Problem 33 - 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: Understanding the Gradient and Normal Vector

The integral \( \oint_{C} abla(r^4) \cdot \mathbf{n} \ ds \) involves the gradient of the function \( r^4 \), where \( r = \sqrt{x^2 + y^2} \). The gradient \( abla(r^4) \) is a vector field. For any scalar field \( f \), the gradient is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Here, we have \( f = r^4 = (x^2 + y^2)^2 \).

Step 2: Calculate the Gradient \(\nabla(r^4)\)

First, differentiate \( r^4 = (x^2 + y^2)^2 \) with respect to \( x \) and \( y \). The partial derivative with respect to \( x \) is \( 4(x^2 + y^2)x \), and with respect to \( y \), it is \( 4(x^2 + y^2)y \). Therefore, the gradient is \( abla(r^4) = (4(x^2 + y^2)x, 4(x^2 + y^2)y) \).

Step 3: Simplify the Expression for Surface Integral

Apply the divergence theorem, which relates a line integral over a closed curve to a double integral over the region it encloses. Since \( abla \cdot abla(r^4) \) represents a divergence of a gradient (or Laplacian), it simplifies as \( abla^2(r^4) \).

Step 4: Apply Green's Theorem

Using Green's Theorem, the line integral \( \oint_{C} abla(r^4) \cdot \mathbf{n} \ ds \) becomes \( \iint_{R} abla^2(r^4) \ dA \). Calculate the Laplacian: \( abla^2(r^4) = \frac{\partial^2 r^4}{\partial x^2} + \frac{\partial^2 r^4}{\partial y^2} \).

Step 5: Compute the Laplacian \(\nabla^2(r^4)\)

Calculate \( \frac{\partial^2 (x^2 + y^2)^2}{\partial x^2} = 12x^2 + 4y^2 \) and \( \frac{\partial^2 (x^2 + y^2)^2}{\partial y^2} = 12y^2 + 4x^2 \), then sum them to find the Laplacian, which gives \( 16(x^2 + y^2) \).

Step 6: Evaluate the Double Integral

The double integral over the region \( R \) is \( \iint_{R} 16(x^2 + y^2) \ dA \). It is equivalent to \( 16 \times (I_x + I_y) \), where \( I_x \) and \( I_y \) are the moments of inertia about the \( x \)- and \( y \)-axes, respectively.

Key Concepts

Laplacian

The Laplacian is an operator that is used quite often in fields like physics and engineering. It is a way to transform a scalar function (like a temperature field or a potential function) into another scalar function that provides useful information. The Laplacian of a function is denoted by \( abla^2 f \), which can be read as "del squared f". This operation essentially measures how much the function "flows" or "curves" at a point.

For a function \( f(x, y) \) in two dimensions, the Laplacian is calculated by taking the partial derivatives with respect to each variable twice. So, \( abla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} \). In our exercise, we applied this to the function \( r^4 = (x^2 + y^2)^2 \). By finding each second partial derivative with respect to \( x \) and \( y \), and then adding them, we obtained \( 16(x^2 + y^2) \).

This calculation is crucial, as the Laplacian can provide us with a concise way of exploring various mathematical and physical scenarios, such as heat flow, wave propagation, and in this case, a connection to moments of inertia.

Divergence Theorem

The Divergence Theorem is a principle that allows us to transform a surface integral into a volume integral. It's particularly useful because it makes certain complex calculations much more manageable by changing the dimension of the space we are working with. It relates the flow of a vector field out of a closed surface to the behavior of the vector field inside that surface.

Mathematically, the theorem can be stated as \( \int \int_S \textbf{F} \cdot \textbf{n} \, dS = \int \int \int_V (abla \cdot \textbf{F}) \, dV \), where \( S \) is the closed surface, \( V \) is the volume inside \( S \), \( \textbf{F} \) is the vector field, and \( \mathbf{n} \) is the outward normal vector. In simpler terms, the theorem says that the amount going out of a surface is equal to whatever is created or gathered inside.

In applying this theorem, we're simplifying the calculation by moving from a complex boundary into simpler internal characteristics. This is fundamental in continuum mechanics and electromagnetism, where understanding how things behave inside a boundary is connected to observations at the boundary.

Moments of Inertia

Moments of inertia are crucial in determining how mass is distributed in a shape with respect to an axis. In physics, it's a way to measure an object's resistance to changes in its rotation. The higher the moment of inertia, the harder it is to change the rotation rate of the object.

For a region \( R \) in the plane, moments of inertia about the \( x \)- and \( y \)-axes are denoted as \( I_x \) and \( I_y \), respectively. Mathematically, these are calculated using the integrals \( I_x = \int \int_R y^2 \, dA \) and \( I_y = \int \int_R x^2 \, dA \). They help to quantify how the mass is distributed in relation to the respective axes.

In our exercise, the moments of inertia \( I_x \) and \( I_y \) play a role in converting a complex line integral into a more manageable double integral. Specifically, using the properties of inertia, we found that the value of the given line integral is expressed in terms of the moments of inertia: \( 16 \times (I_x + I_y) \). This shows the connection between geometry and physics by linking concepts from calculus to physical properties like inertia.

• It reduces to: \( \int \int_R 16(x^2 + y^2)\, dA \)
• This becomes: \( 16 \times (I_x + I_y) \)