Question

Find the following average values. The average of the squared distance between the \(z\) -axis and points in the conical solid \(D=\\{(x, y, z): 2 \sqrt{x^{2}+y^{2}} \leq z \leq 8\\}\)

Short answer

Answer: The average squared distance between the z-axis and points in the conical solid D is \(\frac{3}{\pi}\).

Step-by-step solution

Compute the Volume of the Conical Solid D

In polar coordinates, the cone is given by \(0 \leq r \leq \frac{z}{2}\) and \(0 \leq z \leq 8\). To get the volume, we will integrate the area of the cross section of the cone with respect to the height \(z\). $$Volume = \int_0^8 A(z) dz$$ In polar coordinates, the area \(A(z)\) at height \(z\) can be expressed as the area of a disc with radius \(r=\frac{z}{2}\): $$A(z) = \pi r^2 = \pi \left(\frac{z}{2}\right)^2 = \frac{z^2\pi}{4}$$ Now we can compute the volume by integrating with respect to \(z\). $$Volume = \int_0^8 \frac{z^2\pi}{4} dz = \frac{\pi}{4} \int_0^8 z^2 dz = \frac{\pi}{4}\left[\frac{z^3}{3}\right]_0^8 = \frac{\pi}{4}\cdot\frac{512}{3}$$

Compute the Integral of the Squared Distance

In terms of polar coordinates, the squared distance between a point \((x, y, z)\) and the \(z\)-axis is given by \(x^2 + y^2 = r^2\). To find the integral of this quantity over the conical solid \(D\), we should integrate it with respect to \(r\) and \(z\) using the given limits. $$ S = \int_0^8 \int_0^{\frac{z}{2}} r^2 rdrdz $$ Now we can compute the integral of the squared distance. $$ S = \int_0^8 \left[\frac{r^4}{4}\right]_0^{\frac{z}{2}} dz = \int_0^8 \frac{z^4}{64}dz = \left[\frac{z^5}{320}\right]_0^8 = \frac{8192}{320} $$

Compute the Average Squared Distance

To obtain the average squared distance, we need to divide the integral of the squared distance by the volume of the conical solid \(D\). $$ Average\; Squared\; Distance = \frac{S}{Volume} =\frac{\frac{8192}{320}}{\frac{\pi}{4}\cdot\frac{512}{3}} = \frac{8192}{\frac{512\pi}{3}} = \frac{3}{\pi} $$ Therefore, the average squared distance between the z-axis and points in the conical solid \(D\) is \(\frac{3}{\pi}\).

Key concepts

Polar Coordinates

Understanding polar coordinates is key when solving problems involving symmetrical shapes like a conical solid. In polar coordinates, each point in a plane is determined by a distance from a reference point and an angle from a reference direction. The reference point is usually taken as the origin of the plane where the axis of symmetry, like the z-axis, often becomes more intuitive to deal with.

• The reference point is called the pole.
• The length from the pole is known as the radial coordinate, denoted as \(r\).
• The angle from a reference direction is the angular coordinate, usually measured in radians.

Since conical shapes have a circular symmetry about the z-axis, using polar coordinates simplifies many calculations, including determining the areas and volumes. For the conical solid in this exercise, the transformation between polar and Cartesian coordinates is given by \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). Hence, the conical surface can be expressed in terms of \(r\) and \(z\) through these relationships in polar coordinates.

Average Value

The concept of finding average values can apply to functions or quantities over geometric regions. When you need to find the average value of a function over a certain region, it's essential to sum up the total values across that region (using an integral) and divide by the region's measure (like its volume or area).

• First, calculate the integral of the function over the region, which gives you the total value.
• Second, determine the measure of the region, like volume or area, for normalization.
• The average value is then the total value divided by the measure of the region.

In our exercise, we calculated the average squared distance of points in a conical solid from the z-axis. We integrated over the conical solid and divided by its volume to obtain an average squared distance of \(\frac{3}{\pi}\). This step is crucial for understanding how quantities distribute over a defined shape, offering insights especially when dealing with asymmetrical shapes.

Integral

Integrals are foundational tools in calculus used to calculate areas, volumes, and averages over continuous shapes and functions. The idea of integration is to sum up infinitesimal values across a range—whether it's a line, area, or volume.

• The definite integral calculates the accumulated total over a specified interval.
• For three-dimensional objects, multiple integrals can be used, like double or triple integrals, to account for all dimensions.
• In polar coordinates, conversion of bounds between \(r\), \(\theta\), and \(z\) may be necessary for accurate integration.

In our exercise, we first found the volume of the conical solid by integrating the cross-sectional area along its height. We then calculated the integral of the squared distance \(r^2\) over the entire conical domain. This demonstrated the integral's power in simplifying complex shape calculations down to more manageable formulas, showcasing how integrals allow for precise calculations in physical and theoretical contexts.