4.4: Use of Determinants in Vector Calculus
Consider two vectors \(\mathbf{u}=u_{1} \mathbf{i}+u_{2} \mathbf{j}+u_{3} \mathbf{j}\) and \(\mathbf{v}=v_{1} \mathbf{i}+v_{2} \mathbf{j}+v_{3} \mathbf{j}\) , written as you would find them in Calculus rather than as column matrices in Linear Algebra. The dot product of the two vectors is defined in Calculus as
\[\mathbf{u} \cdot \mathbf{v}=u_{1} v_{1}+u_{2} v_{2}+u_{3} v_{3}, \nonumber \]
and the cross product is defined as
\[\mathbf{u} \times \mathbf{v}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \end{array}\right|=\mathbf{i}\left(u_{2} v_{3}-u_{3} v_{2}\right)-\mathbf{j}\left(u_{1} v_{3}-u_{3} v_{1}\right)+\mathbf{k}\left(u_{1} v_{2}-u_{2} v_{1}\right) \nonumber \]
where the determinant from Linear Algebra is used as a mnemonic to remember the definition. If the angle between the two vectors is given by \(\theta\) , then trigonometry can be used to show that
\[\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta, \quad|\mathbf{u} \times \mathbf{v}|=|\mathbf{u}||\mathbf{v}| \sin \theta \nonumber \]
Now, if \(\mathbf{u}\) and \(\mathbf{v}\) lie in the \(x-y\) plane, then the area of the parallelogram formed from these two vectors, determined from base times height, is given by
\[\begin{aligned} A &=|\mathbf{u} \times \mathbf{v}| \\ &=\left|u_{1} v_{2}-u_{2} v_{1}\right| \\ &=\left|\operatorname{det}\left(\begin{array}{cc} u_{1} & u_{2} \\ v_{1} & v_{2} \end{array}\right)\right| . \end{aligned} \nonumber \]
This result also generalizes to three dimensions. The volume of a parallelopiped formed by the three vectors \(\mathbf{u}, \mathbf{v}\) , and \(\mathbf{w}\) is given by
\[\begin{aligned} V &=|\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})| \\ &=\left|\operatorname{det}\left(\begin{array}{ccc} u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \\ w_{1} & w_{2} & w_{3} \end{array}\right)\right| . \end{aligned} \nonumber \]
An important application of this result is the change-of-variable formula for multidimensional integration. Consider the double integral
\[I=\iint_{A} \ldots d x d y \nonumber \]
over some unspecified function of \(x\) and \(y\) and over some designated area \(A\) in the \(x-y\) plane. Suppose we make a linear transformation from the \(x-y\) coordinate system to some \(u-v\) coordinate system. That is, let
\[u=a x+b y, \quad v=c x+d y, \nonumber \]
or in matrix notation,
\[\left(\begin{array}{l} u \\ v \end{array}\right)=\left(\begin{array}{ll} a & b \\ c & d \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right) . \nonumber \]
Observe that the orthonormal basis vectors \(\mathbf{i}\) and \(\mathbf{j}\) transform into the vectors \(a \mathbf{i}+\) \(c \mathbf{j}\) and \(b \mathbf{i}+d \mathbf{j}\) so that a rectangle in the \(x-y\) coordinate system transforms into a parallelogram in the \(u-v\) coordinate system. The area \(A\) of the parallelogram in the \(u-v\) coordinate system is given by
\[A=\left|\operatorname{det}\left(\begin{array}{ll} a & c \\ b & d \end{array}\right)\right|=\left|\operatorname{det}\left(\begin{array}{ll} a & b \\ c & d \end{array}\right)\right| . \nonumber \]
Notice that because this was a linear transformation, we could have also written the area as
\[A=\left|\operatorname{det}\left(\begin{array}{ll} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{array}\right)\right| \nonumber \]
which is called the Jacobian determinant, or just the Jacobian. This result also applies to infinitesimal areas where a linear approximation can be made, and with
\[u=u(x, y), \quad v=v(x, y), \nonumber \]
the change of variables formula becomes
\[d u d v=\left|\operatorname{det} \frac{\partial(u, v)}{\partial(x, y)}\right| d x d y \nonumber \]
where in general, the Jacobian matrix is defined as
\[\frac{\partial(u, v)}{\partial(x, y)}=\left(\begin{array}{ll} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{array}\right) \nonumber \]
Note that sometimes we define the change of coordinates as
\[x=x(u, v), \quad y=y(u, v), \nonumber \]
and the change of variables formula will be
\[d x d y=\left|\operatorname{det} \frac{\partial(x, y)}{\partial(u, v)}\right| d u d v \nonumber \]
We can give two very important examples. The first in two dimensions is the change of variables from rectangular to polar coordinates. We have
\[x=r \cos \theta, \quad y=r \sin \theta, \nonumber \]
and the Jacobian of the transformation is
\[\left|\operatorname{det} \frac{\partial(x, y)}{\partial(r, \theta)}\right|=\left|\begin{array}{rr} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{array}\right|=r . \nonumber \]
So to find the area of a circle of radius \(R\) , with formula \(x^{2}+y^{2}=R^{2}\) , we have
\[\int_{-R}^{R} \int_{-\sqrt{R^{2}-y^{2}}}^{\sqrt{R^{2}-y^{2}}} d x d y=\int_{0}^{2 \pi} \int_{0}^{R} r d r d \theta=\pi R^{2} . \nonumber \]
The second example in three dimensions is from cartesian to spherical coordinates. Here,
\[x=r \sin \theta \cos \phi, \quad y=r \sin \theta \sin \phi, \quad z=r \cos \theta \nonumber \]
The Jacobian is
\[\left|\operatorname{det} \frac{\partial(x, y, z)}{\partial(r, \theta, \phi)}\right|=\left|\begin{array}{ccc} \sin \theta \cos \phi & r \cos \theta \cos \phi & -r \sin \theta \sin \phi \\ \sin \theta \sin \phi & r \cos \theta \sin \phi & r \sin \theta \cos \phi \\ \cos \theta & -r \sin \theta & 0 \end{array}\right|=r^{2} \sin \theta . \nonumber \]
So to find the area of a sphere of radius \(R\) , with formula \(x^{2}+y^{2}+z^{2}=R^{2}\) , we have
\[\int_{-R}^{R} \int_{-\sqrt{R^{2}-z^{2}}}^{\sqrt{R^{2}-z^{2}}} \int_{-\sqrt{R^{2}-y^{2}-z^{2}}}^{\sqrt{R^{2}-y^{2}-z^{2}}} d x d y d z=\int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^{R} r^{2} \sin \theta d r d \theta d \phi=\frac{4}{3} \pi R^{3} \nonumber \]