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Section 3.2 Change of Variables

A multiple integral often becomes much easier after rewriting the region and the integrand in a better coordinate system. The change of variables theorem explains exactly how this works. The extra factor is the absolute value of a determinant, which measures how volumes are locally stretched by the coordinate map.

Definition 3.2.1. Jacobian Determinant.

Let \(U \subseteq \R^n\) be open, and let \(\Phi=\langle \Phi_1,\ldots,\Phi_n \rangle \colon U \to \R^n\) be differentiable. For \(1 \le i,j \le n\text{,}\) the partial derivative \(D_j\Phi_i(\vu)\) is the \((i,j)\)-entry of the derivative matrix \(D\Phi(\vu)\text{.}\) The determinant
\begin{equation*} J_{\Phi}(\vu) := \det\bigl(D_j\Phi_i(\vu)\bigr)_{i,j=1}^n \end{equation*}
is called the Jacobian determinant of \(\Phi\) at \(\vu\text{.}\)
In practice, the theorem is usually combined with Fubini’s theorem. One first describes the region in the new variables, then computes the resulting iterated integral.
The coordinate changes below are not literally one-to-one on certain boundary sets, such as \(r=0\) or \(\theta=0\) and \(\theta=2\pi\text{.}\) In applications, one applies the theorem on the interior and then includes the boundary afterward. These lower-dimensional boundary pieces do not affect the value of the integral.

Subsection 3.2.1 Polar Coordinates

To make the coordinate map a \(C^1\) bijection, we restrict to
\begin{equation*} U_{\mathrm{pol}} := \{(r,\theta): r \gt 0,\ 0 \lt \theta \lt 2\pi\} \end{equation*}
and define
\begin{equation*} \Phi(r,\theta)=\langle r\cos\theta,\; r\sin\theta \rangle. \end{equation*}
Then \(\Phi\) is a \(C^1\) bijection from \(U_{\mathrm{pol}}\) onto the slit plane
\begin{equation*} V_{\mathrm{pol}} := \R^2 \setminus \{(x,0): x \ge 0\}. \end{equation*}
The derivative matrix is
\begin{equation*} \bigl(D_j\Phi_i(r,\theta)\bigr)_{i,j=1}^2 = \begin{bmatrix} \cos\theta \amp -r\sin\theta \\ \sin\theta \amp r\cos\theta \end{bmatrix}, \end{equation*}
so
\begin{equation*} \det\bigl(D_j\Phi_i(r,\theta)\bigr)_{i,j=1}^2 = r. \end{equation*}
Therefore, in polar coordinates,
\begin{equation*} dA = r\,dr\,d\theta. \end{equation*}
A disk centered at the origin in the xy-plane with a radius from the origin to a point on.
Figure 3.2.3. A disk described in polar coordinates.

Example 3.2.4.

Compute
\begin{equation*} \iint_D (x^2+y^2)\,dA, \end{equation*}
where \(D=\{(x,y):x^2+y^2\le 4\}\text{.}\)
Solution.
In polar coordinates, \(x^2+y^2=r^2\text{,}\) and the disk becomes \(0 \le r \le 2\text{,}\) \(0 \le \theta \le 2\pi\text{.}\) Thus
\begin{equation*} \iint_D (x^2+y^2)\,dA = \int_0^{2\pi}\int_0^2 r^3\,dr\,d\theta. \end{equation*}
Therefore
\begin{equation*} \iint_D (x^2+y^2)\,dA = \int_0^{2\pi}\left(\frac{r^4}{4}\Big|_0^2\right)d\theta = \int_0^{2\pi} 4\,d\theta = 8\pi. \end{equation*}

Subsection 3.2.2 Cylindrical Coordinates

To make the coordinate map a \(C^1\) bijection, we restrict to
\begin{equation*} U_{\mathrm{cyl}} := \{(r,\theta,z): r \gt 0,\ 0 \lt \theta \lt 2\pi,\ z \in \R\} \end{equation*}
and define
\begin{equation*} \Phi(r,\theta,z)=\langle r\cos\theta,\; r\sin\theta,\; z \rangle. \end{equation*}
Then \(\Phi\) is a \(C^1\) bijection from \(U_{\mathrm{cyl}}\) onto
\begin{equation*} V_{\mathrm{cyl}} := \R^3 \setminus \{(x,0,z): x \ge 0,\ z \in \R\}. \end{equation*}
The derivative matrix is
\begin{equation*} \bigl(D_j\Phi_i(r,\theta,z)\bigr)_{i,j=1}^3 = \begin{bmatrix} \cos\theta \amp -r\sin\theta \amp 0 \\ \sin\theta \amp r\cos\theta \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}, \end{equation*}
so
\begin{equation*} \det\bigl(D_j\Phi_i(r,\theta,z)\bigr)_{i,j=1}^3 = r. \end{equation*}
Therefore, in cylindrical coordinates,
\begin{equation*} dV = r\,dr\,d\theta\,dz. \end{equation*}
A solid right circular cylinder around the z-axis, with radius 2 and height 3.
Figure 3.2.5. A solid cylinder described in cylindrical coordinates.

Example 3.2.6.

Compute
\begin{equation*} \iiint_E (x^2+y^2)\,dV, \end{equation*}
where \(E=\{(x,y,z):x^2+y^2\le 4,\ 0\le z\le 3\}\text{.}\)
Solution.
In cylindrical coordinates, \(x^2+y^2=r^2\text{,}\) and the cylinder is described by \(0 \le r \le 2\text{,}\) \(0 \le \theta \le 2\pi\text{,}\) \(0 \le z \le 3\text{.}\) Therefore
\begin{equation*} \iiint_E (x^2+y^2)\,dV = \int_0^3 \int_0^{2\pi} \int_0^2 r^3\,dr\,d\theta\,dz. \end{equation*}
Hence
\begin{equation*} \iiint_E (x^2+y^2)\,dV = \int_0^3 \int_0^{2\pi} 4\,d\theta\,dz = \int_0^3 8\pi\,dz = 24\pi. \end{equation*}

Subsection 3.2.3 Spherical Coordinates

We use the same convention as in the parameterization section: spherical coordinates are given on the range
\begin{equation*} U_{\mathrm{sph}} := \{(\rho,\theta,\phi): \rho \gt 0,\ 0 \lt \theta \lt 2\pi,\ 0 \lt \phi \lt \pi\} \end{equation*}
by the formula
\begin{equation*} \Phi(\rho,\theta,\phi) = \langle \rho\sin\phi\cos\theta,\; \rho\sin\phi\sin\theta,\; \rho\cos\phi \rangle, \end{equation*}
Then \(\Phi\) is a \(C^1\) bijection from \(U_{\mathrm{sph}}\) onto
\begin{equation*} V_{\mathrm{sph}} := \R^3 \setminus \{(x,0,z): x \ge 0\}. \end{equation*}
A direct calculation gives
\begin{equation*} \left|\det\bigl(D_j\Phi_i(\rho,\theta,\phi)\bigr)_{i,j=1}^3\right| = \rho^2\sin\phi. \end{equation*}
Therefore, in spherical coordinates,
\begin{equation*} dV = \rho^2\sin\phi\,d\rho\,d\theta\,d\phi. \end{equation*}
A sphere centered at the origin with a radius from the origin to the surface.
Figure 3.2.7. A ball described in spherical coordinates.

Example 3.2.8.

Compute the volume of the ball \(B_a=\{(x,y,z):x^2+y^2+z^2\le a^2\}\text{.}\)
Solution.
Here the integrand is \(1\text{,}\) and the ball is described by \(0 \le \rho \le a\text{,}\) \(0 \le \theta \le 2\pi\text{,}\) and \(0 \le \phi \le \pi\text{.}\) Thus
\begin{equation*} \iiint_{B_a} 1\,dV = \int_0^\pi \int_0^{2\pi} \int_0^a \rho^2\sin\phi\,d\rho\,d\theta\,d\phi. \end{equation*}
Evaluating gives
\begin{equation*} \iiint_{B_a} 1\,dV = \left(\int_0^a \rho^2\,d\rho\right) \left(\int_0^{2\pi} d\theta\right) \left(\int_0^\pi \sin\phi\,d\phi\right) = \frac{a^3}{3}(2\pi)(2) = \frac{4}{3}\pi a^3. \end{equation*}

Subsection 3.2.4 A Classical Gaussian Integral

The following computation is a classical application of polar coordinates, often attributed to Lord Kelvin.
A bell-shaped surface over the xy-plane, highest at the origin and decaying rapidly away.
Figure 3.2.9. The graph of \(z=e^{-(x^2+y^2)}\text{.}\)

Example 3.2.10.

Compute
\begin{equation*} \iint_{\R^2} e^{-(x^2+y^2)}\,dA. \end{equation*}
Solution.
Let \(D_R=\{(x,y):x^2+y^2\le R^2\}\text{.}\) Then
\begin{equation*} \iint_{\R^2} e^{-(x^2+y^2)}\,dA = \lim_{R\to\infty}\iint_{D_R} e^{-(x^2+y^2)}\,dA. \end{equation*}
On \(D_R\text{,}\) polar coordinates give \(x^2+y^2=r^2\) and \(dA=r\,dr\,d\theta\text{.}\) Hence
\begin{equation*} \iint_{D_R} e^{-(x^2+y^2)}\,dA = \int_0^{2\pi}\int_0^R e^{-r^2}r\,dr\,d\theta. \end{equation*}
With the substitution \(u=r^2\text{,}\) \(du=2r\,dr\text{,}\)
\begin{equation*} \int_0^R e^{-r^2}r\,dr = \frac12 \int_0^{R^2} e^{-u}\,du = \frac12\left(1-e^{-R^2}\right). \end{equation*}
Therefore
\begin{equation*} \iint_{D_R} e^{-(x^2+y^2)}\,dA = \int_0^{2\pi} \frac12\left(1-e^{-R^2}\right)\,d\theta = \pi\left(1-e^{-R^2}\right). \end{equation*}
Letting \(R \to \infty\text{,}\) we obtain
\begin{equation*} \iint_{\R^2} e^{-(x^2+y^2)}\,dA = \pi. \end{equation*}
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