Problem 25 - Thomas Calculus

Question

A parabolic rain gauge \(A\) bowl is in the shape of the graph of \(z=x^{2}+y^{2}\) from \(z=0\) to \(z=10\) in. You plan to calibrate the bowl to make it into a rain gauge. What height in the bowl would correspond to 1 in. of \(\operatorname{rain} ? 3\) in. of rain?

Step-by-Step Solution

Answer

For 1 inch of rain, the height is approximately 1.41 inches. For 3 inches, it is about 4.24 inches.

Step 1: Understand the Problem

First, we recognize that the bowl is in the shape of a paraboloid defined by the equation \( z = x^2 + y^2 \). We need to find the volume corresponding to 1 inch of rain and 3 inches of rain.

Step 2: Formula for Volume

The volume \( V \) of the paraboloid from \( z=0 \) to a height \( h \) is given by the triple integral over the region \( 0 \leq z \leq h \), where \( x^2 + y^2 = z \). The integral becomes \[ V = \int_{0}^{h} \int_{0}^{2\pi} \int_{0}^{\sqrt{z}} r \, dr \, d\theta \, dz. \]

Step 3: Integrate with Respect to \( r \)

After transforming to cylindrical coordinates, integrate with respect to \( r \):\[\int_{0}^{\sqrt{z}} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{\sqrt{z}} = \frac{z}{2}.\]

Step 4: Integrate with Respect to \( \theta \)

Substitute the result of the \( r \) integral into the remaining integrals:\[\int_{0}^{2\pi} d\theta = 2\pi.\]Multiply this by the previous step's result gives:\[\int_{0}^{2\pi} \frac{z}{2} \, d\theta = \pi z.\]

Step 5: Integrate with Respect to \( z \)

Finally, integrate with respect to \( z \):\[\int_{0}^{h} \pi z \, dz = \left[ \frac{\pi z^2}{2} \right]_{0}^{h} = \frac{\pi h^2}{2}.\]This expression gives the volume from \( z=0 \) to \( z=h \).

Step 6: Solve for 1 inch of Rain

We know 1 inch of rain should correspond to an equivalent volume. Set up the expression:\[\frac{\pi h_1^2}{2} = \pi (1)^2 \rightarrow \frac{\pi h_1^2}{2} = \pi \rightarrow h_1^2 = 2 \rightarrow h_1 = \sqrt{2} \approx 1.41.\]

Step 7: Solve for 3 inches of Rain

Similarly, for 3 inches of rain:\[\frac{\pi h_3^2}{2} = \pi (3)^2 \rightarrow \frac{\pi h_3^2}{2} = 9\pi \rightarrow h_3^2 = 18 \rightarrow h_3 = \sqrt{18} \approx 4.24.\]

Step 8: Interpret the Results

For 1 inch of rain, the depth in the bowl is about 1.41 inches, and for 3 inches of rain, it is about 4.24 inches when converted into the corresponding volume of water that fits the rain gauge's paraboloid shape.

Key Concepts

Understanding Paraboloids

A paraboloid is a three-dimensional shape that can be thought of as a circular parabola extended around an axis. Imagine taking the shape of a standard parabola, such as \( y = x^2 \), and rotating it around the y-axis. This kind of geometric figure is known as a paraboloid.
There are various types of paraboloids. Here, we are dealing with an elliptical paraboloid expressed by the equation \( z = x^2 + y^2 \). This surface opens upwards in the z-direction, forming a bowl-like shape.
In our rain gauge problem, the paraboloid runs from \( z=0 \) to \( z=10 \) inches, effectively shaping the bowl that collects rainwater. Understanding this form is crucial for visualizing the region dedicated to the integral calculations.

Volume Integral Calculation

The concept of finding volume via an integral, particularly a triple integral, is essential in calculus. When considering volume integrals, we accumulate an infinite number of infinitesimally small volume elements over a defined region.
For our paraboloid bowl, we're interested in the volume from \( z=0 \) to a certain height \( h \). This is computed using a triple integral. This involves integrating the infinitesimal pieces from a base up to the given height.
The triple integral \[ V = \int_{0}^{h} \int_{0}^{2\pi} \int_{0}^{\sqrt{z}} r \, dr \, d\theta \, dz \]integrates over \( r \) from 0 to \( \sqrt{z} \), over \( \theta \) from 0 to \( 2\pi \) (full circle), and then over \( z \) from 0 to \( h \), collectively covering the entire volume beneath the paraboloid cap.

Introduction to Cylindrical Coordinates

Cylindrical coordinates offer a smart way to deal with three-dimensional problems where rotational symmetry around an axis is evident.
Compared to Cartesian coordinates \((x, y, z)\), cylindrical coordinates \((r, \theta, z)\) use a radial distance \( r \), an angle \( \theta \), and the height \( z \) to determine a point in space.
This system greatly simplifies the description of volumes with circular symmetry. For our paraboloid, the equation \( z = x^2 + y^2 \) transforms smoothly into this system, making the integral more manageable. Radially extending circles in planes \( z = c \) translate into \( r = \sqrt{z} \) here, reducing the complexity of volume calculations.

Detailed Triple Integral Calculation

A triple integral calculation is the process of integrating a function over a three-dimensional region. It involves three nested integrals, each corresponding to the dimensions we're integrating over.
In our case, we begin with integrating \( r \) from 0 to \( \sqrt{z} \), represented as \[ \int_{0}^{\sqrt{z}} r \, dr = \frac{z}{2} \].
Next, we integrate over \( \theta \) from 0 to \( 2\pi \): \[ \int_{0}^{2\pi} \, d\theta = 2\pi \].
After multiplying these results, we tackle the integral with respect to \( z \), \[ \int_{0}^{h} \pi z \, dz = \frac{\pi h^{2}}{2} \].
This compounded integral helps us find the volume of a paraboloid cap, crucial to determining how much water corresponds to each inch of rain in our gauge.