Unit 5: Multivariable integration and applications

I. Foundations of Multivariable Integration

Multivariable integration extends the definite integral from intervals to regions in two or three dimensions. Its governing principle is continuous accumulation: a region is divided into small elements, the integrand is evaluated on each element, and the resulting contributions are summed through a limiting process.

• Integrand: A function (f(x,y)) or (f(x,y,z)) whose values represent density, height, mass per unit area, mass per unit volume, or another quantity being accumulated.
• Domain of integration: A planar region (R\subseteq\mathbb{R}^2) for a double integral or a solid region (V\subseteq\mathbb{R}^3) for a triple integral.
• Differential element: The infinitesimal area or volume over which accumulation occurs:
  
  MATH

    dA=dx\,dy,\qquad dV=dx\,dy\,dz.

  
  Here, (dA) is an area element and (dV) is a volume element in Cartesian coordinates.
• Iterated integration: A multiple integral is normally evaluated as successive single-variable integrals. Variables not currently being integrated are treated as constants.
• Geometric interpretation: If (f(x,y)\geq0), then (\iint_R f\,dA) gives the volume beneath (z=f(x,y)) and above (R).
• Integrability assumption: Continuous functions on closed, bounded regions are integrable. Piecewise-continuous functions are also integrable when their discontinuities are sufficiently limited.
• Limits and order: Integration limits describe the domain. The innermost limits may depend on the outer variables, but outer limits cannot depend on variables already integrated out.

II. Double Integrals — Accumulation over Planar Regions

A. Double integrals

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

• Definition: For a function (f(x,y)) on a region (R), the double integral is the limit of a two-dimensional Riemann sum:
  

  MATH

    \iint_R f(x,y)\,dA
    =\lim_{\max\Delta A_i\to0}\sum_{i=1}^{n}f(x_i^*,y_i^*)\Delta A_i.

  
  Here, (\Delta A_i) is the area of the (i)-th subregion and ((x_i^,y_i^)) is a sample point in it.

• Rectangular region: If (R=[a,b]\times[c,d]), Fubini’s theorem permits either order:
  

  MATH

    \iint_R f(x,y)\,dA
    =\int_a^b\int_c^d f(x,y)\,dy\,dx
    =\int_c^d\int_a^b f(x,y)\,dx\,dy.

  
  The constants (a,b) are (x)-bounds, while (c,d) are (y)-bounds.

• General regions:

  1. Type I region: (a\leq x\leq b), with (g_1(x)\leq y\leq g_2(x)).
  2. Type II region: (c\leq y\leq d), with (h_1(y)\leq x\leq h_2(y)).

• Iterated form: For a Type I region,
  

  MATH

    \iint_R f(x,y)\,dA
    =\int_a^b\left[\int_{g_1(x)}^{g_2(x)}f(x,y)\,dy\right]dx.

• Worked example: Over (R=[0,1]\times[0,2]),
  

  MATH

    \iint_R(x+y)\,dA
    =\int_0^1\int_0^2(x+y)\,dy\,dx
    =\int_0^1(2x+2)\,dx=3.

III. Reversing Iterated Integrals — Describing the Same Region Differently

A. Change of order of integration

Changing the order of integration rewrites an integral so that the same region is traversed in the opposite coordinate direction.

• Purpose: Reversal may simplify difficult limits or make the antiderivative of the integrand obtainable. The region itself and the value of the integral remain unchanged.

• Procedure:

  1. Read the original limits and express the corresponding inequalities.
  2. Sketch or identify the boundary curves and their intersection points.
  3. Project the region onto the new outer-variable axis.
  4. Determine the new inner-variable bounds.
  5. Split the region if one pair of bounds cannot describe it.

• Original description:
  

  MATH

    I=\int_0^1\int_x^1 f(x,y)\,dy\,dx.

  
  These limits represent (0\leq x\leq1) and (x\leq y\leq1), a triangle bounded by (x=0), (y=x), and (y=1).

• Reversed description: For a fixed (y), the coordinate (x) runs from (0) to (y):
  

  MATH

    I=\int_0^1\int_0^y f(x,y)\,dx\,dy.

• Worked example: Reversing the order makes the following integral elementary:
  

  MATH

    \int_0^1\int_x^1 e^{y^2}\,dy\,dx
    =\int_0^1\int_0^y e^{y^2}\,dx\,dy
    =\int_0^1 y e^{y^2}\,dy
    =\frac{e-1}{2}.

• Common constraint: Limits must describe the complete region exactly once. A curved or non-monotone boundary may require two or more iterated integrals after reversal.

IV. Triple Integrals — Accumulation throughout Solid Regions

A. Triple integrals

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

• Definition: For an integrable function (f(x,y,z)) on a solid (V),
  

  MATH

    \iiint_V f(x,y,z)\,dV
    =\lim_{\max\Delta V_i\to0}
    \sum_{i=1}^{n}f(x_i^*,y_i^*,z_i^*)\Delta V_i.

  
  Here, (\Delta V_i) is a small volume element and ((x_i^,y_i^,z_i^*)) is a sample point.

• Rectangular box: If (V=[a,b]\times[c,d]\times[p,q]), then
  

  MATH

    \iiint_V f\,dV
    =\int_a^b\int_c^d\int_p^q f(x,y,z)\,dz\,dy\,dx.

  
  For continuous (f), the order may be chosen from the six permutations of (x,y,z).

• General solid: If (R) is the projection of (V) onto the (xy)-plane and
  (u(x,y)\leq z\leq v(x,y)), then
  

  MATH

    \iiint_V f\,dV
    =\iint_R\int_{u(x,y)}^{v(x,y)}
    f(x,y,z)\,dz\,dA.

• Physical meaning: With density (\rho(x,y,z)), total mass is
  

  MATH

    M=\iiint_V\rho(x,y,z)\,dV,

  
  where (\rho) has units of mass per unit volume, so (M) has units of mass.

• Worked example: For (V=[0,1]^3),
  

  MATH

    \iiint_V(x+y+z)\,dV
    =\int_0^1\int_0^1\int_0^1(x+y+z)\,dz\,dy\,dx
    =\frac{3}{2}.

V. Coordinate Transformations — Simplifying Domains and Integrands

A. Change of variables

A change of variables replaces the original coordinates with coordinates adapted to the geometry of the domain.

• Two-dimensional rule: If (x=x(u,v)) and (y=y(u,v)), then
  

  MATH

    \iint_R f(x,y)\,dx\,dy
    =\iint_S f(x(u,v),y(u,v))
    \left|\frac{\partial(x,y)}{\partial(u,v)}\right|du\,dv.

  
  Here, (S) is the transformed region and the absolute Jacobian determinant measures local area scaling.

• Jacobian:
  

  MATH

    \frac{\partial(x,y)}{\partial(u,v)}
    =
    \begin{vmatrix}
    \partial x/\partial u & \partial x/\partial v\\
    \partial y/\partial u & \partial y/\partial v
    \end{vmatrix}.

  
  The absolute value is required because area and volume elements are non-negative.

• Polar coordinates: Circular regions are simplified by
  

  MATH

    x=r\cos\theta,\qquad y=r\sin\theta,\qquad dA=r\,dr\,d\theta.

  
  Here, (r\geq0) is radial distance and (\theta) is the polar angle.

• Three-dimensional coordinates:

  1. Cylindrical:
     
     MATH

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

  2. Spherical:
     
     MATH

          x=\rho\sin\phi\cos\theta,\quad
          y=\rho\sin\phi\sin\theta,\quad
          z=\rho\cos\phi,\quad
          dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.

     
     Here, (\rho) is distance from the origin and (\phi) is measured from the positive (z)-axis.

• Validity conditions: The transformation should be continuously differentiable, essentially one-to-one on the domain, and have a non-zero Jacobian except possibly on negligible boundary sets.

VI. Geometric Applications of Double Integration — Planar Area and Surface-Based Volume

A. Application of double integrals to calculate area and volume

Double integrals calculate planar area by integrating (1) and volume by integrating vertical height.

• Area of a region:
  

  MATH

    A=\iint_R 1\,dA.

  
  For (R={(x,y):a\leq x\leq b,\ g_1(x)\leq y\leq g_2(x)}),
  

  MATH

    A=\int_a^b[g_2(x)-g_1(x)]\,dx.

• Polar area: If (R) is described by
  (\alpha\leq\theta\leq\beta) and (0\leq r\leq R(\theta)), then
  

  MATH

    A=\int_\alpha^\beta\int_0^{R(\theta)}r\,dr\,d\theta
    =\frac12\int_\alpha^\beta R(\theta)^2\,d\theta.

• Volume below a surface: If (z=f(x,y)\geq0) above (R), then
  

  MATH

    V=\iint_R f(x,y)\,dA.

• Volume between surfaces: If (z=u(x,y)) is above (z=l(x,y)), then
  

  MATH

    V=\iint_R[u(x,y)-l(x,y)]\,dA.

  
  The integrand is the non-negative vertical height of the solid.

• Worked example: The volume under (z=4-x-y) above (0\leq x\leq1), (0\leq y\leq1) is
  

  MATH

    V=\int_0^1\int_0^1(4-x-y)\,dy\,dx=3.

  
  Since lengths are cubed, the answer is (3) cubic units.

VII. Geometric Applications of Triple Integration — Direct Solid Volume

A. Application of triple integrals to calculate volume

A triple integral calculates volume by accumulating infinitesimal volume elements throughout a solid.

• Cartesian formula:
  

  MATH

    \operatorname{Vol}(V)=\iiint_V 1\,dx\,dy\,dz.

  
  The integration limits must encode every bounding plane or surface of (V).

• Choice of coordinates: Coordinate systems should match symmetry:

  1. Cartesian coordinates suit boxes and plane-bounded solids.
  2. Cylindrical coordinates suit cylinders, cones, and solids of revolution about an axis.
  3. Spherical coordinates suit spheres and regions bounded by spheres or cones.

• Comparison with double integration: A double integral computes volume using vertical height over a planar projection, whereas a triple integral fills the entire solid with elements (dV). Both methods give the same value when their limits describe the same solid.

• Worked example: For a sphere of radius (a>0), spherical bounds are
  (0\leq\rho\leq a), (0\leq\phi\leq\pi), and (0\leq\theta\leq2\pi):
  

  MATH

    \operatorname{Vol}(V)
    =\int_0^{2\pi}\int_0^\pi\int_0^a
    \rho^2\sin\phi\,d\rho\,d\phi\,d\theta
    =\frac{4}{3}\pi a^3.

• Dimensional check: The element (dV) has dimensions (L^3). In spherical coordinates, (\rho^2\,d\rho) contributes (L^3), while (\sin\phi\,d\phi\,d\theta) is dimensionless.