Problem 54 - Advanced Engineering Mathematics

Question

Use triple integrals and cylindrical coordinates. Find the volume of the solid that is bounded by the graphs of the given equations. $$ y=x^{2}+z^{2}, \quad 2 y=x^{2}+z^{2}+4 $$

Step-by-Step Solution

Answer

The volume of the solid is \(4\pi\).

Step 1: Understand the Solid

We are given two quadratic equations in three dimensions. The first equation, \(y = x^2 + z^2\), represents a paraboloid opening upwards with its vertex at the origin. The second equation, \(2y = x^2 + z^2 + 4\), can be rewritten as \(y = \frac{x^2 + z^2 + 4}{2}\), which is another paraboloid, but shifted upwards and expanded. We want to find the volume of the solid between these two surfaces.

Step 2: Convert to Cylindrical Coordinates

In cylindrical coordinates, \(x = r\cos\theta\), \(z = r\sin\theta\), and \(x^2 + z^2 = r^2\). Convert both equations as follows: \(y = r^2\) and \(y = \frac{r^2 + 4}{2}\). This simplifies the problem in polar coordinates.

Step 3: Determine Limits of Integration

The first paraboloid is \(y = r^2\) and the second is \(y = \frac{r^2 + 4}{2}\). Solving \(r^2 = \frac{r^2 + 4}{2}\) gives the intersection of the surfaces when \(r = 2\). Therefore, \(0 \leq r \leq 2\), \(0 \leq \theta < 2\pi\), and \(r^2 \leq y \leq \frac{r^2 + 4}{2}\).

Step 4: Set Up the Integral

The volume integral in cylindrical coordinates is given by \(V = \int_{0}^{2\pi} \int_{0}^{2} \int_{r^2}^{\frac{r^2 + 4}{2}} r \, dy \, dr \, d\theta\). The factor of \(r\) accounts for the Jacobian determinant from Cartesian to cylindrical coordinates.

Step 5: Integrate with Respect to y

First, integrate with respect to \(y\):\[ \int_{r^2}^{\frac{r^2 + 4}{2}} r \, dy = r \left[ y \right]_{r^2}^{\frac{r^2 + 4}{2}} = r \left( \frac{r^2 + 4}{2} - r^2 \right) = r \left( \frac{-r^2 + 4}{2} \right). \]

Step 6: Simplify Expression

So, the integral simplifies to:\[ \frac{r(4 - r^2)}{2} = 2r - \frac{r^3}{2}. \]

Step 7: Integrate with Respect to r

Now integrate \[ \int_0^2 \left( 2r - \frac{r^3}{2} \right) \, dr = \left[ r^2 - \frac{r^4}{8} \right]_{0}^{2} = (4 - 2) - (0 - 0) = 2. \]

Step 8: Integrate with Respect to θ

Finally, integrate with respect to \(\theta\):\[ \int_0^{2\pi} 2 \, d\theta = 2\theta \bigg|_0^{2\pi} = 4\pi. \]

Step 9: Calculate the Volume

Multiply the results of the integrations to get the total volume. The solution to the triple integral gives us the volume of the solid:\[ V = 4. \]

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional extension of the two-dimensional polar coordinates. This system is helpful when dealing with problems having symmetry around a vertical axis (often the z-axis). Instead of using Cartesian coordinates \(x, y, z\), cylindrical coordinates are expressed as \(r, \theta, z\), where:
\begin{itemize} \item \(r\) is the radial distance from the origin to the point's projection in the \(xy\)-plane. \item \(\theta\) is the angle between the positive x-axis and the line connecting the origin with the point's projection in the \(xy\)-plane. \item \(z\) remains the vertical distance (or height) from the \(xy\)-plane. \end{itemize}
Using cylindrical coordinates simplifies the integration process, especially for problems with symmetrical shapes around an axis. In our exercise, converting the equations to cylindrical coordinates expressed the relationships as \(y = r^2\) and \(y = \frac{r^2+4}{2}\), allowing for straightforward limits of integration: \(0 \leq r \leq 2\), \(0 \leq \theta < 2\pi\), and appropriate bounds for \(y\). This change makes evaluating triple integrals more efficient by aligning with both surface and volume symmetries.

Paraboloids

Paraboloids are a type of quadratic surface that can be visualized as a three-dimensional parabola. They take the form of either elliptic or hyperbolic, but in this context, elliptic paraboloids are considered. A typical elliptic paraboloid, for example, would have the formula \(z = x^2 + y^2\).
The exercise involves two paraboloids:

• \(y = x^2 + z^2\): This represents a paraboloid opening in the positive \(y\)-direction, centered at the origin.
• \(2y = x^2 + z^2 + 4\): By rearranging, it becomes \(y = \frac{x^2 + z^2 + 4}{2}\), which is another paraboloid placed higher above the first.

These surfaces intersect, creating a symmetrical solid between them. Understanding the geometric nature of paraboloids is essential for setting up these kinds of integration problems. The key is identifying that these surfaces form when a quadratic relationship is present in the equations.

Volume Calculation

Volume calculation using triple integrals requires determining the proper limits of integration and setting up the integral expression correctly. The integration is conducted over three variables, generally denoted as \(x\), \(y\), and \(z\) or their cylindrical counterparts \(r\), \(\theta\), and \(y\).
For the given problem, the volume between two paraboloids was calculated by establishing:

• Cylindrical coordinates provided radial symmetry to simplify the integration process.
• The integral is set up as: \ \int_{0}^{2\pi} \int_{0}^{2} \int_{r^2}^{\frac{r^2 + 4}{2}} r \, dy \, dr \, d\theta \.

The limits specify the domain of the solid object, while the factor \(r\) in the integral accounts for the conversion from Cartesian to cylindrical coordinates. Calculating in steps involved:

• Integrating with respect to \(y\).
• Simplifying and integrating with respect to \(r\).
• Finally, integrating with respect to \(\theta\).

The steps led to a total volume of \(4\pi\), encapsulating the region between the paraboloids.