Problem 16 - Thomas Calculus

Question

Polar inertia of triangular plate Find the polar moment of inertia about the origin of a thin triangular plate of constant density \(\delta=3\) bounded by the \(y\) -axis and the lines \(y=2 x\) and \(y=4\) in the \(x y\) -plane.

Step-by-Step Solution

Answer

The polar moment of inertia about the origin is 104.

Step 1: Determine the Area of Integration

The given triangular plate is bounded by the lines \(y = 2x\), \(y = 4\), and the \(y\)-axis. To find the area, we first find the points of intersection of these lines. The intersection of \(y = 2x\) and \(y = 4\) occurs when \(2x = 4\), giving \(x = 2, y = 4\). The intersections with the \(y\)-axis are \((0, 0)\) for \(y = 2x\) and \((0, 4)\) for \(y = 4\). Thus, the vertices of the triangle are \((0, 0)\), \((0, 4)\), and \((2, 4)\).

Step 2: Setup the Polar Moment of Inertia Formula

The polar moment of inertia \(J\) about the origin for a region \(R\) in the \(xy\)-plane is given by \[ J = \int\int_R (x^2 + y^2) \, dx \, dy \] with density \(\delta = 3\). Thus, the polar moment of inertia becomes \[ J = 3 \int\int_R (x^2 + y^2) \, dx \, dy. \]

Step 3: Establish Integration Limits

For integration, consider the region in terms of \(y\), which varies from the \(x\)-axis (\(y=0\)) to the line \(y=4\). For a given \(y\), \(x\) ranges from \(x=0\) on the \(y\)-axis to the line \(y=2x\), which gives \(x = \frac{y}{2}\). Thus, the limits for \(x\) are from \(0\) to \(\frac{y}{2}\) and \(y\) from \(0\) to \(4\).

Step 4: Integrate Over the Defined Region

Substitute the limits and perform the double integration: \[ J = 3 \int_{0}^{4} \int_{0}^{\frac{y}{2}} (x^2 + y^2) \, dx \, dy. \] First, integrate with respect to \(x\): \[ \int_{0}^{\frac{y}{2}} (x^2 + y^2) \, dx = \left[ \frac{x^3}{3} + y^2x \right]_{0}^{\frac{y}{2}} = \frac{(\frac{y}{2})^3}{3} + y^2(\frac{y}{2}). \]This simplifies to \[ \frac{y^3}{24} + \frac{y^3}{2} = \frac{13y^3}{24}. \]

Step 5: Final Integration

Integrate with respect to \(y\): \[ J = 3 \int_{0}^{4} \frac{13y^3}{24} \, dy = 3 \times \left[ \frac{13y^4}{96} \right]_{0}^{4}. \]Evaluate: \[ 3 \times \frac{13 \times 256}{96} = 3 \times \frac{3328}{96} = 3 \times 34.6667 = 104. \]

Step 6: Verify the Result

The final integration result gives the polar moment of inertia as \(J = 104\). Verify the calculations for errors by ensuring that each integration step corresponds to linear algebraic transformations and evaluate using the defined limits correctly.

Key Concepts

Triangular Plate

Understanding the concept of a triangular plate is essential in solving problems related to the polar moment of inertia. A triangular plate is a flat, two-dimensional shape with three straight sides and the three vertex points where these sides meet.
In this particular exercise, the triangular plate is defined by three boundaries:

• The y-axis.
• The line \(y = 2x\).
• The horizontal line \(y = 4\).

The vertices of the triangle, found by identifying the intersection points, are at \(0,0\), \(0,4\), and \(2,4\). These vertices help in setting the limits for the integration needed to calculate the polar moment of inertia. This specific form actually resembles a right triangle, where clear boundaries simplify the integration process.

Double Integration

Double integration is a statistical method used to calculate the volume under a surface defined by a function over a two-dimensional region.
For the polar moment of inertia problem, the function is \((x^2 + y^2)\) multiplied by the constant density \(\delta = 3\), resulting in an integral of the form:

• \[ J = 3 \int\int_R (x^2 + y^2) \, dx \, dy \]

The purpose of double integration in this exercise is to sum up all the infinitesimal contributions of inertia from each tiny area within the triangular region to find the total polar moment of inertia about the origin.
First, you integrate over \(x\), keeping \(y\) somewhat constant, then another integration over \(y\) for the whole region. This approach handles all shifts in both directions within the limits defined by the triangle.

Density

Density is an important concept when dealing with mass distribution across a body. It gives an insight into how mass is spread over an area or volume.
In this exercise, the plate has a constant density denoted by \(\delta = 3\). This means the mass per unit area for each part of the plate is uniform.
The role of density in calculating polar moment of inertia (\(J\)) is crucial because it scales the moment of each infinitesimal part of the region.
Mathematically, this is represented as:

• \[ J = \int\int_R \delta (x^2 + y^2) \, dx \, dy \], where \(\delta = 3\).

The uniform density simplifies the integration as you can factor it out of the integral, directly affecting the final calculation of the inertia, creating a straightforward multiplication factor across the region.

Polar Coordinate System

The polar coordinate system offers a different way to describe locations of points within a plane, using a radius and an angle rather than rectangular coordinates \((x, y)\). Although the calculation here is approached using Cartesian coordinates, understanding polar coordinates is beneficial due to their common use in rotational systems.
If this problem were approached in polar coordinates, the conversion would require:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)
• \(x^2 + y^2 = r^2\)

This transformation can sometimes simplify the integration of circular or rotational systems, making them invaluable in deeper studies.
The polar moment of inertia is particularly connected to this coordinate system, as it inherently measures resistance to rotation about a point, aligning conceptually with how polar coordinates map systems based on a central pivot.