Problem 34 - Thomas Calculus

Question

Find the mass of the solid region bounded by the parabolic surfaces \(z=16-2 x^{2}-2 y^{2}\) and \(z=2 x^{2}+2 y^{2}\) if the density of the solid is \(\delta(x, y, z)=\sqrt{x^{2}+y^{2}}\)

Step-by-Step Solution

Answer

The mass is \(\frac{256\pi}{15}\).

Step 1: Understand the bounding surfaces

Identify the bounding surfaces given by the equations \(z=16-2x^2-2y^2\) and \(z = 2x^2 + 2y^2\). These describe two parabolic surfaces. The region we are interested in is where these surfaces intersect each other.

Step 2: Find the region of integration

To find the region of integration, set the two surfaces equal: \[16-2x^2-2y^2 = 2x^2+2y^2\]Solving this gives \(16 = 4x^2 + 4y^2\), or \(x^2 + y^2 = 4\). This describes a circle of radius 2 in the xy-plane.

Step 3: Set up the integral for mass

The mass of the solid is computed by integrating the density over the volume of the solid bounded by the surfaces. Set up the triple integral:\[M = \int \int \int_{V} \delta(x, y, z) \, dz \, dx \, dy\]

Step 4: Change to cylindrical coordinates

Use cylindrical coordinates where \(x = r \cos \theta\), \(y = r \sin \theta\). The density becomes \(\delta(r) = r\). The bounds for \(r\) are from 0 to 2, and \(\theta\) from 0 to \(2\pi\). The zz in cylindrical coordinates range from \(z = 2r^2\) to \(z = 16 - 2r^2\). The mass becomes:\[M = \int_0^{2\pi} \int_0^2 \int_{2r^2}^{16-2r^2} r \, dz \, r \, dr \, d\theta\]

Step 5: Evaluate the integral

Evaluate the integrals starting with the innermost integral with respect to \(z\):\[M = \int_0^{2\pi} \int_0^2 \left[ r z \right]_{2r^2}^{16-2r^2} \, r \, dr \, d\theta\]Simplify:\[M = \int_0^{2\pi} \int_0^2 r \left( (16-2r^2) - (2r^2) \right) \, r \, dr \, d\theta\]\[M = \int_0^{2\pi} \int_0^2 r (16 - 4r^2) \, r \, dr \, d\theta\]\[M = \int_0^{2\pi} \int_0^2 (16r^2 - 4r^4) \, dr \, d\theta\]

Step 6: Integrate with respect to r

Integrate with respect to \(r\):\[M = \int_0^{2\pi} \left( \left[ \frac{16}{3} r^3 - \frac{4}{5} r^5 \right]_0^2 \right) \, d\theta\]\[M = \int_0^{2\pi} \left( \frac{16}{3} (8) - \frac{4}{5} (32) \right) \, d\theta\]\[M = \int_0^{2\pi} \left( \frac{128}{3} - \frac{128}{5} \right) \, d\theta\]\[M = \int_0^{2\pi} \left( \frac{320}{15} - \frac{192}{15} \right) \, d\theta\]\[M = \int_0^{2\pi} \frac{128}{15} \, d\theta\]

Step 7: Integrate with respect to \(\theta\)

Complete the integration with respect to \(\theta\):\[M = \left[ \frac{128}{15} \theta \right]_0^{2\pi}\]\[M = \frac{128}{15} (2\pi)\]\[M = \frac{256\pi}{15}\]

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates are an alternative coordinate system to the traditional Cartesian coordinates. They are especially useful for problems involving symmetry around an axis, like circular or cylindrical shapes. This system uses three components: radius \(r\), angle \(\theta\), and height \(z\).

• \(r\) is the distance from the \(z\)-axis to the projection of the point in the xy-plane.
• \(\theta\) is the angle from the positive \(x\)-axis to the projection of the point in the xy-plane.
• \(z\) is simply the height above the xy-plane.

When doing triple integrals, converting to cylindrical coordinates simplifies expressions, particularly when the region has rotational symmetry. To convert a point from Cartesian to cylindrical coordinates, use \(x = r \cos \theta\) and \(y = r \sin \theta\).
This is handy in our exercise since the boundary defined by \(x^2 + y^2 = 4\) (a circle) fits naturally in a cylindrical framework, enabling us to easily set up the integral bounds for \(r\) and \(\theta\).

Density Function

A density function \(\delta(x, y, z)\) describes how mass is distributed throughout a volume. In our exercise, the density function provided is \(\delta(x, y, z) = \sqrt{x^2 + y^2}\), which essentially represents the distance of the point \((x, y)\) from the origin in the xy-plane.

• This means that the density increases as you move away from the center of the base of the region.
• Density functions can vary depending on spatial variables, making them non-uniform, as is the case here.

When computing the mass of a solid region with a given density, integrate the density function over the volume of the solid. In this case, cylindrical coordinates make the transition smoother: \(\delta(r, \theta, z) = r\), as \(\sqrt{x^2 + y^2} = r\). Thus, the density directly influences the mass integral as shown in our calculations.

Parabolic Surfaces

Parabolic surfaces are 3D regions shaped like parabolas. They can be open upwards, downwards, or sideways, depending on the equation. Here, the solid's boundaries are defined by two paraboloids:

• The upper surface: \(z = 16 - 2x^2 - 2y^2\) is a downward-opening paraboloid.
• The lower surface: \(z = 2x^2 + 2y^2\) is an upward-opening paraboloid.

These equations tell us that the vertical cross-sections of these surfaces are parabolas. By setting them equal \(16 - 2x^2 - 2y^2 = 2x^2 + 2y^2\), we find the boundary defining the base of the region in the xy-plane: a circle \(x^2 + y^2 = 4\). This circle is the projection of the intersection of the two surfaces, crucial for setting the integration limits.

Volume of Solid

To find the volume of a solid region bounded by surfaces, we use triple integrals. These integrals evaluate the extent of a volume by layering infinitesimally thin slices of the region and summing them up.

• The bounds for integration are determined by where the given surfaces intersect.
• The region in our exercise is between two paraboloids, so setup involves calculating the difference between their \(z\)-coordinates.

In our exercise, the integral starts from the lower surface \(z = 2r^2\) and goes up to the upper surface \(z = 16 - 2r^2\). With cylindrical coordinates and incorporating a given density function, we arrive at a volume calculation integral. Solving this triple integral provides the mass, linking the volume to the physical mass of the solid when combined with the density. The process outlined ensures precise computation accounting for geometric and density complexities.