Chapter 16: Problem 33
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 \(I_{y}\)
Short Answer
Step by step solution
Understanding the Gradient and Normal Vector
Calculate the Gradient \(\nabla(r^4)\)
Simplify the Expression for Surface Integral
Apply Green's Theorem
Compute the Laplacian \(\nabla^2(r^4)\)
Evaluate the Double Integral
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Laplacian
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
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
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: \( 16 \times (I_x + I_y) \)