Converting Integrals to Cylindrical and Spherical Coordinates: A Practical Example

In this article, we delve into the process of converting and evaluating a three-dimensional integral using both cylindrical and spherical coordinate systems. Specifically, we will consider an integral defined over a particular region in the first quadrant, bounded by the xy-plane and a given surface. Let's explore this process step-by-step.

Introduction to the Problem

The region of integration is defined by the surface (z a^2 - x^2 - y^2) and the constraint (x^2 y^2 leq a^2) in the first quadrant. The goal is to convert this integral into cylindrical and spherical coordinates, and then evaluate it.

Step 1: Conversion to Cylindrical Coordinates

To convert the given integral into cylindrical coordinates, we use the following transformations:

(x r cos theta), (, y r sin theta), and (, z z)

Using these transformations, the region of integration changes as follows:

The constraint (x^2 y^2 leq a^2) transforms into (r leq a). The equation of the surface (z a^2 - x^2 - y^2) becomes (z a^2 - r^2).

The integral in the given region in Cartesian coordinates is:

(int_0^a int_0^{sqrt{a^2 - x^2}} int_0^{a^2 - x^2 - y^2} x^2 dz dy dx)

After converting to cylindrical coordinates, it becomes:

(int_0^{frac{pi}{2}} int_0^a int_0^{a^2 - r^2} r^3 cos^2 theta dz dr dtheta)

Step 2: Evaluating the Integral

We begin by evaluating the innermost integral:

(int_0^{a^2 - r^2} r^3 cos^2 theta dz r^3 cos^2 theta left[ z right]_0^{a^2 - r^2} r^3 cos^2 theta (a^2 - r^2))

Substituting this back into the integral, we get:

(int_0^{frac{pi}{2}} int_0^a r^3 cos^2 theta (a^2 - r^2) dr dtheta)

We can split this into two separate integrals:

(int_0^{frac{pi}{2}} int_0^a (a^2 r^3 cos^2 theta - r^5 cos^2 theta) dr dtheta)

We now evaluate each integral separately. First, evaluate the integral with respect to (r):

(int_0^a (a^2 r^3 - r^5) dr a^2 left[ frac{r^4}{4} right]_0^a - left[ frac{r^6}{6} right]_0^a a^2 cdot frac{a^4}{4} - frac{a^6}{6} frac{a^6}{4} - frac{a^6}{6} frac{a^6}{12})

Substitute this result back into the integral with respect to (theta):

(int_0^{frac{pi}{2}} frac{1}{2} (a^6 cos^2 theta - a^6 cos^4 theta) dtheta)

Using the identity (cos^2 theta frac{1 cos 2theta}{2}) and (cos^4 theta left( frac{1 cos 2theta}{2} right)^2 frac{1 2cos 2theta cos^2 2theta}{4}), we have:

(int_0^{frac{pi}{2}} frac{1}{2} (a^6 cdot frac{1 cos 2theta}{2} - a^6 cdot frac{1 2cos 2theta cos^2 2theta}{4}) dtheta)

( int_0^{frac{pi}{2}} frac{a^6}{4} (cos 2theta - cos^2 2theta) dtheta)

( int_0^{frac{pi}{2}} frac{a^6}{4} (cos 2theta - frac{1 cos 4theta}{2}) dtheta)

( int_0^{frac{pi}{2}} frac{a^6}{4} (cos 2theta - frac{1}{2} - frac{cos 4theta}{2}) dtheta)

( int_0^{frac{pi}{2}} frac{a^6}{4} (cos 2theta - frac{1}{2}) dtheta - int_0^{frac{pi}{2}} frac{a^6}{8} cos 4theta dtheta)

( frac{a^6}{4} left[ frac{1}{2} sin 2theta - frac{1}{2}theta right]_0^{frac{pi}{2}} - frac{a^6}{8} left[ frac{1}{4} sin 4theta right]_0^{frac{pi}{2}})

Conclusion

The result of the integration is:

(I frac{a^6 pi}{48})

This process demonstrates the use of coordinate transformations to simplify complex integrals and highlights the importance of using appropriate coordinate systems for different problems.

Keywords: cylindrical coordinates, spherical coordinates, integral evaluation