Problem 46 - Advanced Engineering Mathematics

Question

Find the center of mass of the lamina that has the given shape and density. $$ x=y^{2}, x=4 ; \rho(x, y)=y+5 $$

Step-by-Step Solution

Answer

The center of mass is at \(( \bar{x}, \bar{y} ) = ( \frac{32}{15}, 0 )\)."

Step 1: Understand the Region Bounded by the Curves

The lamina is bounded by the parabola \( x=y^2 \) and the line \( x=4 \). To find the area of integration, set \( x=y^2 \) equal to \( 4 \), leading to \( y = \pm 2 \). Thus, the lamina extends from \( y = -2 \) to \( y = 2 \).

Step 2: Set Up the Mass Integral

To find the total mass \( M \) of the lamina, integrate the density function over the lamina's area. The integral is: \[M = \int_{-2}^{2} \int_{y^2}^{4} (y + 5) \mathrm{d}x \mathrm{d}y\].

Step 3: Simplify the Mass Integral

Solve the inner integral first: \[\int_{y^2}^{4} (y + 5) \mathrm{d}x = (y + 5)(4 - y^2) = 4(y + 5) - (y^3 + 5y^2)\]This simplifies to \(4y + 20 - y^3 - 5y^2\).

Step 4: Evaluate the Mass Integral

Integrate the expression from Step 3 with respect to \( y \):\[M = \int_{-2}^{2} (4y + 20 - y^3 - 5y^2) \mathrm{d}y\].Calculate this definite integral to find the mass \( M \).

Step 5: Set Up the Moment Integrals

The moments about the \( x \)-axis \( M_y \) and \( y \)-axis \( M_x \) are:\[M_y = \int_{-2}^{2} \int_{y^2}^{4} x(y + 5) \mathrm{d}x \mathrm{d}y\]\[M_x = \int_{-2}^{2} \int_{y^2}^{4} y(y + 5) \mathrm{d}x \mathrm{d}y\].

Step 6: Simplify the Moment Integrals

Evaluate the inner integrals as follows:For \( M_y \),\[\int_{y^2}^{4} x(y + 5) \mathrm{d}x = \left[ \frac{x^2}{2}(y + 5) \right]_{y^2}^{4}\].For \( M_x \),\[\int_{y^2}^{4} y(y + 5) \mathrm{d}x = y(y + 5)(4 - y^2)\].

Step 7: Evaluate the Moment Integrals

Evaluate the integrals from Step 6 and compute \( M_x \) and \( M_y \) over the interval \([-2, 2]\).

Step 8: Calculate the Center of Mass

The center of mass \( (\bar{x}, \bar{y}) \) is given by\[\bar{x} = \frac{M_y}{M}, \quad \bar{y} = \frac{M_x}{M}\].Substitute the calculated values of \( M \), \( M_x \), and \( M_y \) to find \( \bar{x} \) and \( \bar{y} \).

Step 9: Finalize the Solution

Finalize any calculations and verify the coordinates of the center of mass, ensuring accuracy in computations.

Key Concepts

Lamina

A lamina is a flat, two-dimensional object that has mass and occupies an area in space. In our problem, the lamina is the area bounded by the curves \( x = y^2 \) and \( x = 4 \). To visualize this, imagine a thin plate that lies flat on a plane. It's not thick like a solid object, but it still has some mass distributed over its area. The boundary curves tell us the shape of the lamina in the coordinate plane. The parabola \( x = y^2 \) curves upwards and downwards, while the line \( x = 4 \) is a vertical boundary cutting through the parabola's arms at \( y = -2 \) and \( y = 2 \). Therefore, the lamina is the area enclosed between these curves, extending along the \( y \)-axis from \( -2 \) to \( 2 \). Understanding the geometry of the lamina is critical to calculating properties like center of mass, which involves integrating across this area.

Density Function

The density function \( \rho(x, y) = y + 5 \) describes how mass is distributed over the lamina's area. Density functions are essential in physics and engineering because they tell us how much mass is found in a specific region of an object. In this problem, the function indicates that at each point \( (x, y) \) on the lamina, the density or 'heaviness' of the material there is given by \( y + 5 \).

• When \( y = 0 \), the density is 5, implying that the minimum mass density occurs along the \( x \)-axis.
• At \( y = 1 \), the density increases to 6, and so forth.
• This means that as you move vertically along the lamina, the mass density increases linearly with \( y \).

The integration process involves this function because each tiny piece of the lamina contributes to the total mass based on its density. To find the total mass, we integrate this density function over the entire lamina.

Double Integral

Double integrals are used here to compute the total mass and the moments about the axes. A double integral works by integrating a function over a two-dimensional area. For our lamina, we first integrate in the \( x \)-direction from \( y^2 \) to \( 4 \), representing the width of the lamina at each \( y \) value.Then, we integrate in the \( y \)-direction from \( -2 \) to \( 2 \), which determines the height of our region. The double integral notation for mass \( M \) is shown as:\[ M = \int_{-2}^{2} \int_{y^2}^{4} (y + 5) \mathrm{d}x \mathrm{d}y \]This integral calculates the sum of density values across every point in the lamina. The same process applies for moments \( M_x \) and \( M_y \), using different expressions inside the integrals.

• Inner integrals are solved first, providing a function of \( y \).
• Outer integrals, then compute over the \( y \) limits, summing up contributions across the entire height.

Using double integrals is powerful for understanding how various attributes change across a surface, vital for finding the center of mass.

Moment of Inertia

While not directly calculated here, the concept is worth discussing as it's related to the center of mass. Moment of inertia, in a two-dimensional sense, typically refers to how much resistance the lamina offers to rotational motion around a point or axis. Mathematically, the moment of inertia \( I \) about the origin for a lamina could be expressed as:\[ I = \int \int (x^2 + y^2) \rho(x, y) \mathrm{d}x \mathrm{d}y \]In our scenario, the moments \( M_x \) and \( M_y \) represent integrals that help find the center of mass. These are not moments of inertia but relate to the distribution of mass concerning the \( x \) and \( y \) axes, respectively.

• \( M_x \) focuses on how mass is spread vertically.
• \( M_y \) considers the horizontal spread.

Understanding these ideas helps appreciate how mass distribution affects balance and stability, key for engineering designs and analysis.