Question

As needed, use a computer to plot graphs and to check values of integrals. Show that the \(z\) coordinate of the centroid of the volume inside the elliptic cone $$\frac{z^{2}}{h^{2}}=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}, \quad 0

Short answer

The coordinate \( \bar{z} \) of the centroid is \( \frac{3}{4} h \)

Step-by-step solution

- Coordinates Transformation

Perform the change of variables as suggested: let \( z = hz' \), \( x = ax' \), and \( y = by' \). The elliptic cone equation becomes \( \frac{(hz')^2}{h^2} = \frac{(ax')^2}{a^2} + \frac{(by')^2}{b^2} \). Simplifying, we get \( z'^2 = x'^2 + y'^2 \), which is a circular cone.

- Changing to Circular Coordinates

Now change to cylindrical coordinates in the primed system: \( x' = r' \cos( \theta ) \), \( y' = r' \sin( \theta ) \), and \( z' = z' \). The volume element in cylindrical coordinates is \( dV' = r' \, dr' \, d \theta \, dz' \).

- Integrate to Find the Volume

The limits for \( r' \) will be from 0 to \( z' \), \( \theta \) from 0 to \( 2 \pi \), and \( z' \) from 0 to 1. Therefore, the volume integral is \( V' = \int_{0}^{1} \int_{0}^{2 \pi} \int_{0}^{z'} r' \ dr' \, d \theta \, dz' \). The volume of the cone in these normalized coordinates is \( V' = \frac{\pi}{3} \).

- Setting the Integral for \( \bar{z} \)

Now, find the integral for the \( \bar{z} \) coordinate: \( \bar{z} = \frac{1}{V'} \int_V z \, dV \). Substituting \( z = h z' \) and the volume element \( dV = a b h r' \, dr' \, d \theta \, dz' \), we write the integral as \( \bar{z} = \frac{1}{V} \int_{0}^{h} \int_{0}^{2 \pi} \int_{0}^{\frac{z}{h}} z r a b h \, dr \, d \theta \, dz \), where \( V \) is the actual volume of the cone.

- Transforming and Simplifying the Integral

The integral \( \bar{z} = \frac{3}{V} \int_{0}^{h} \int_{0}^{2 \pi} \int_{0}^{z} a b r^2 z \, dr \theta \, dz \) simplifies by integrating over \( r \). Simplifying further, we get \( \bar{z} = \frac{3}{4} h \).

Key concepts

Triple Integrals

Triple integrals are used to calculate volumes and other quantities in three-dimensional space. They extend the concept of integrating functions of one variable or double integrals (in two dimensions) to three dimensions.
Using triple integrals, you can integrate a function over a volume, which is particularly useful in physics and engineering.
The general form is: \( \text{Integral:} \ \ \ \bigintss_V f(x,y,z) \,dx \,dy \,dz \), where \( V \) is the volume over which you are integrating.
Calculations involve nested integrations, often starting with the innermost integral and progressing outward.
To solve, define the integral bounds clearly and work step-by-step.

Coordinate Transformation

Coordinate transformations are pivotal in simplifying the computation of integrals. This involves changing the variables in which a function is expressed, making the region of integration easier to describe.
In this exercise, we transformed the original coordinates using: \ \ \( z = hz' \ x = ax' \ y = by' \ \).
This transformation converts the elliptic cone into a circular cone, greatly simplifying integration.
Proper transformation ensures that integral bounds are simpler and calculations more manageable.

Cylindrical Coordinates

Cylindrical coordinates simplify the description of regions with rotational symmetry around the \( z \)-axis. Use the following to convert Cartesian coordinates: \ \ \( x = r \ \cos( \theta ) \ \)
\( y = r \ \sin( \theta ) \ \)
\( z = z \).
The volume element in cylindrical coordinates is expressed as \( dV = r \, dr \, d\theta \, dz \).
For our transformed cone, we use these coordinates to rewrite the integral for volume: \( \bigintss_{0}^{1} \bigintss_{0}^{2\pi} \bigintss_{0}^{z'} r' \, dr' \, d\theta \, dz' \).
This helps organizing and solving easily.

Volume of a Cone

The volume of a cone is derived by integrating over its height and base area. For a circular cone, the volume formula is \( V = \frac{1}{3} \pi r^2 h \), where \( r \) is the radius and \( h \) is the height.
For the elliptic cone, normalization simplifies the volume calculation to an equation independent of \( a \) and \( b \): \ \ \( V' = \frac{\pi}{3} \).
To find the centroid \( \bar{z} \), integrate \( z \) over the volume, using \( x', y', z' \) coordinates: \ \ \( \bar{z} = \frac{3}{V} \bigintss_{0}^{h} \bigintss_{0}^{2\pi} \bigintss_{0}^{z} ab \,r^2 z \, dr \, d\theta \, dz. \).
Resulting in \( \bar{z} = \frac{3}{4}h \).