Question

Find the following average values. The average of the squared distance between the origin and points in the solid paraboloid \(D=\left\\{(x, y, z): 0 \leq z \leq 4-x^{2}-y^{2}\right\\}\)

Short answer

The average squared distance between the origin and points in the solid paraboloid D is 2/3.

Step-by-step solution

Compute the volume of the region D

We can calculate the volume of the paraboloid using a triple integral in polar coordinates: \(\displaystyle V = \iiint\limits_D dV = \int\limits_0^{2\pi}\int\limits_0^{\sqrt{4-z}}\int\limits_0^{4-r^2} r\,dz\,dr\,d\theta\) Now, we can solve the innermost integral with respect to \(z\): \(\displaystyle V = \int\limits_0^{2\pi}\int\limits_0^{\sqrt{4-z}}(4-r^2)r\,dr\,d\theta\) Now we'll integrate with respect to \(r\). Let \(u=r^2\), then \(du = 2r\,dr\). We'll have to update the integral bounds: when \(r=0\), \(u=0\), and when \(r=\sqrt{4-z}\), \(u=4-z\). \(\displaystyle V = \frac{1}{2}\int\limits_0^{2\pi}\int\limits_0^{4-z}(4-u) du\,d\theta\) Now integrate with respect to \(u\): \(\displaystyle V = \frac{1}{2}\int\limits_0^{2\pi}(4(4-z)-(4-z)^2)\,d\theta\) Now integrate with respect to \(\theta\): \(\displaystyle V = \pi(16z-2z^2)\Big|_0^4=\pi(64-32)=32\pi\)

Compute the integral of the squared distance

Now let's calculate the integral for the squared distance. The squared distance from the origin for a point \((x, y, z)\) is given by \(x^2 + y^2 + z^2\). In polar coordinates, since \(x^2 + y^2 = r^2\), this becomes \(r^2 + z^2\). We need to calculate \(\displaystyle I = \iiint\limits_D (r^2+z^2)dV = \int\limits_0^{2\pi}\int\limits_0^{\sqrt{4-z}}\int\limits_0^{4-r^2}(r^2+z^2)r\,dz\,dr\,d\theta\) We'll integrate with respect to \(z\) first: \(\displaystyle I = \int\limits_0^{2\pi}\int\limits_0^{\sqrt{4-z}}\frac{1}{3}(4-r^2)^2(r^2+z^2)r\,dr\,d\theta\) Next, we'll integrate with respect to \(r\), using the substitution \(u = r^2\), \(du = 2r\,dr\): \(\displaystyle I = \frac{1}{6}\int\limits_0^{2\pi}\int\limits_0^{4-z}(3u^2+2uz-z^2) du\,d\theta\) Now integrate with respect to \(u\): \(\displaystyle I = \frac{1}{6}\int\limits_0^{2\pi}\left(u^3+uz^2-\frac{1}{3}z^3\right)\Big|_0^{4-z}d\theta\) Now we can integrate with respect to \(\theta\): \(\displaystyle I = \frac{\pi}{6}(192-64z-24z^2+4z^3)\Big|_0^4=\frac{\pi}{6}(768-640)=\frac{\pi}{6}(128)\)

Compute the average squared distance

Now that we have both the integral of the squared distance and the volume of the region, we can compute the average squared distance by dividing the former by the latter: \(\displaystyle \text{Average squared distance} = \frac{I}{V} = \frac{\left(\frac{\pi}{6}(128)\right)}{32\pi} = \frac{128}{192} = \frac{2}{3}\) Thus, the average squared distance between the origin and points in the solid paraboloid \(D\) is \(\frac{2}{3}\).

Key concepts

Triple Integrals

Triple integrals provide a way to integrate over three-dimensional regions. We use them to calculate volumes, mass, or other quantities over a 3D space. When using triple integrals, we represent the volume or quantity of interest as \[ \iiint_R f(x,y,z) \, dV, \]where \( f(x,y,z) \) is a function of the three variables and \( dV \) is the differential volume element, often given as \( dx \, dy \, dz \) in Cartesian coordinates. One key application is finding the volume of a region by setting \( f(x,y,z) = 1 \) and computing \[ \iiint_R 1 \, dV. \] Triple integrals can be evaluated using different coordinate systems for convenience, such as Cartesian, cylindrical, or spherical coordinates. The choice depends on the symmetry and boundaries of the region. In our exercise, we employ polar coordinates for ease in dealing with circular regions.

Polar Coordinates

Polar coordinates are especially useful when dealing with regions having circular symmetry or boundaries. Particularly in two dimensions, polar coordinates \((r, \theta)\) translate Cartesian coordinates \((x, y)\) using the relationships \( x = r \cos \theta \) and \( y = r \sin \theta \).When extending to three dimensions, polar coordinates become part of cylindrical or spherical coordinates. For cylindrical coordinates, we work with \((r, \theta, z)\) where \( z \) remains as the vertical axis. For spherical coordinates, we typically use \((\rho, \theta, \varphi)\), but in our exercise for a paraboloid, cylindrical coordinates are appropriate.Integrals in polar forms use differentials \( r \, dr \, d\theta \) for calculating area or volume, adjusting naturally to the geometric nature of circular or radial regions. In scenarios like our paraboloid problem, this adaptation enables a simpler computation by aligning with the inherent symmetry of the region involved.

Volume of Paraboloid

The volume of a paraboloid can be found using triple integrals. In our case, the solid paraboloid defined by \( D = \left\{(x, y, z): 0 \leq z \leq 4-x^{2}-y^{2}\right\} \) is bounded vertically by the plane \( z = 0 \) and curved surface \( z = 4 - x^2 - y^2 \). By converting the spatial boundaries into polar coordinates, the integration over \( z \) is performed first, within the curve's height at each radius \( r \). The bounds for integration with respect to \( r \) are from 0 to \( \sqrt{4-z} \), and \( \theta \) spans from 0 to \( 2\pi \).We compute the integral \[ V = \int\limits_0^{2\pi} \int\limits_0^{\sqrt{4 - z}} \int\limits_0^{4 - r^2} r \, dz \, dr \, d\theta, \] solving first with respect to \( z \), then \( r \), and finally \( \theta \). The resulting volume provides insight into the extent of the paraboloid.

Average Value in Calculus

The average value of a function over a region helps describe a representative value of the function across that space. In our context, the objective is to find the average of the squared distance from the origin, \( r^2 + z^2 \), within a solid paraboloid \[ D = \left\{(x, y, z): 0 \leq z \leq 4-x^{2}-y^{2}\right\}. \]To find this, we need two things: the total volume of the region and the integral of the squared distance.The average is found by dividing the total value of the function over the region by the volume:\[ \text{Average} = \frac{1}{V} \iiint\limits_D (r^2+z^2) \, dV. \]The numerator, \[ \iiint\limits_D (r^2+z^2) \, dV, \]represents the accumulation of all squared distances from the origin to every point in \( D \), while \( V \) is the total volume calculated as part of the exercise.Ultimately, the resulting average of \( \frac{2}{3} \) gives us insight into the typical distance of points from the origin throughout this spatial structure.