Problem 90 - Thomas Calculus

Question

Centroid Find the centroid of the region in the first octant that is bounded above by the cone \(z=\sqrt{x^{2}+y^{2}}\) , below by the plane \(z=0,\) and on the sides by the cylinder \(x^{2}+y^{2}=4\) and the planes \(x=0\) and \(y=0\)

Step-by-Step Solution

Answer

The centroid is at \((\frac{4}{3}, \frac{4}{3}, \frac{2}{3})\)."

Step 1: Describe the Region and Set Bounds

The region is in the first octant, meaning all coordinates \((x, y, z)\) are positive. It is bounded by the cone \(z = \sqrt{x^2 + y^2}\) and the cylinder \(x^2 + y^2 = 4\). The boundaries \(x = 0\) and \(y = 0\) further restrict the region to the first quadrant of the \(xy\)-plane.

Step 2: Set Up the Integration for the Volume

To find the centroid, we first need the volume of the region. In cylindrical coordinates, where \(x = r \cos \theta\), \(y = r \sin \theta\), and \(z = z\), the region is defined by \(0 \leq r \leq 2\), \(0 \leq \theta \leq \frac{\pi}{2}\), and \(0 \leq z \leq r\). The differential volume element is \(dV = r \, dz \, dr \, d\theta\). The volume is given by: \[ V = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} \int_{0}^{r} r \, dz \, dr \, d\theta \]

Step 3: Calculate the Volume

Integrate with respect to \(z\) first: \[ \int_{0}^{r} r \, dz = r \cdot z \bigg|_0^r = r^2 \] Next, integrate with respect to \(r\): \[ \int_{0}^{2} r^2 \, dr = \frac{r^3}{3} \bigg|_0^2 = \frac{8}{3} \] Finally, integrate with respect to \(\theta\): \[ \int_{0}^{\frac{\pi}{2}} \frac{8}{3} \, d\theta = \frac{8}{3} \cdot \frac{\pi}{2} = \frac{4\pi}{3} \] Thus, the volume \(V\) of the region is \(\frac{4\pi}{3}\).

Step 4: Set Up and Calculate Integrals for Centroid Coordinates

The centroid \((\overline{x}, \overline{y}, \overline{z})\) is given by:\[ \overline{x} = \frac{1}{V} \int \int \int x \, dV \]\[ \overline{y} = \frac{1}{V} \int \int \int y \, dV \]\[ \overline{z} = \frac{1}{V} \int \int \int z \, dV \]We will use cylindrical coordinates:- For \(\overline{x}\), replace \(x = r \cos \theta\):\[ \overline{x} = \frac{1}{\frac{4\pi}{3}} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} \int_{0}^{r} (r \cos \theta) r \, dz \, dr \, d\theta \]- For \(\overline{y}\), replace \(y = r \sin \theta\):\[ \overline{y} = \frac{1}{\frac{4\pi}{3}} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} \int_{0}^{r} (r \sin \theta) r \, dz \, dr \, d\theta \]- For \(\overline{z}\):\[ \overline{z} = \frac{1}{\frac{4\pi}{3}} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} \int_{0}^{r} z \, r \, dz \, dr \, d\theta \]

Step 5: Evaluate Integrals for Each Coordinate

For \(\overline{x}\):\[ \overline{x} = \frac{1}{\frac{4\pi}{3}} \cdot \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r (r \cos \theta) \cdot r \, dr \, d\theta = \frac{3}{4\pi} \cdot \left( \frac{16\pi}{9} \right) \cos \theta = \frac{4}{3} \]For \(\overline{y}\), similarly,\[ \overline{y} = \frac{4}{3} \]For \(\overline{z}\):\[ \overline{z} = \frac{1}{\frac{4\pi}{3}} \cdot \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} \int_{0}^{r} z r \, dz \, dr \, d\theta = \frac{2}{3} \]Thus, the coordinates of the centroid are \((\frac{4}{3}, \frac{4}{3}, \frac{2}{3})\).

Key Concepts

Cylindrical Coordinates

In mathematical problems that involve three dimensions, cylindrical coordinates can be incredibly useful. They are particularly helpful when working with symmetric shapes like cylinders and cones. Instead of using the usual Cartesian coordinates \((x, y, z)\), cylindrical coordinates express location using the parameters \(r, \theta, z\):

• \(r\) is the radial distance from the origin to the point's projection in the \(xy\)-plane.
• \(\theta\) is the angle between the positive \(x\)-axis and the line from the origin to the point's projection in the \(xy\)-plane.
• \(z\) is the same as in Cartesian coordinates, indicating height.

So a point \((x, y, z)\) in Cartesian coordinates converts to cylindrical coordinates as \((r \cos \theta, r \sin \theta, z)\). This transformation is often simpler when boundaries, like cones or cylinders, are involved.

Volume Integration

Volume integration allows us to calculate the volume of a three-dimensional region, especially when boundaries can be complex. In this exercise, we used cylindrical coordinates to set up the integral. Cylindrical coordinates simplify the process because they naturally align with the shapes involved.

To find the volume of the given region:

• The limits for \(r\) are from 0 to 2, indicating the radial boundary of the cylinder.
• The limits for \(\theta\) are from 0 to \(\frac{\pi}{2}\), which is limited by the first quadrant.
• The limits for \(z\) are from 0 to \(r\), confined by the cone \(z = \sqrt{x^2 + y^2}\).

The differential volume element in cylindrical coordinates is \(dV = r \, dz \, dr \, d\theta\). By integrating this over the specified limits, we calculate the region's volume, which is essential in determining the centroid.

Triple Integrals

Triple integrals are used to evaluate a three-dimensional region by integrating a function over the entire space, combining the integration over all three variables. In the original problem, we used triple integrals to calculate the volume as well as the coordinates of the centroid.

Here's a stepwise breakdown:

• The integration over \(z\) is done first for each cylindrical slice. Since \(z\) is bounded by 0 and \(r\), it involves substituting the limits in the integral.
• Next, the integration over \(r\) proceeds from 0 to 2, which gives us the radial integration of the volume.
• Finally, the integration over \(\theta\) goes from 0 to \(\frac{\pi}{2}\), completing the triple integral.

By following these steps, triple integrals help to piece together the contribution of infinitesimally small elements into a comprehensive result.

Centroid Coordinates

Finding the centroid of a given region requires knowing how to compute its average location. The centroid \((\overline{x}, \overline{y}, \overline{z})\) indicates the balance point of the shape, where it could be balanced on a pin, so to speak, without tipping over.

To find these coordinates, calculate:

• \(\overline{x}\), which involves integrating \(x \cdot dV\) (in cylindrical, this is \(r \cos \theta \cdot dV\)) over the region and then averaging by the volume.
• \(\overline{y}\), integrating \(y \cdot dV\) (\(r \sin \theta \cdot dV\)) in a similar fashion.
• \(\overline{z}\), simply integrating \(z \cdot dV\) regardless of geometry specifics.

Each coordinate integral provides a measure of how its respective variable extends across the region, and collectively, they pinpoint the centroid location.