Problem 27 - Thomas Calculus

Question

Let \(R\) be a region in the \(x y\) -plane that is bounded by a piecewise smooth simple closed curve \(C\) and suppose that the moments of inertia of \(R\) about the \(x-\) and \(y\) -axes are known to be \(I_{x}\) and \(I_{y}\) .Evaluate the integral $$\oint_{C} \nabla\left(r^{4}\right) \cdot \mathbf{n} d s$$ where \(r=\sqrt{x^{2}+y^{2}},\) in terms of \(I_{x}\) and \(I_{y}.\)

Step-by-Step Solution

Answer

The integral evaluates to \( 16(I_x + I_y) \).

Step 1: Understand the problem

We have a closed curve \( C \) around a region \( R \). We need to evaluate the line integral \( \oint_{C} abla\left(r^{4}\right) \cdot \mathbf{n} \, ds \), where \( r = \sqrt{x^2 + y^2} \). This involves using Green's Theorem in some form.

Step 2: Express \( \nabla(r^4) \)

Compute the gradient of \(r^4\): since \(r = \sqrt{x^2 + y^2}\), we have \(r^4 = (\sqrt{x^2 + y^2})^4 = (x^2 + y^2)^2\). The gradient is:\[ abla(r^4) = abla((x^2+y^2)^2) = 4(x^2+y^2)(x\mathbf{i}+y\mathbf{j}) \].

Step 3: Interpret \( \mathbf{n} \, ds \)

The vector \( \mathbf{n} \) is a unit normal to the curve \( C \), related to Green's or Gauss' theorems for flux integrals, and \( ds \) is an element of arc length along \( C \). For an outward normal, the integral \( \oint_C abla \cdot \mathbf{n} \, ds = \iint_R abla \cdot abla \left( r^4 \right) \, dA \) by Divergence Theorem.

Step 4: Calculate \( \nabla \cdot \nabla(r^4) \)

The divergence of a gradient, \( abla \cdot abla f \), is the Laplacian \( \Delta f \). Therefore, we find the Laplacian of \( r^4 = (x^2+y^2)^2 \):\[ \Delta(r^4) = \frac{\partial^2}{\partial x^2}(x^2+y^2)^2 + \frac{\partial^2}{\partial y^2}(x^2+y^2)^2 \].First term: \( \frac{\partial^2}{\partial x^2}(x^2+y^2)^2 = 12x^2 + 4y^2 \).Second term: \( \frac{\partial^2}{\partial y^2}(x^2+y^2)^2 = 12y^2 + 4x^2 \).So, \( \Delta(r^4) = 16(x^2 + y^2) \).

Step 5: Apply Green's Theorem

Using the divergence format of Green’s Theorem. The integral \( \oint_C abla(r^4) \cdot \mathbf{n} \, ds \) becomes \( \iint_R \Delta(r^4) \, dA = 16 \iint_R (x^2 + y^2) \, dA \). This integral gives us \( 16 \left(I_x + I_y \right) \) since moments about axes for position (\( x^2+y^2 \)) relate to the moments of inertia.

Key Concepts

Understanding the Gradient of a Function

The gradient of a function is a central concept in vector calculus. Think of it as a vector that points in the direction of the steepest ascent of a scalar field. If you have a function like \( f(x, y) \), its gradient is denoted \( abla f \), and it provides a way to measure how much \( f \) changes as you move in different directions in space.
A gradient is composed of partial derivatives. For a function \( r^4 \), where \( r = \sqrt{x^2 + y^2} \), its gradient \( abla(r^4) \) becomes \( 4(x^2+y^2)(x\mathbf{i}+y\mathbf{j}) \).
This shows how the function changes with respect to each variable, with \( \mathbf{i} \) and \( \mathbf{j} \) being unit vectors in the x and y directions, respectively. Essentially, a gradient vector tells you both **magnitude and direction** of the greatest rate of change at any given point.

Exploring Line Integrals

Line integrals are used to calculate the value of a field along a curve. They are essential when you need to integrate a function not just over an area, but along a path.
The integral \( \oint_C abla f \cdot \mathbf{n} \, ds \) involves a curve \( C \) and a unit normal vector \( \mathbf{n} \) that is perpendicular to the curve. Here, the dot product \( abla f \cdot \mathbf{n} \) captures how much of the gradient is pointing across the curve.
A line integral along a closed curve like \( C \) plays roles in physics, notably in computing work done by a force field on a particle as it moves around a path. When working with line integrals, it's crucial to consider both the path and the field’s properties.

Utilizing Green's Theorem

Green's Theorem relates line integrals around a closed curve to area integrals over the region enclosed by the curve. This powerful tool in vector calculus provides a bridge between an area in the plane and what happens along its boundary.
For example, using Green's Theorem, the line integral \( \oint_C abla(r^4) \cdot \mathbf{n} \, ds \) converts to an area integral over region \( R \) of the divergence, i.e., \( \iint_R \Delta(r^4) \, dA \).

• This conversion simplifies calculations significantly, especially when direct computation of a line integral is complex.
• It underscores the deep relationship between circulation around a boundary and flux through the interior area.

By turning boundary problems into area problems, Green's Theorem provides elegant solutions in fluid dynamics and electromagnetism.

Moments of Inertia Made Simple

Moments of inertia measure how much an object resists rotational motion around an axis. They are vital in mechanics and involve mass distribution with respect to rotations.
Mathematically, the moment of inertia about the x-axis, \( I_x \), and the y-axis, \( I_y \), are integrals that consider both the area's shape and mass distribution.
For closed curve regions, these moments are related directly to expressions involving \( x^2 \) and \( y^2 \) values inside the region. In our problem, the integral simplifies to \( 16 (I_x + I_y) \), showing a clear link between the physical property (resistance to rotation) and mathematical expression.
Moments of inertia play key roles in predicting motion and designing systems in engineering, ensuring stable and efficient performance.