the result has the same units as (F).