Question

Consider the region \(R\) bounded by three pairs of parallel planes: \(a x+b y=0, a x+b y=1, c x+d z=0\) \(c x+d z=1, e y+f z=0,\) and \(e y+f z=1,\) where \(a, b, c, d, e\) and \(f\) are real numbers. For the purposes of evaluating triple integrals, when do these six planes bound a finite region? Carry out the following steps. a. Find three vectors \(\mathbf{n}_{1}, \mathbf{n}_{2},\) and \(\mathbf{n}_{3}\) each of which is normal to one of the three pairs of planes. b. Show that the three normal vectors lie in a plane if their triple scalar product \(\mathbf{n}_{1} \cdot\left(\mathbf{n}_{2} \times \mathbf{n}_{3}\right)\) is zero. c. Show that the three normal vectors lie in a plane if ade \(+\) bcf \(=0\) d. Assuming \(\mathbf{n}_{1}, \mathbf{n}_{2},\) and \(\mathbf{n}_{3}\) lie in a plane \(P,\) find a vector \(\mathbf{N}\) that is normal to \(P\). Explain why a line in the direction of \(\mathbf{N}\) does not intersect any of the six planes, and thus the six planes do not form a bounded region. e. Consider the change of variables \(u=a x+b y, v=c x+d z\) \(w=e y+f z .\) Show that $$ J(x, y, z)=\frac{\partial(u, v, w)}{\partial(x, y, z)}=-a d e-b c f $$ What is the value of the Jacobian if \(R\) is unbounded?

Short answer

Answer: The region is bounded when \(ade + bcf \neq 0\). In this case, the Jacobian determinant of the transformation \((u, v, w) = (ax + by, cx + dz, ey + fz)\) is non-zero, indicating that the normal vectors do not lie in the same plane, and the region is thus bounded.

Step-by-step solution

Finding normal vectors

For a plane given by an equation of the form \(ax + by = c\) or \(ax + bz =c\) or \(ey + f z= c\) (with these being the generic forms of all the given planes), the vector normal to the plane can be obtained by taking the coefficients of \(x, y\) and \(z\). So, for the three pairs of parallel planes the normal vectors are: \(\mathbf{n}_{1} = [a, b, 0]\), \(\mathbf{n}_{2} = [c, 0, d]\), \(\mathbf{n}_{3} = [0, e, f]\).

Checking if the normal vectors lie in the same plane

The three normal vectors defined above will lie in the same plane if their triple scalar product is zero, as it means that they are linearly dependent. The triple scalar product can be calculated using the expression \(\mathbf{n}_{1} \cdot (\mathbf{n}_{2} \times \mathbf{n}_{3})\). Calculating this gives \(a(c\cdot0 - 0\cdot f) - b(0\cdot0 - c\cdot e) + 0 (0\cdot f - 0\cdot0) = 0\) which simplifies to \(-bce = 0\). If this is true, then either \(b=0\), \(c=0\), or \(e=0\) which would directly imply \(ade + bcf = 0\).

Finding a vector normal to the plane of the normal vectors

Assuming \(\mathbf{n}_{1}, \mathbf{n}_{2},\) and \(\mathbf{n}_{3}\) do indeed lie in a plane \(P\), we can find a vector that is normal to \(P\). Since we have established that these vectors are linearly dependent, we can use any two to form the normal vector to \(P\) using the cross product. So if we choose \(\mathbf{n}_{1}\) and \(\mathbf{n}_{2}\), \(\mathbf{N} = \mathbf{n}_{1} \times \mathbf{n}_{2}\). In this case, \(\mathbf{N} = [ad, -ac, bc]\). A line in the direction of ∥𝑁𝑁∥ does not intersect the planes because it is orthogonal to the plane defined by the normal vectors, and since the normal vectors in turn are orthogonal to their respective planes, the line of ∥𝑁𝑁∥ is parallel to all six planes. Thus, no intersection is possible.

Evaluating the Jacobian

The Jacobian \(J\) of the transformation \((u, v, w) = (ax + by, cx + dz, ey + fz)\) is given as \(J = |det|\frac{\partial(u, v, w)}{\partial(x, y, z)}| = -ade - bcf\). From previous steps, we know that if \(ade + bcf = 0\), the three normal vectors lie in the same plane and the region \(R\) is unbounded. Hence, if the region is unbounded, the Jacobian is zero.

Key concepts

Normal Vectors

In geometry and vector calculus, a normal vector is a vector that is perpendicular to a given surface or plane. When dealing with a plane defined by equations like \(ax + by = c\), the normal vector can be directly found by taking the coefficients of the variables. For example, the plane \(ax + by = 0\) has the normal vector \([a, b, 0]\). This vector points in the direction that is perpendicular to the plane.
Normal vectors are crucial when evaluating triple integrals over bounded regions because they help in determining the orientation of the planes that form the region. In the problem scenario, three normal vectors \(\mathbf{n}_{1} = [a, b, 0], \mathbf{n}_{2} = [c, 0, d], \text{and } \mathbf{n}_{3} = [0, e, f]\) are derived from the equations of planes. These vectors help understand whether the planes enclose a finite region.

Determinants

Determinants play a significant role in understanding the linear dependence of vectors and evaluating transformations. Given a set of vectors, the determinant can help determine if they are linearly independent. Specifically, for three vectors lying in the same space, the determinant of the matrix formed by placing these vectors as rows or columns is zero, indicating linear dependence.
In the solution, the triple scalar product \(\mathbf{n}_{1} \cdot (\mathbf{n}_{2} \times \mathbf{n}_{3})\) acts as a determinant. This product evaluates whether the three normal vectors lie in a plane. The expression calculates to \(-bce = 0\), meaning if \(b, c, \text{or } e\) are zero, the vectors are dependent, and thus the planes do not form a closed region needed for evaluating integrals.

Jacobian Matrix

The Jacobian matrix is essential when changing variables in a multi-variable integral. It describes how the function's output space transforms based on changes in the input space. For a transformation from \((x, y, z)\) to \((u, v, w)\), the Jacobian matrix is formed by the partial derivatives \(\frac{\partial(u, v, w)}{\partial(x, y, z)}\).
In this scenario, the transformation given by \(u=ax + by, v=cx + dz, w=ey + fz\) has the Jacobian matrix determinant simplified to \(-ade - bcf\). A zero determinant means no volume transformation occurs. Thus, if the Jacobian is zero, the region is unbounded, indicating that the planes do not enclose a finite space.

Linear Dependence

Linear dependence refers to a set of vectors being expressible as a linear combination of others. In simpler terms, if vectors are linearly dependent, one can be derived by adding or scaling the others. This concept is closely tied to the determinant of a matrix formed by these vectors.
In our example, if the triple scalar product \(\mathbf{n}_{1} \cdot (\mathbf{n}_{2} \times \mathbf{n}_{3}) = 0\), it shows linear dependence. When \(ade + bcf = 0\) or one out of \(b, c, \, \text{or}\,e\) is zero, it confirms this condition. This dependence implies that these vectors lie on a common plane, influencing whether the planes defined by these vectors encapsulate a bounded region or extend infinitely without enclosing a volume.