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Chapter 15: Problem 31

Use a table of integrals or a computer algebra system (if necessary) to find the moment of inertia around the z-axis of the given surface \(S\). Assume that \(S\) has constant density \(\delta \equiv 1\). \(S\) is the part of the parabolic cylinder \(z=4-y^{2}\) that lies inside the rectangular cylinder \(-1 \leqq x \leqq 1,-2 \leqq y \leqq 2 .\)

Short Answer

Expert verified
Use a computer algebra system to evaluate the integral for numerical result.

Step by step solution

01

Understanding Moment of Inertia Calculation

The moment of inertia around the z-axis for a surface with a density \( \delta = 1 \) is given by \( I_z = \iint_S (x^2 + y^2) \, dS \). Our task is to integrate the function \( x^2 + y^2 \) over the surface \( S \) defined by the given boundaries.
02

Describe the Surface S

The surface \( S \) is bounded by the parabolic cylinder given by \( z = 4 - y^2 \) for points \( (x, y, z) \) inside the rectangular cylinder \(-1 \le x \le 1\) and \(-2 \le y \le 2\). The surface in question is a part of this parabolic cylinder capped by limits on \( x \) and \( y \).
03

Setup the Integral for Moment of Inertia

Given the constant density, we need to evaluate \( I_z = \int_{-1}^{1} \int_{-2}^{2} (x^2 + y^2) \sqrt{1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2} \, dy \, dx \), where \( z = 4-y^2 \). Here, \( \frac{\partial z}{\partial x} = 0 \) and \( \frac{\partial z}{\partial y} = -2y \). Thus, the integrand becomes \( \sqrt{1 + 4y^2} \).
04

Solve the Integral

Substitute into the integral: \[ I_z = \int_{-1}^{1} \int_{-2}^{2} (x^2 + y^2) \sqrt{1 + 4y^2} \, dy \, dx \]. This integral can be solved by first integrating with respect to \( y \) and then with respect to \( x \). However, due to symmetry and complexity in direct integration, numerical methods or a computer algebra system like Mathematica can be useful.
05

Numerical Evaluation

Using a computer algebra system (CAS) or numerical integration tool, evaluate the double integral: \( \int_{-1}^{1} \) \( \int_{-2}^{2} (x^2 + y^2) \sqrt{1 + 4y^2} \, dy \, dx \). The result provides the moment of inertia around the z-axis for the given surface.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Surface Integration
Understanding surface integration is key to finding the moment of inertia for a surface. This involves calculating a double integral over a defined surface area. In the context of our exercise, we need to integrate the expression \( x^2 + y^2 \) over the parabolic cylinder defined by the surface \( S \). To do this efficiently:
  • We first describe the boundaries of the surface \( S \). For this problem, \( S \) is captured by the parabolic cylinder \( z = 4 - y^2 \) and confined within the limits \( -1 \le x \le 1 \) and \( -2 \le y \le 2 \).
  • Next, we establish the integral for the moment of inertia, which incorporates the geometry of the surface. This means involving the surface's characteristics such as slope, as described by the partial derivatives.
  • For our surface, since \( \frac{\partial z}{\partial x} = 0 \) and \( \frac{\partial z}{\partial y} = -2y \), the integration becomes \( \int_{-1}^{1} \int_{-2}^{2} (x^2 + y^2) \sqrt{1 + 4y^2} \, dy \, dx \).
By solving this double integral, we effectively sum the weighted distribution of mass across the defined surface, yielding the desired moment of inertia.
Parabolic Cylinder
A parabolic cylinder is a type of surface created by translating a parabola along an axis perpendicular to the plane of the parabola. In this exercise:
  • Our parabolic cylinder is given by the equation \(z = 4 - y^2 \). This indicates a parabola opening downward along the \( z \)-axis and varying along \( y \).
  • The bounding cylinder constrains \( z \) to the xy-plane, maintaining it between the limits of \( x \) from \(-1\) to \(1\) and \( y \) from \(-2\) to \(2\).
  • This creates a finite section of the parabola that interacts with specified X and Y limits, helping to determine the portion of the surface where integration happens.
This geometrical setup is crucial for determining how the surface contributes to the moment of inertia. The parabolic shape impacts the distribution of the surface elements and thereby the integral's value.
Density Function
In this exercise, the density function is set to a constant value, \( \delta = 1 \), simplifying the moment of inertia calculation. The concept of density function is important in these scenarios because it provides:
  • A measure of how mass is distributed across the surface.
  • In more complex scenarios, density might vary across the surface, which would need integration of \( \delta(x, y) \times (x^2 + y^2) \) instead of just \( (x^2 + y^2) \).
  • With a constant density, calculations become more straightforward since the density does not require further calculation or adjustment during the integration.
Thus, assuming a constant density simplifies our task, making the moment of inertia solely dependent on the geometry of the surface defined by the parabolic cylinder.
Numerical Methods
Numerical methods are invaluable when solving complex integrals that are difficult or impossible to evaluate analytically. For our given integral:
  • The expression \( \int_{-1}^{1} \int_{-2}^{2} (x^2 + y^2) \sqrt{1 + 4y^2} \, dy \, dx \) might have an analytic solution that's cumbersome to derive.
  • Using numerical methods or computer algebra systems like Mathematica can simplify the process, providing an accurate approximation of the integral.
  • These methods involve breaking down the integration into sums over small, manageable increments, allowing a computer to efficiently calculate and tally them.
Choosing numerical methods offers a practical approach, especially in educational contexts, where the focus is on understanding principles rather than manual computation of tedious calculations.

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Most popular questions from this chapter

Find the moment of inertia \(\iint_{S}\left(x^{2}+y^{2}\right) d S\) of the given surface \(S .\) Assume that \(S\) has constant density \(\delta \equiv 1\). \(S\) is the part of the cylinder \(x^{2}+z^{2}=1\) that lies between the planes \(y=-1\) and \(y=1\). As parameters on the cylinder use \(y\) and the polar angular coordinate in the \(x z\) -plane.

Apply Green's theorem to evaluate the integral $$\oint_{C} P d x+Q d y$$ around the specified closed curve \(C\). \(P(x, y)=y+e^{x} \cdot Q(x, y)=2 x^{2}+\cos y ; \quad C\) is the bound- ary of the triangle with vertices \((0,0),(1,1)\), and \((2,0)\).

The Laplacian of the twice-differentiable scalar function 7 is defined to be \(\boldsymbol{r}^{2} f=\operatorname{div}(\operatorname{grad} f)=\boldsymbol{\nabla} \cdot \boldsymbol{\nabla} f .\) Show that $$\boldsymbol{\nabla}^{2} f=\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial^{2} f}{\partial z^{2}}$$

Let \(f\) and \(g\) be functions with continuous second-order partial derivatives in the region \(R\) bounded by the piecewise smooth simple closed curve \(C\). Apply Green's theorem in vector form to show that $$\oint_{C} f \nabla g \cdot \mathbf{n} d s=\iint_{R}[f \nabla \cdot \nabla g+\nabla f \cdot \nabla g] d A$$ It was this formula rather than Green's theorem itself that appeared in Green's book of 1828 .

Determine whether the vector fields are conservative. Find potential functions for those that are conservative (either by inspection or by using the method of Example 4 ). \(\mathbf{F}(x, y)=\left(3 x^{2} y^{3}+y^{4}\right) \mathbf{i}+\left(3 x^{3} y^{2}+y^{4}+4 x y^{3}\right) \mathbf{j}\)
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