Problem 54 - Thomas Calculus

Question

Let \(D\) be the region bounded below by the cone \(z=\sqrt{x^{2}+y^{2}}\) and above by the plane \(z=1 .\) Set up the triple integrals in spherical coordinates that give the volume of \(D\) using the following orders of integration. $$ a.d \rho d \phi d \theta \quad \text { b. } d \phi d \rho d \theta $$

Step-by-Step Solution

Answer

Triple integrals are set as \(d\rho\ d\phi\ d\theta\) and \(d\phi\ d\rho\ d\theta\).

Step 1: Understanding the Boundaries

The given solid region is bounded below by the cone described by the equation \(z = \sqrt{x^2 + y^2}\) and above by the plane \(z = 1\). In spherical coordinates, the conversion for these variables are: \(x = \rho \sin \phi \cos \theta, y = \rho \sin \phi \sin \theta, z = \rho \cos \phi\).

Step 2: Identify Spherical Coordinate Limits

The plane \(z=1\) corresponds to \(\rho \cos \phi = 1\). Since we know that \(z = \sqrt{x^2 + y^2}\) corresponds to \(\phi = \frac{\pi}{4}\) for the cone, we have the bounds: \(\phi\) is from \(0\) to \(\frac{\pi}{4}\), \(\theta\) is from \(0\) to \(2\pi\), and \(\rho\) from \(0\) to \(\frac{1}{\cos \phi}\).

Step 3: Set up Integral for Order d\(\rho\) d\(\phi\) d\(\theta\)

Given the order \(d\rho\ d\phi\ d\theta\), the triple integral for the volume is set up as: \[ \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{1}{\cos \phi}} \rho^2 \sin \phi\ d\rho\ d\phi\ d\theta \] where \(\rho^2 \sin \phi\) is the Jacobian for converting from Cartesian to spherical coordinates.

Step 4: Set up Integral for Order d\(\phi\) d\(\rho\) d\(\theta\)

For the order \(d\phi\ d\rho\ d\theta\), the limits remain the same: \(\phi\) from \(0\) to \(\frac{\pi}{4}\), \(\rho\) from \(0\) to \(\frac{1}{\cos \phi}\), and \(\theta\) from \(0\) to \(2\pi\). So, the integral is: \[ \int_{0}^{2\pi}\int_{0}^{\frac{1}{\cos \phi}}\int_{0}^{\frac{\pi}{4}} \rho^2 \sin \phi\ d\phi\ d\rho\ d\theta \].

Key Concepts

Triple Integrals

Triple integrals are used to find the volume under a surface in three-dimensional space. Imagine this as layering flat sheets over each other to fill an entire volume.
We perform integration thrice—once in each of the three spatial dimensions:

• The first integral captures the innermost dimension (such as depth).
• The second integral adds up layers.
• The third brings the volume together across the widest dimension.

In our exercise, we're interested in how these integrals apply to spherical coordinates! This is super helpful for solving problems relating to spheres, cones, and cylinders. By knowing how to properly set up the integral using given bounds, you can find exactly how much space a region occupies.
The integral uses the Jacobian determinant of spherical coordinates to account for the conversion, leading us into precisely locating and summing volumes.

Volume of Region

Finding the volume of a region bound by surfaces can be tricky without visualization. Think of the region as slices of fruit in a bowl—each slice layers onto another to complete a form.
Within our problem, we have a cone below and a plane on top. By finding specific boundaries in each direction, we can carefully layer spherical slices.

• The plane at the top caps the region at a height of 1.
• A cone underneath gives shape going outwards from a point at the origin.

Flipping between integral orders can help solve these kinds of problems more efficiently! It's often a strategic choice that depends on the unique shape of the bounds.

Spherical Coordinate Conversion

Converting from Cartesian to spherical coordinates allows us to incorporate symmetry, which makes integration simpler for some geometries. Accessing a coordinate system based on spheres makes especially good sense for spheres and cones.
In spherical coordinates:

• \( x = \rho \sin \phi \cos \theta \) maps a point’s distance horizontally.
• \( y = \rho \sin \phi \sin \theta \) does so vertically.
• \( z = \rho \cos \phi \) measures upwards, towards zenith.

By switching into these equations during triple integration, it takes advantage of the natural properties of spherical shapes. Choosing spherical coordinates here lets us handle the cone and plane shapes with far more straightforward terms. This reduces complex geometry back into neat limits, like slicing a cone in perfect circles!

Jacobian Determinant

The Jacobian determinant is a crucial factor that accounts for how much volume scales during transformation between coordinate systems. It provides the extra weights needed when converting from Cartesian coordinates into spherical, cylindrical, or any format.
When you recalculate a volume change through a transformation:

• It adjusts for non-linear stretching—or squishing—of your space.
• Calculating it right supports accurate integration as you shift systems.

In the spherical sense, the Jacobian for coordinates like ours gives us \( \rho^2 \sin \phi \). It's a product of adjusting to both radial and angular spread. The term \( \rho^2 \) scales outward in a radial sweep, while \( \sin \phi \) captures compression closer to the poles of a sphere.
This weight and shape adjustment is why using Jacobians in transformation integrations is indispensable!