Recipe

Perform a change of variables in a triple integral using the Jacobian

Use when: E's boundary is awkward in x,y,z, such as an ellipsoid, but becomes simple under a substitution x=x(u,v,w),\ y=y(u,v,w),\ z=z(u,v,w)

Goal: the integral rewritten over a simpler solid B in u,v,w, with the 3\times3 Jacobian paying for the change

Taught in Change of variables in triple integrals · extends Perform a change of variables in a double integral using the Jacobian

1. 1

   Check what the substitution turns E's boundary equation into; a diagonal scaling x=au,\ y=bv,\ z=cw turns an ellipsoid into the plain unit ball.

   Example

   Find \displaystyle\iiint_E z^2\,dV where E is the solid ellipsoid x^2+\dfrac{y^2}{4}+\dfrac{z^2}{9}\le1. Let x=u,\ y=2v,\ z=3w: substituting turns the boundary into u^2+v^2+w^2\le1, the unit ball B.

   Why: exactly as in the double-integral version, this substitution is worth using only when it turns an awkward boundary into one with simple, constant-style bounds, here the unit ball's \rho\le1.

2. 2

   Invert the substitution algebraically; a diagonal scaling inverts directly, one variable at a time.

   Example

   x=u,\ y=2v,\ z=3w is already solved for x,y,z in terms of u,v,w; no further algebra is needed here.

   Why: the Jacobian and the integrand both need to be written as functions of u,v,w, exactly as in the two-variable case.

3. 3

   Form the 3\times3 determinant of partials \dfrac{\partial(x,y,z)}{\partial(u,v,w)} and take its absolute value.

   Example

   Here the Jacobian matrix is \text{diag}(1,2,3), so \left|\dfrac{\partial(x,y,z)}{\partial(u,v,w)}\right|=1\cdot2\cdot3=6.

   • the Jacobian matrix is diagonal, as any pure axis-by-axis scaling gives — its determinant is just the product of the diagonal entries, no cofactor expansion needed
   • the Jacobian matrix has nonzero off-diagonal entries — expand the 3\times3 determinant by cofactors along whichever row or column has the most zeros

   Why: a 3\times3 determinant measures how much a small box of volume du\,dv\,dw scales under the transformation, the same volume-scaling role the 2\times2 determinant played for area.

4. 4

   Substitute x(u,v,w),y(u,v,w),z(u,v,w) into the integrand, and integrate over B with dV=\left|\dfrac{\partial(x,y,z)}{\partial(u,v,w)}\right|du\,dv\,dw.

   Example

   z^2=(3w)^2=9w^2, so \displaystyle\iiint_E z^2\,dV = \iiint_B 9w^2\cdot6\,dV = 54\iiint_B w^2\,dV.

   Why: this is the three-variable version of the exact same change of variables formula, and it's what makes the ellipsoid disappear from the problem in favor of the plain unit ball.

5. 5

   A ball is exactly the shape spherical coordinates were built for, so switch to w=\rho\cos\phi, dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta to finish.

   Example

   \displaystyle54\iiint_B w^2\,dV = 54\int_0^{2\pi}\!\!\int_0^\pi\!\!\int_0^1 (\rho\cos\phi)^2\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 54\left(\int_0^1\rho^4\,d\rho\right)\!\left(\int_0^\pi\cos^2\phi\sin\phi\,d\phi\right)\!\left(\int_0^{2\pi}d\theta\right). The three factors are \dfrac15, \dfrac23, and 2\pi, giving 54\cdot\dfrac15\cdot\dfrac23\cdot2\pi=\boxed{\dfrac{72\pi}{5}}.

   Why: the point of trading the ellipsoid for the ball was exactly to unlock this kind of coordinate system, which has no native description of an ellipsoid at all.