Question

Find the center of mass of the solid represented by the indicated space region \(D\) with density function \(\delta(x, y, z)\). \(D\) is bounded by the coordinate planes and \(z=2-2 x / 3-2 y ; \quad \delta(x, y, z)=10 \mathrm{gm} / \mathrm{cm}^{3}\).

Short answer

The center of mass is calculated using triple integrals over the specified region, considering symmetry if possible.

Step-by-step solution

Determine the Boundaries of the Region

Since the solid is bounded by the coordinate planes and the plane \(z = 2 - \frac{2x}{3} - 2y\), we know that \(x, y, z \geq 0\). Also, from the equation of the plane, the region is bounded such that \(z\) ranges from 0 to \(2 - \frac{2x}{3} - 2y\). Additionally, to ensure realistic bounds, solve for the region on the xy-plane: \(z=0\) gives \(2 - \frac{2x}{3} - 2y = 0\). Simplifying, the constraints become \(3y = 1 - x\)。 Thus, \(0 \leq x \leq 3\) and \(0 \leq y \leq \frac{3-3x}{2}\).

Express Mass and Calculate Total Mass

The total mass of the solid can be calculated by integrating the density function over the region \(D\). The mass \(m\) is given by:\[ m = \int \int \int_D \delta(x, y, z)\, dz\, dy\, dx. \]Substitute \(\delta(x, y, z) = 10\):\[ m = \int_0^3 \int_0^{\frac{3-3x}{2}} \int_0^{2-\frac{2x}{3}-2y} 10\, dz\, dy\, dx. \]First, compute the inner integral over \(z\):\[ \int_0^{2-\frac{2x}{3}-2y} 10\, dz = 10z \Bigg|_0^{2-\frac{2x}{3}-2y} = 10\left(2-\frac{2x}{3}-2y\right). \]

Integrate over y and x for Mass

Now integrate over \(y\):\[ \int_0^{\frac{3-3x}{2}} 10\left(2-\frac{2x}{3}-2y\right)\, dy = \int_0^{\frac{3-3x}{2}} [20 - \frac{20x}{3} - 20y]\, dy. \]Compute the integral:\[ \left[ 20y - \frac{20x}{3}y - 10y^2 \right]_0^{\frac{3-3x}{2}} = 20\left(\frac{3-3x}{2}\right) - \frac{20x}{3}\left(\frac{3-3x}{2}\right) - 10\left(\frac{3-3x}{2}\right)^2. \]Simplify this expression and perform the integration over \(x\).

Compute Mass (Continue Integration over x)

Now, integrate over \(x\):\[ \int_0^3 \text{expression in } x\, dx. \]Solve this integration to find the total mass \(m\).

Find Center of Mass Coordinates

The coordinates of the center of mass \((\bar{x}, \bar{y}, \bar{z})\) are determined using:\[ \bar{x} = \frac{1}{m} \int \int \int_D x\delta(x, y, z)\, dz\, dy\, dx, \]\[ \bar{y} = \frac{1}{m} \int \int \int_D y\delta(x, y, z)\, dz\, dy\, dx, \]\[ \bar{z} = \frac{1}{m} \int \int \int_D z\delta(x, y, z)\, dz\, dy\, dx. \]Substitute and evaluate these integrals similarly by repeating the integration process over the specified bounds. The symmetry of the problem might simplify these calculations.

Key concepts

Calculating Mass of a Solid

Determining the mass of a solid in three-dimensional space is equivalent to summing up all the small "chunks" of mass within the solid, each characterized by a density function. Imagine you have countless tiny cubes filling up the entire solid, each with its own mass density that can vary by position. The mass of each little cube is the density times its volume.
To find the total mass of the solid, especially when the density varies, we use a mathematical operation called integration. For solids represented in three-dimensional space, we specifically use triple integrals. This involves integrating over the region the solid occupies and the density function of the solid. This method ensures that every part of the solid is considered, contributing to the total mass calculation.

Triple Integrals in Calculus

Triple integrals are a powerful tool in multivariable calculus, used to evaluate volume and mass in three-dimensional spaces. When you set up a triple integral, you're essentially stacking multiple integrals together to account for the variations in all three axes: x, y, and z.
In the context of calculating mass, triple integrals allow us to integrate the density function over the entire solid.

• The innermost integral typically deals with the z-axis, followed by the y-axis, and finally the x-axis. This nested integration ensures thorough coverage of the space region.
• It's crucial to carefully determine the bounds for each of these integrals, reflecting the physical dimensions and limits of the solid's region.
• Performing a triple integral step by step allows breaking down complex interactions across the three dimensions into simpler, solvable segments.

All these aspects help to compute not just volume, but more importantly, the mass of objects, given their density aspects.

Density Functions in Physics

Density functions play a critical role in both physics and calculus when calculating the center of mass. Density describes how much mass exists within a specific volume. It is denoted commonly by the symbol \( \delta(x, y, z) \), indicating it can vary in different locations across the xyz-space.
A constant density function, like \( \delta(x, y, z) = 10 \) gm/cm³ from the exercise, implies uniform density throughout the solid. However, the density function might depend on variables x, y, and z, which can dramatically affect how mass is distributed.

• Understanding how density affects mass distribution is essential as it dictates how integrals are set up for mass calculations.
• Physically, if certain parts of the region have higher densities, they will contribute more to the total mass, shifting the center of mass.

In the exercise, knowing that density is constant simplifies the calculations, focusing on integrating over the region directly without additional complexities.

Coordinate Planes in Three-Dimensional Space

Three-dimensional space is defined with respect to three coordinate planes: the XY-plane, the YZ-plane, and the ZX-plane. Understanding these planes is essential for solving problems involving volumes and integrals.

• The coordinate planes are formed by holding one of the coordinates constant at zero, e.g., the XY-plane is where z=0.
• Each plane acts as a boundary in various problems, including the one provided, where the solid is constrained by these planes.
• When dealing with integration limits, the coordinate planes frequently set the lower or upper bounds for the integrals.

By visualizing the intersection of these planes with the solid's region, one gains a clear understanding of the limits and boundaries, which helps in setting up the integrals correctly and solving the problem efficiently.