Calculating the Moment of Inertia of a Solid of Revolution: An Example with x2 - y2 3, y 2, and y 0

In this article, we will explore the process of calculating the moment of inertia for a solid generated by rotating a specific area about an axis. Our example will involve the curves defined by the hyperbola x2 - y2 3, the horizontal lines y 2 and y 0, and the line of revolution x 0. The moment of inertia of a solid about an axis is a fundamental concept in physics and engineering, often used in the analysis of rotating objects.

Understanding the Region

To begin, let us understand the area bounded by the given curves:

The hyperbola x2 - y2 3 can be rewritten as x2 y2 3. The lines y 2 and y 0 are horizontal and define the bounds of our region.

Finding the Intersection Points

First, we need to find the points of intersection between the hyperbola and the line y 2:

[ x^2 - 2^2 3 implies x^2 - 4 3 implies x^2 7 implies x pm sqrt{7}]

Thus, the points of intersection are ( (sqrt{7}, 2) ) and ( (-sqrt{7}, 2) ).

Setting Up the Moment of Inertia Integral

The moment of inertia ( I ) about the line ( x 0 ) can be calculated using the formula: [ I int_A y^2 , dA]

Here, dA is the area element. We will use vertical strips to find the area, with the height of the strip as y and the width as the distance between the hyperbola and the y-axis.

Expressing x in Terms of y

From the hyperbola equation ( x^2 - y^2 3 ), we have:

[ x sqrt{y^2 3}]

The area element ( dA ) can be expressed in terms of y as:

[ dA 2sqrt{y^2 3} , dy]

Setting Up the Limits of Integration

The limits for y are from 0 to 2. Thus, the integral becomes:

[ I int_0^2 y^2 2sqrt{y^2 3} , dy]

Simplifying, we get:

[ I 2 int_0^2 y^2 sqrt{y^2 3} , dy]

Integrating the Expression

To compute this integral, we can use a substitution:

Let ( u y^2 3 ), then ( du 2y , dy ) or ( y , dy frac{du}{2} ). The limits change as follows: When y 0, ( u 3 ). When y 2, ( u 7 ).

Thus, the integral becomes:

[ I 2 int_3^7 left(frac{u - 3}{2}right) sqrt{u} frac{du}{2}]

Simplifying, we get:

[ I frac{1}{2} int_3^7 u - 3 sqrt{u} , du]

Now, we need to compute:

[ int u - 3 sqrt{u} , du int u^{3/2} - 3u^{1/2} , du]

This can be integrated:

[ int u^{3/2} , du frac{2}{5} u^{5/2} quad int u^{1/2} , du frac{2}{3} u^{3/2}]

Thus:

[ I frac{1}{2} left[ frac{2}{5} u^{5/2} - 3 cdot frac{2}{3} u^{3/2} right]_3^7]

Evaluating the Integral

Evaluate the integral:

[ I frac{1}{2} left[ frac{2}{5} 7^{5/2} - 3^{5/2} - 2 cdot 7^{3/2} - 3^{3/2} right]]

By performing the calculations, you will arrive at the moment of inertia of the solid generated by rotating the given area about the line x 0.