Question

A triple integral in spherical coordinates is given. Describe the region in space defined by the bounds of the integral. $$ \int_{0}^{2 \pi} \int_{0}^{\pi / 6} \int_{0}^{a \sec \varphi} \rho^{2} \sin (\varphi) d \rho d \theta d \varphi $$

Short answer

The region is a truncated spherical sector with a cone of 30 degrees and upper bound at the plane \(z = a\).

Step-by-step solution

Understand the Spherical Coordinates

Spherical coordinates are defined as \[\begin{align*}x &= \rho \sin(\varphi) \cos(\theta), \y &= \rho \sin(\varphi) \sin(\theta), \z &= \rho \cos(\varphi).\end{align*}\]Here, \(\rho\) is the radial distance from the origin, \(\varphi\) is the angle from the positive z-axis, and \(\theta\) is the azimuthal angle in the xy-plane.

Analyze the Boundaries

The integral is given as \[\int_{0}^{2 \pi} \int_{0}^{\pi / 6} \int_{0}^{a \sec \varphi} \rho^{2} \sin(\varphi) d \rho d \theta d \varphi\].- The \(\theta\) bounds, from \(0\) to \(2\pi\), cover a full rotation around the z-axis.- The \(\varphi\) bounds, from \(0\) to \(\pi/6\), limit the angle starting from the positive z-axis to 30 degrees down.- The \(\rho\) bounds, from \(0\) to \(a \sec \varphi\), extend up to \(a \csc(\varphi)\), which describes a plane since \(z = \rho \cos(\varphi) = a\).

Interpret the Region

The region described is a truncated spherical sector. It extends from the origin in every direction around the z-axis (full sweep in \(\theta\)), extends from the positive z-axis to a 30-degree cone, and is limited by the plane \(z = a\). Beyond a height \(z = a\), the region is cut off by this plane.

Key concepts

Spherical Coordinates

Spherical coordinates are a three-dimensional coordinate system that can be very handy in describing points in a spherical manner. Unlike Cartesian coordinates which use

• a flat xy-plane and z-axis, the spherical system uses radial and angular measures.
• The three coordinates are:
  • \( \rho \) (radial distance) - the distance from the origin to the point.
  • \( \varphi \) (polar angle, or colatitude) - the angle between the point and the positive z-axis.
  • \( \theta \) (azimuthal angle) - the angle from the positive x-axis in the xy-plane.

They offer a great way to handle problems involving spheres or circular regions. Using spherical coordinates in integrals allows for better alignment with symmetries of the problem, often simplifying calculations considerably.

Radial Distance

The radial distance \( \rho \) is one of the primary coordinates in spherical systems. It signifies how far a point is from the origin. Imagine a sphere with radius \( \rho \); the larger the \( \rho \), the larger the sphere. In the given integral, the limits for \( \rho \) extend from 0 to \( a \sec(\varphi) \). This means that points are confined within a sphere that stretches outward to a boundary determined by the angle \( \varphi \).
Its importance lies in capturing the volumetric element of a region we wish to describe or calculate, thus playing a crucial role in triple integrals when calculating volumes in spherical coordinates.

Azimuthal Angle

The azimuthal angle \( \theta \) is akin to the way a compass might describe directions in a flat plane, but here it revolves around the origin. It specifies how far you have rotated from the positive x-axis within the xy-plane. In spherical coordinates, \( \theta \) provides rotational symmetry. For this integral, \( \theta \) ranges from 0 to \( 2\pi \), encompassing a full 360-degree sweep around the z-axis.
By allowing this full rotation, we consider every potential azimuthal direction, thereby exploring the entirety of a circular section in our spatial region. Understanding \( \theta \) helps in visualizing and calculating properties of regions that are invariant under such rotations.

Truncated Spherical Sector

A truncated spherical sector is a fascinating geometric region. It can be visualized as a sector of a sphere that is 'cut off' or truncated by a plane. In this context, the plane is described by \( z = a \), thereby limiting the height of the sector. The boundaries of our integral reflect this truncation:
- **\( \theta \)** takes us in a full circle around the z-axis;- **\( \varphi \)** starts from the top of the z-axis and swings down to a 30-degree angle (\( \pi/6 \) radians);- **\( \rho \)** stretches outward up to the plane \( z = a \). This gives us a portion of a spherical sector but halts it at \( z = a \). These constraints form a wedge-like shape when cut through the center with a vertical plane. Notably, this concept simplifies spatial analysis, especially in physics and engineering problems that involve symmetrical regions.