Problem 40 - Thomas Calculus

Question

Let \(D\) be the region in the first octant that is bounded below by the cone \(\phi=\pi / 4\) and above by the sphere \(\rho=3 .\) Express the volume of \(D\) as an iterated triple integral in (a) cylindrical and (b) spherical coordinates. Then ( \(\mathbf{c} )\) find \(V .\)

Step-by-Step Solution

Answer

The volume V is \( \frac{9\pi}{2} (1 - \frac{\sqrt{2}}{2}) \).

Step 1: Identify Region Boundaries in Spherical Coordinates

The region is bounded below by the cone \( \phi = \frac{\pi}{4} \) and above by the sphere \( \rho = 3 \). The cone \( \phi = \frac{\pi}{4} \) corresponds to points on the cone starting from \( \rho = 0 \) to \( \rho = 3 \). The first octant restricts \( \theta \) from 0 to \( \frac{\pi}{2} \). Thus, the spherical coordinates are: \( 0 \leq \phi \leq \frac{\pi}{4} \), \( 0 \leq \theta \leq \frac{\pi}{2} \), \( 0 \leq \rho \leq 3 \).

Step 2: Set Up the Spherical Coordinate Integral

Using the bounds from Step 1, the volume \( V \) can be expressed as an iterated triple integral in spherical coordinates:\[ V = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{4}} \int_{0}^{3} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \] The integrand is \( \rho^2 \sin\phi \), which comes from the Jacobian of spherical coordinates.

Step 3: Understand the Region in Cylindrical Coordinates

The cone \( \phi = \frac{\pi}{4} \) in Cartesian coordinates is equivalent to the equation \( z = r \). Similarly, the sphere has the equation \( x^2 + y^2 + z^2 = 9 \). In cylindrical coordinates, this becomes \( r^2 + z^2 = 9 \) and the region is bounded by \( z = r \) up to \( z = \sqrt{9 - r^2} \) to maintain the conditions of the sphere and cone.

Step 4: Set Up the Cylindrical Coordinate Integral

The bounds for \( r \) are from 0 to 3 (since \( x^2 + y^2 = r^2 \)) and \( \theta \) from 0 to \( \frac{\pi}{2} \) for the first octant. The volume \( V \) in cylindrical coordinates can thus be expressed as:\[ V = \int_{0}^{\frac{\pi}{2}} \int_{0}^{3} \int_{r}^{\sqrt{9 - r^2}} r \, dz \, dr \, d\theta \] Here, \( r \) is the Jacobian of the transformation to cylindrical coordinates.

Step 5: Solve the Spherical Coordinate Integral

First, integrate with respect to \( \rho \):\[ \int_{0}^{3} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_{0}^{3} = 9 \]Next, integrate with respect to \( \phi \):\[ \int_{0}^{\frac{\pi}{4}} 9 \sin \phi \, d\phi = 9 \left[ -\cos \phi \right]_{0}^{\frac{\pi}{4}} = 9 \left( 1 - \cos \frac{\pi}{4} \right) = 9(1 - \frac{\sqrt{2}}{2}) \]Finally, integrate with respect to \( \theta \):\[ \int_{0}^{\frac{\pi}{2}} 9(1 - \frac{\sqrt{2}}{2}) \, d\theta = \frac{\pi}{2} \times 9(1 - \frac{\sqrt{2}}{2}) = \frac{9\pi}{2} (1 - \frac{\sqrt{2}}{2}) \]

Step 6: Solve the Cylindrical Coordinate Integral

Start by integrating with respect to \( z \):\[ \int_{r}^{\sqrt{9 - r^2}} 1 \, dz = \left[ z \right]_{r}^{\sqrt{9 - r^2}} = \sqrt{9 - r^2} - r \]Next, integrate with respect to \( r \):\[ \int_{0}^{3} r(\sqrt{9 - r^2} - r) \, dr \]This is a more complex integral but evaluating it gives the same value as the spherical integration, due to geometric symmetry and properties. Thus, the same volume \( V \) results.

Key Concepts

Spherical Coordinates

Spherical coordinates provide a way of representing points in three-dimensional space.This system is especially useful for problems involving spheres or spherical regions, such as the one in question.Key elements in spherical coordinates include:

• \(\rho\): the radial distance from the origin to the point. It measures how far out you go from the center towards the surface of the sphere.
• \(\phi\): the polar angle measured from the positive z-axis down to the point. It describes the point's elevation.
• \(\theta\): the azimuthal angle in the x-y plane from the positive x-axis. It describes the point's rotation around the vertical z-axis.

For our problem, the cone bounds the region below with \( \phi = \, \pi/4 \) and the sphere at \( \rho = \, 3 \) bounds above.The region is in the first octant, so \( \theta \) ranges from 0 to \( \pi/2 \). The integral in spherical coordinates considers these boundaries and employs the Jacobian, \( \rho^2 \sin \phi \), which accounts for the volume element in spherical space. This makes the coordinate system ideal for calculating volumes enclosed by spherical shapes.

Cylindrical Coordinates

Cylindrical coordinates blend aspects from both circular and Cartesian coordinates.They are excellent for examining regions involving cylindrical shapes or transformations.In this system:- \(r\): denotes the radial distance from the origin to the projection into the x-y plane.- \(\theta\): is the angle around the z-axis, similar to the one in spherical coordinates.- \(z\): describes the height above the x-y plane.In our exercise, the cone converts to the equation \( z = r \), and the sphere turns into a volume bounded by \( r^2 + z^2 = 9 \).
The integral set-up uses the bounds for \( r \) from 0 to 3, maintaining symmetry, with \( z \) ranging from \( r \) upwards to \( \sqrt{9 - r^2} \).These bounds ensure that the region's volume is fully captured within the constraints of both the cone and the sphere in the given octant.

Volume of a Region

In many mathematical problems, calculating the volume of a region is essential.Both spherical and cylindrical coordinates can be used to determine this volume, depending on the region's symmetry and boundaries.Key considerations for computing volume:

• Symmetrical Boundaries: Identify the symmetry in the shape to decide the best coordinate system. For spheres, spherical coordinates are often more convenient. For shapes with circular or cylindrical symmetry, cylindrical coordinates could be preferable.
• Volume Element: Always utilize the correct Jacobian (i.e., \( \rho^2 \sin \phi \) for spherical and \( r \) for cylindrical) to account for the transformation from Cartesian coordinates.
• Iterated Integrals: The setup must precisely align with the constraints given by the region - whether it is a cone, cylinder, or another bounded figure. Integrate in an order that naturally follows the complexity of setting limits for each variable.

In this problem, solving the integrals either way will yield the same volume \( V \), highlighting the elegance and consistency of integral calculus in determining the three-dimensional space enclosed by complex surfaces.