Problem 55 - Advanced Engineering Mathematics

Question

Find the centroid of the homogeneous solid that is bounded by the hemisphere \(z=\sqrt{a^{2}-x^{2}-y^{2}}\) and the plane \(z=0\)

Step-by-Step Solution

Answer

The centroid of the hemisphere is at \(\left(0, 0, \frac{3a}{8}\right)\).

Step 1: Understand the Problem

We are given a hemisphere defined by the equation \(z = \sqrt{a^2 - x^2 - y^2}\) with its base on the plane \(z=0\). A centroid is the point that represents the geometric center of a solid.

Step 2: Symmetry Considerations

The hemisphere is symmetric with respect to the x-axis and y-axis. Therefore, the x-coordinate and y-coordinate of the centroid will be zero:\[\bar{x} = 0,\quad \bar{y} = 0\]

Step 3: Set Up the Formula for \(\bar{z}\)

The formula for the z-coordinate of the centroid \(\bar{z}\) of a solid in the first octant is \[\bar{z} = \frac{1}{V} \int_V z \, dV\]where \(V\) is the volume of the hemisphere. Since the hemisphere is centered at the origin, we consider the whole solid and symmetry simplifies calculations.

Step 4: Calculate the Volume of the Hemisphere

The volume \(V\) of a hemisphere with radius \(a\) is given by:\[V = \frac{2}{3}\pi a^3\]

Step 5: Express dV in Cylindrical Coordinates

Convert to cylindrical coordinates: \(x = r\cos\theta\), \(y = r\sin\theta\), \(z = z\). The differential volume element \(dV\) becomes \(dV = r \, dr \, d\theta \, dz\).

Step 6: Set Up the Integral for \(\bar{z}\)

The limits for \(r\) are from 0 to \(a\), \(\theta\) from 0 to \(2\pi\), and \(z\) from 0 to \(\sqrt{a^2 - r^2}\). The integral is:\[\bar{z} = \frac{1}{V} \int_0^{2\pi} \int_0^a \int_0^{\sqrt{a^2 - r^2}} z \, r \, dz \, dr \, d\theta\]

Step 7: Evaluate the Integral

Evaluate the innermost integral with respect to \(z\):\[\int_0^{\sqrt{a^2 - r^2}} z \, dz = \left[ \frac{z^2}{2} \right]_0^{\sqrt{a^2 - r^2}} = \frac{a^2 - r^2}{2}\]Next, evaluate the integral with respect to \(r\):\[\int_0^a \frac{a^2 - r^2}{2} r \, dr = \left[ \frac{a^2r^2}{4} - \frac{r^4}{8} \right]_0^a = \frac{a^4}{8}\]

Step 8: Integrate with Respect to \(\theta\)

Integrate the result with respect to \(\theta\):\[\int_0^{2\pi} \frac{a^4}{8} \, d\theta = \frac{a^4}{8} \times 2\pi = \frac{\pi a^4}{4}\]

Step 9: Find \(\bar{z}\)

Divide by the volume to find \(\bar{z}\):\[\bar{z} = \frac{1}{\frac{2}{3}\pi a^3} \cdot \frac{\pi a^4}{4} = \frac{3a}{8}\]

Step 10: Final Result

Combine the results for \(\bar{x}\), \(\bar{y}\), and \(\bar{z}\) to get the centroid of the hemisphere:\[\left( 0, 0, \frac{3a}{8} \right)\]

Key Concepts

Hemisphere Volume

The volume of a hemisphere is an essential element when finding the centroid of a solid object like a hemisphere. To calculate this volume, imagine cutting a sphere in half along its equatorial plane. The result is a hemisphere.
The formula for the volume of a hemisphere with radius \(a\) is given by:

• \[ V = \frac{2}{3}\pi a^3 \]

This formula is derived from the volume of a complete sphere, which is \(\frac{4}{3}\pi a^3\), divided by 2 since we are only considering half of a sphere. Calculating this volume is crucial in centroid computations, as the centroidal z-coordinate includes dividing by this volume.

Cylindrical Coordinates

For complex three-dimensional shapes, converting standard Cartesian coordinates into cylindrical coordinates simplifies calculations significantly.
Cylindrical coordinates represent a point in space as \( (r, \theta, z) \), where:

• \(r\) is the radial distance from the origin in the xy-plane.
• \(\theta\) is the angle around the z-axis.
• \(z\) is the height above the xy-plane.

To convert Cartesian coordinates \((x, y, z)\) into cylindrical coordinates, you use:

• \( x = r\cos\theta \)
• \( y = r\sin\theta \)
• \( z = z \)

This conversion is especially helpful in integration, as it often simplifies double or triple integrals over regions with circular symmetry, such as hemispheres.

Symmetric Geometry

Symmetric geometry is an important concept in simplifying centroid calculations. The given hemisphere is symmetric across both x and y axes because its shape and density are uniform.
Due to this symmetry, the x and y coordinates of the centroid are zero:

• \( \bar{x} = 0 \)
• \( \bar{y} = 0 \)

This leaves finding only the z-coordinate, which is the vertical axis of symmetry in this shape. Working with symmetric objects greatly reduces the complexity of centroid computations because symmetry inherently balances out the solid's structure, concentrating the centroid along the axis of symmetry.

Integral Calculus

Integral calculus is the mathematical tool that allows us to find the centroid's z-coordinate in this problem.
We use it to sum up infinitesimally small volumes across the entire hemisphere. The z-coordinate of the centroid, \(\bar{z}\), is calculated using the formula:

• \[ \bar{z} = \frac{1}{V} \int_{V} z \, dV \]

This process involves setting up a triple integral in cylindrical coordinates. Taking advantage of the radial symmetry, the integration becomes more manageable.

• First, integrate over \(z\) from 0 to \(\sqrt{a^2 - r^2}\), giving an expression involving \(r\).
• Next, integrate this result over \(r\) from 0 to \(a\).
• Finally, integrate over the angular parameter \(\theta\) from 0 to \(2\pi\).

This step-by-step integration showcases how integral calculus handles continuous sums over considerable spatial domains, leading to precise results like the centroid \( (0, 0, \frac{3a}{8}) \).