Question

By expressing both the integrand and the surface element in spherical polar coordinates, show that the surface integral $$ \int \frac{x^{2}}{x^{2}+y^{2}} d S $$ over the surface \(x^{2}+y^{2}=z^{2}, 0 \leq z \leq 1\), has the value \(\pi / \sqrt{2}\).

Short answer

The value of the surface integral is \( \frac{\pi}{\sqrt{2}} \).

Step-by-step solution

- Convert Cartesian Coordinates to Spherical Coordinates

Using spherical coordinates, express the Cartesian coordinates as follows: \(x = \rho \, \sin \theta \, \cos \phi\), \(y = \rho \, \sin \theta \, \sin \phi\), and \(z = \rho \, \cos \theta\).

- Identify the Surface Equation

The surface is given by \(x^2 + y^2 = z^2\). In spherical coordinates, this becomes \(\rho^2 \, \sin^2 \theta = \rho^2 \, \cos^2 \theta\). Thus, \( \tan^2 \theta = 1 \) or \( \theta = \frac{\pi}{4} \).

- Set Up the Integral in Spherical Coordinates

Given \( \theta = \frac{\pi}{4} \) on the surface, express the integrand and surface element accordingly: Integrand: \( \frac{x^2}{x^2 + y^2} = \frac{\rho^2 \, \sin^2 \theta \, \cos^2 \phi}{\rho^2 \, \sin^2 \theta} = \cos^2 \phi \). Surface element: \( dS = \rho^2 \, \sin \theta \, d\phi \, d\rho\).

- Evaluate the Surface Integral

Express the integral in terms of \( \phi \) and \( \rho \): \[ \int_0^{2\pi} \int_0^1 \cos^2 \phi \, \rho^2 \, \sin \ \frac{\pi}{4} \, d\rho \, d\phi \]. Since \( \sin \ \frac{\pi}{4} = \frac{\sqrt{2}}{2} \): \[ \int_0^{2\pi} \int_0^1 \cos^2 \phi \, \frac{\rho^2 \sqrt{2}}{2} \, d\rho \, d\phi \].

- Simplify and Integrate

Simplify the integral and integrate separately over \( \rho \) and \( \phi \): \[ \frac{\sqrt{2}}{2} \int_0^{2\pi} \cos^2 \phi \, d\phi \int_0^1 \rho^2 \, d\rho \]. Integrate \( \rho^2 \): \[ \int_0^1 \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_0^1 = \frac{1}{3} \]. Integrate \( \cos^2 \phi \): \( \int_0^{2\pi} \cos^2 \phi \, d\phi = \pi \). The final value is: \( \frac{\sqrt{2}}{2} \times \frac{\pi}{3} = \frac{\pi \sqrt{2}}{6} \).

Key concepts

spherical coordinates

Spherical coordinates are an alternative to Cartesian coordinates for describing points in three-dimensional space. They are particularly useful when dealing with problems that have symmetry about a point, such as spheres and cones. In spherical coordinates, a point is represented by three values:

• \rho (rho) - the radial distance from the origin to the point
• \theta (theta) - the polar angle, measured from the positive z-axis
• \theta (phi) - the azimuthal angle, measured from the positive x-axis in the xy-plane

The coordinates are converted by: \(x = \rho \, \sin \theta \, \cos \phi\), \(y = \rho \, \sin \theta \, \sin \phi\), and \(z = \rho \, \cos \theta\). These transformations help simplify the equations and integrals when dealing with three-dimensional objects.

surface integrals

Surface integrals extend the concept of integrals to functions over a surface in three-dimensional space. They are essential in physics and engineering to calculate quantities like flux and surface area. To evaluate a surface integral, you often need to:

• Describe the surface parametrically
• Determine the differential surface area element, \(dS\)
• Express the integrand in the chosen coordinates

For example, given a surface defined by \(x^2 + y^2 = z^2\), the surface integral of a function \(f(x, y, z)\) can be calculated by converting it into spherical coordinates and then integrating over the surface.

coordinate transformations

Coordinate transformations change the system of coordinates used to describe a point or a function. This can simplify equations and integrals in many cases. The most common transformations are:

• Cartesian to spherical coordinates
• Cartesian to cylindrical coordinates

For spherical coordinates, the transformations are:

• \( x = \rho \, \sin \theta \, \cos \phi \)
• \( y = \rho \, \sin \theta \, \sin \phi \)
• \( z = \rho \, \cos \theta \)

These transformations are very useful in simplifying the evaluation of integrals over curved surfaces, such as spheres and cones. Coordinate transformations are key in multivariable calculus because they adapt the integration process to better match the problem's geometry and symmetry properties.

multivariable calculus

Multivariable calculus extends the principles of calculus to functions of multiple variables. It involves:

• Partial derivatives
• Multiple integrals
• Vector calculus

Surface integrals are a critical component of multivariable calculus, allowing for the integration of functions over surfaces in three dimensions. To work effectively with these integrals, one often uses coordinate transformations to switch between Cartesian, cylindrical, and spherical coordinates.
For example, in the exercise, the integration over the surface \( x^2 + y^2 = z^2 \) was simplified using spherical coordinates. Understanding how to switch coordinate systems and work with differentials in these systems is essential for solving complex integrals in multivariable calculus.