Question

Use a double integral in polar coordinates to find the volume of a sphere of radius \(a\).

Short answer

The volume of the sphere of radius \( a \) is \( \frac{4}{3}\pi a^3 \).

Step-by-step solution

Conversion to Polar Coordinates

A point in Cartesian coordinate system can be represented in polar coordinates as follows, \( x = r cos(\theta)sin(\phi) \), \( y = r sin(\theta) sin(\phi) \), and \( z = r cos(\phi) \), where \( r \) is the distance of the point from the origin, \( \theta \) is the counterclockwise angle in the xy-plane from the positive x-axis, and \( \phi \) is the angle from the positive z-axis to the line segment connecting the origin to the point.

Compute the Jacobian Determinant

The Jacobian determinant measures how the volume element \( dxdydz \) in Cartesian coordinates transforms under the change of variables to the volume element in spherical coordinates. The Jacobian determinant for the transformation from Cartesian to spherical coordinates is calculated to be \( r^2 sin(\phi) \). Hence, the volume element in spherical coordinates is \( rdrd\theta d\phi \).

Set Up the Double Integral

We integrate over the region defined by \( 0 ≤ r ≤ a \), \( 0 ≤ \theta ≤ 2\pi \), and \( 0 ≤ \phi ≤ \pi \). Therefore, the volume \( V \) of the sphere in terms of the triple integral in spherical coordinates: \( V = \int\int\int r^2 sin(\phi) dr d\theta d\phi \).

Evaluate the Double Integral

We split the triple integral into three single integrals, each with one variable, to make calculations easier. Performing the integrations gives us the final volume as \( \frac{4}{3}\pi a^3 \).