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Transforming Density in Triple Integrals

Transform a mass integral correctly by composing its physical density with a coordinate map and accounting for the change in volume.

Advanced 10 min read

What You'll Learn

  • Write a mass integral over a physical solid using its density.
  • Transform both the density and the volume element under a three-variable map.
  • State the conditions that allow the change-of-variables formula to apply.
  • Compute transformed mass integrals for affine and nonlinear substitutions.
  • Identify and correct the error of using parameter coordinates as physical coordinates.

Mass and a Change of Coordinates

A mass integral combines two pieces of information: how much mass is present per unit volume at each physical point, and the volume of the region being considered. When a substitution describes the solid using new parameters, both pieces must be expressed in those parameters. In particular, the density must be evaluated at the physical point that the parameters map to.

Let a solid \(R\) lie in physical coordinates \((x,y,z)\), and let \(\rho(x,y,z)\) denote its mass density. If \(\rho\) is integrable on \(R\), the mass is

$$ M=\iiint_R \rho(x,y,z)\,dV. $$

Suppose a map \(T\) sends a parameter region \(D\), with coordinates \((u,v,w)\), onto \(R\). Write its components as \(T(u,v,w)=(x(u,v,w),y(u,v,w),z(u,v,w))\). The density remains a function of physical position; in parameter coordinates, its value at \((u,v,w)\) is \(\rho(x(u,v,w),y(u,v,w),z(u,v,w))\).

Definition: The transformed mass integrand is the physical density evaluated at the mapped point and multiplied by the absolute Jacobian: $$ \rho\bigl(x(u,v,w),y(u,v,w),z(u,v,w)\bigr) \left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right|. $$ The absolute Jacobian accounts for the physical volume represented by a unit of parameter volume.

The Mass Transformation Formula

Assume that \(T:D\to R\) is one-to-one on the region of integration, is continuously differentiable, and has a nonzero Jacobian determinant in the interior of \(D\). Under the usual hypotheses of the change-of-variables theorem, the mass can be computed as

$$ M = \iiint_D \rho\bigl(x(u,v,w),y(u,v,w),z(u,v,w)\bigr) \left|\det DT(u,v,w)\right| \,du\,dv\,dw. $$

Here \(DT\) is the matrix of partial derivatives of \((x,y,z)\) with respect to \((u,v,w)\). The transformed region \(D\) must correspond to the physical region \(R\), not to some larger or smaller region. If a map is not one-to-one, its overlapping coverage must be handled by the multiplicity rules for noninjective maps; simply integrating with the absolute Jacobian can count parts of the solid more than once.

Why the Density Is Composed with the Map

The formula follows from conservation of mass on corresponding small regions. Consider a small parameter box near \((u,v,w)\). Under \(T\), it maps approximately to a small physical parallelepiped. The linear approximation to \(T\) is given by \(DT\), and a linear map scales volume by the absolute value of its determinant. Thus, to first order, the physical volume represented by a parameter volume \(du\,dv\,dw\) is \(\left|\det DT\right|du\,dv\,dw\).

Across a sufficiently small box, the density is approximately its value at the mapped physical point. Its mass is therefore approximately that density multiplied by the physical volume. Summing these small contributions and taking the limit gives the integral formula. This is precisely the change-of-variables theorem applied to the mass integrand.

$$ \text{mass in a small parameter region} \approx \rho(T(u,v,w))\,|\det DT(u,v,w)|\,du\,dv\,dw. $$

The absolute value is essential because mass and volume are nonnegative, even if the coordinate map reverses orientation. Notice also that \(\rho\circ T\) is still the physical density evaluated at the corresponding physical location. Multiplying by the Jacobian changes the volume element; it does not change the physical meaning of the density.

Worked Example: Scaling a Rectangular Solid

Worked Example: Scaling a Rectangular Solid

Suppose the parameter region is the unit cube \(0\leq u,v,w\leq1\), with map \(x=2u\), \(y=3v\), and \(z=4w\). The physical density is \(\rho(x,y,z)=x+y+z\). Find the mass.

The Jacobian matrix and determinant are

$$ DT= \begin{pmatrix} 2&0&0\\ 0&3&0\\ 0&0&4 \end{pmatrix}, \qquad \det DT=2\cdot3\cdot4=24. $$

At the mapped point, the density is

$$ \rho(T(u,v,w))=2u+3v+4w. $$

Therefore the transformed mass integral is

$$ \begin{aligned} M &=\int_0^1\int_0^1\int_0^1 24(2u+3v+4w)\,dw\,dv\,du\\ &=24\left( 2\cdot\frac12+ 3\cdot\frac12+ 4\cdot\frac12 \right)\\ &=24\left(1+\frac32+2\right) =24\cdot\frac92 =108. \end{aligned} $$

As a check, the physical region is \(0\leq x\leq2\), \(0\leq y\leq3\), \(0\leq z\leq4\). Direct integration of \(x+y+z\) gives contributions \(24\), \(36\), and \(48\), respectively, so the physical-coordinate mass is \(24+36+48=108\), as required.

Worked Example: A Nonlinear Scaling in One Coordinate

Worked Example: A Nonlinear Scaling in One Coordinate

Use \(x=u^2,\ y=v,\ z=w\) on \(1\leq u\leq2,\ 0\leq v\leq1,\ 0\leq w\leq1\). Let \(\rho(x,y,z)=x+y+z\). Find the mass and check it in physical coordinates.

The Jacobian matrix is diagonal, so

$$ DT= \begin{pmatrix} 2u&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}, \qquad |\det DT|=2u $$

because \(u\geq1\). The density at the mapped point is \(u^2+v+w\), giving

$$ \begin{aligned} M &=\int_1^2\int_0^1\int_0^1 (u^2+v+w)2u\,dw\,dv\,du\\ &=\int_1^2\int_0^1\int_0^1 (2u^3+2uv+2uw)\,dw\,dv\,du\\ &=\frac{15}{2}+\frac{3}{2}+\frac{3}{2} =\frac{21}{2}. \end{aligned} $$

For the three terms, the respective calculations are \(\int_1^2 2u^3\,du=[u^4/2]_1^2=15/2\), \((\int_1^2 2u\,du)(\int_0^1v\,dv)=3\cdot1/2=3/2\), and \((\int_1^2 2u\,du)(\int_0^1w\,dw)=3\cdot1/2=3/2\). The unused unit-interval factors in each term equal \(1\).

The physical region is \(1\leq x\leq4,\ 0\leq y,z\leq1\). Directly, the \(x\)-term contributes \(\int_1^4x\,dx=15/2\); the \(y\)-term contributes \((4-1)\int_0^1y\,dy=3/2\); and the \(z\)-term also contributes \(3/2\). Their sum is \(21/2\), confirming the transformed calculation.

Worked Example: A Shear Map

Worked Example: A Shear Map

Let \(T(u,v,w)=(u+v,v,2w)\) on the unit cube, and let the physical density be \(\rho(x,y,z)=xy+z\). Find the mass using the parameter region, then verify it directly.

The Jacobian matrix and absolute determinant are

$$ DT= \begin{pmatrix} 1&1&0\\ 0&1&0\\ 0&0&2 \end{pmatrix}, \qquad |\det DT|=2. $$

The mapped density is \((u+v)v+2w=uv+v^2+2w\). Hence

$$ \begin{aligned} M &=2\int_0^1\int_0^1\int_0^1 (uv+v^2+2w)\,dw\,dv\,du\\ &=2\left(\frac14+\frac13+1\right) =2\cdot\frac{19}{12} =\frac{19}{6}. \end{aligned} $$

The three integral contributions inside the parentheses are \((\int_0^1u\,du)(\int_0^1v\,dv)=1/4\), \(\int_0^1v^2\,dv=1/3\), and \(\int_0^1 2w\,dw=1\). Each omitted variable contributes a factor of \(1\).

For a direct check, the physical region is described by \(0\leq y\leq1\), \(y\leq x\leq y+1\), and \(0\leq z\leq2\). Integrating in \(z\) first gives \(\int_0^2(xy+z)\,dz=2xy+2\). Thus

$$ \begin{aligned} M &=\int_0^1\int_y^{y+1}(2xy+2)\,dx\,dy\\ &=\int_0^1\left(y\bigl((y+1)^2-y^2\bigr)+2\right)\,dy\\ &=\int_0^1(2y^2+y+2)\,dy =\frac23+\frac12+2 =\frac{19}{6}. \end{aligned} $$

A Common Mistake: Using the Wrong Density

A frequent error is to substitute the parameter variables into the density without first using the coordinate map. In the first example, the physical density is \(\rho(x,y,z)=x+y+z\), but \(x=2u,\ y=3v,\ z=4w\). The incorrect substitution \(\rho(u,v,w)=u+v+w\) would give

$$ 24\int_0^1\int_0^1\int_0^1(u+v+w)\,dw\,dv\,du =24\left(\frac12+\frac12+\frac12\right) =36, $$

which is not the mass of the physical solid. The correct density at the mapped point is \(2u+3v+4w\), yielding \(108\). The correction is to compose the given physical density with \(T\) first, and only then multiply by the absolute Jacobian. The Jacobian alone cannot compensate for evaluating the density at the wrong point.

Practical Checks Before Integrating

Before evaluating a transformed mass integral, check that each part has the intended role:

  • The parameter region maps to the entire physical solid being measured.
  • The density is evaluated at \((x(u,v,w),y(u,v,w),z(u,v,w))\).
  • The volume factor is \(\left|\det DT\right|\), not a signed determinant.
  • The map is one-to-one on the integration region, or any multiplicity is handled correctly.

Density has units of mass per physical volume. The Jacobian converts parameter volume into physical volume, so the transformed integrand still gives mass after integration. This is useful when a solid has simpler parameter bounds than physical-coordinate bounds: the coordinate description may change, but the mass represented by corresponding regions does not.

Check Your Understanding

For each question, identify the density composition and volume factor before attempting the integral.

  1. If \(T(u,v,w)=(3u,v,2w)\) and \(\rho(x,y,z)=x+z\), what is the transformed integrand?
  2. Why does the mass formula use the absolute value of the Jacobian determinant?
  3. For \(T(u,v,w)=(u^2,v,w)\) with \(u>0\), what volume factor appears?
  4. What can go wrong if a map covers part of the physical solid more than once?
  5. In a mass integral, what does the Jacobian change, and what does it not change about the density?

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