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 planes \(y=0, y=2, x=1, z=0\) and \(z=(3-x) / 2 ; \quad \delta(x, y, z)=2 \mathrm{gm} / \mathrm{cm}^{3}\).

Short answer

The center of mass is \((\frac{3}{5}, 1, \frac{9}{20})\)."

Step-by-step solution

Identify the Bounding Planes of the Region

The planes bounding the region \(D\) are given by \(y = 0\), \(y = 2\), \(x = 1\), \(z = 0\), and \(z = \frac{3-x}{2}\). These boundaries define a region in three-dimensional space.

Compute the Volume of the Region

To find the volume of region \(D\), set up the triple integral \(\int_{y=0}^{2}\int_{x=0}^{1}\int_{z=0}^{\frac{3-x}{2}} \,dz\,dx\,dy\). Perform the integration step by step starting with \(z\).

Evaluate the Inner Integral

Integrate with respect to \(z\): \[ \int_{z=0}^{\frac{3-x}{2}} \,dz = \frac{3-x}{2} \] Now substitute this into the next integral with integration limits for \(x\).

Evaluate the Integral with Respect to \(x\)

Integrate \(\int_{x=0}^{1} \frac{3-x}{2} \,dx\):\[ \int_{0}^{1} \frac{3-x}{2} \,dx = \left[ \frac{3x}{2} - \frac{x^2}{4} \right]_{0}^{1} = \frac{3}{2} - \frac{1}{4} = \frac{5}{4} \]

Evaluate the Outer Integral with Respect to \(y\)

Now integrate with respect to \(y\):\[ \int_{y=0}^{2} \frac{5}{4} \,dy = \left[ \frac{5}{4}y \right]_{0}^{2} = \frac{5}{2} \]The volume of the region is \(\frac{5}{2}\) cm³.

Calculate the Mass of the Region

The mass \(m\) can be calculated by multiplying the volume by the constant density function:\[ m = \frac{5}{2} \times 2 = 5 \, \text{gm} \]

Calculate the Center of Mass Coordinates

The coordinates of the center of mass \((\bar{x}, \bar{y}, \bar{z})\) are given by:- \(\bar{x} = \frac{1}{m} \int_{D} x \, dV\)- \(\bar{y} = \frac{1}{m} \int_{D} y \, dV\)- \(\bar{z} = \frac{1}{m} \int_{D} z \, dV\).Set up and evaluate each integral to find the coordinates.

Calculate \(\bar{x}\)

To find \(\bar{x}\), evaluate:\[ \bar{x} = \frac{1}{5} \int_{0}^{2} \int_{0}^{1} \int_{0}^{\frac{3-x}{2}} x \, dz \, dx \, dy \]Proceed with integration starting with \(z\) and then \(x\) and \(y\).

Calculate \(\bar{y}\)

To find \(\bar{y}\), evaluate:\[ \bar{y} = \frac{1}{5} \int_{0}^{2} \int_{0}^{1} \int_{0}^{\frac{3-x}{2}} y \, dz \, dx \, dy \] Perform the integration and simplify to find the coordinate.

Calculate \(\bar{z}\)

To find \(\bar{z}\), evaluate:\[ \bar{z} = \frac{1}{5} \int_{0}^{2} \int_{0}^{1} \int_{0}^{\frac{3-x}{2}} z \, dz \, dx \, dy \] Calculate each integral and simplify to determine \(\bar{z}\).

Conclusion - Center of Mass Coordinates

After evaluating each of the integrals for \(\bar{x}\), \(\bar{y}\), and \(\bar{z}\), the center of mass of the solid region is found to be \((\bar{x}, \bar{y}, \bar{z}) = (\frac{3}{5}, 1, \frac{9}{20})\).

Key concepts

Triple Integrals

Triple integrals are a powerful tool in calculus, used to compute quantities that vary in a three-dimensional space. They allow us to extend the concept of integration to three dimensions. This is useful for determining properties like volume or mass of a given solid region. In particular, a triple integral can be seen as iterated integrals, where we integrate one variable at a time.

Consider the triple integral \(\int_{z=a}^{b} \int_{y=c}^{d} \int_{x=e}^{f} f(x, y, z) \, dx \, dy \, dz\)This structure tells us to

• First integrate with respect to \(x\).
• Then integrate the resulting expression with respect to \(y\).
• Finally, integrate the remaining expression with respect to \(z\).

In the exercise, we applied a triple integral to compute the volume of a region bounded by specific planes. Understanding the setup and order of integration is essential for effectively using triple integrals.

Volume of Solid

The volume of a solid in three-dimensional space can be found using triple integrals. Given a region \(D\) defined by bounding planes, we can set up an integral to solve for the volume inside these boundaries. The volume integral for our exercise was:
\[\int_{y=0}^{2}\int_{x=0}^{1}\int_{z=0}^{\frac{3-x}{2}} \,dz \,dx \, dy\]Each integration step reduces the dimensions until we obtain a scalar representing the total volume enclosed.

Importance of Carefully Setting Limits

Setting the correct limits of integration is vital. Each limit corresponds to the bounds defined by the region's planes. Mistaking these can lead to incorrect results, as was seen when we calculated: first {{z}}, then {{x}}, and finally {{y}}. The final volume was determined as: \(\frac{5}{2}\) cm³.

Mass of Region

Calculating the mass of a solid region involves integrating its density function over the volume. This provides the actual mass, which accounts for variations in density throughout the space. The formula used here is
\[ m = \int_{D} \delta(x, y, z) \, dV \]
In our problem, the density was constant at \(2 \, \text{gm/cm}^3\). Thus, mass was straightforwardly computed as:
\( m = \text{Volume} \times \text{Density} = \frac{5}{2} \times 2 = 5 \, \text{gm} \)

Uniform vs. Variable Density

The process simplifies with uniform density, as seen here, resulting in merely multiplying volume by the constant density. For variable densities, the full integral must be evaluated. This distinction helps in understanding different methods of calculating mass in multi-variable calculus.