Introduction

When dealing with mathematical integrals, particularly two-dimensional ones, it is often beneficial to convert them into a more manageable form, especially when Cartesian coordinates (x, y) might complicate the calculation. This article will guide you through the process of transforming a two-dimensional integral into polar coordinates, emphasizing a specific example for clarity. We will demonstrate this technique step-by-step, ensuring that every detail and transformation is explained thoroughly.

Understanding the Integral

Consider the integral

[int_{0}^{1} int_{0}^{1} f(x, y), dx, dy]

The limits provided in this example are invalid since they specify a single limit along one axis, which is not sufficient for a two-dimensional integral. Hence, we need to reconsider the problem with an appropriate setup. For the sake of this article, let's assume we are working with a specific function, like an integrand that simplifies the problem in polar coordinates.

Transformation to Polar Coordinates

For a standard two-dimensional integral, the transformation to polar coordinates involves the following:

Polar coordinates: [x r cos theta, y r sin theta] Area element: [dA dx , dy r , dr , dtheta]

However, our specific example will focus on a simpler transformation, where we assume a constant function that simplifies the integral, or a function that results in a straightforward form in polar coordinates.

Example Problem

Given a specific function, let's consider the integral in polar coordinates:

[I int_{0}^{1} int_{0}^{1} g(x, y), dx, dy]

For this example, let's use a function where the transformation simplifies the integral. Let's assume the function simplifies to:

[g(x, y) x]

Now, transforming to polar coordinates:

[x r cos theta, quad y r sin theta, quad dA r , dr , dtheta]

Substitute (x) and (dA):

[I int_{0}^{pi/2} int_{0}^{1} r cos theta cdot r , dr , dtheta int_{0}^{pi/2} cos theta , dtheta int_{0}^{1} r^2 , dr]

Evaluate the inner integral:

[int_{0}^{1} r^2 , dr left[frac{r^3}{3}right]_{0}^{1} frac{1}{3}]

Evaluate the outer integral:

[int_{0}^{pi/2} cos theta , dtheta left[sin thetaright]_{0}^{pi/2} 1]

Combine the results:

[I 1 cdot frac{1}{3} frac{1}{3}]

Transformation and Simplification

Alternatively, let's consider a more complex transformation, such as:

[g(x, y) r sqrt{cos^2 theta - sin^2 theta} r sqrt{cos 2theta}]

Substitute this into the integral:

[I int_{0}^{pi/2} int_{0}^{1} r sqrt{cos 2theta} cdot r , dr , dtheta int_{0}^{pi/2} sqrt{cos 2theta} , dtheta int_{0}^{1} r^2 , dr]

Evaluate the inner integral as before:

[int_{0}^{1} r^2 , dr frac{1}{3}]

Evaluate the outer integral:

[int_{0}^{pi/2} sqrt{cos 2theta} , dtheta]

This integral can be solved using integration techniques, but for simplicity, let's denote the result as (A):

[I A cdot frac{1}{3}]

Therefore, the final simplified result is:

[I frac{A}{3}]

Conclusion

Evaluating a two-dimensional integral using polar coordinates can significantly simplify problems, especially when dealing with circular or radial symmetry. Through careful substitution and integration, the original problem can be transformed and solved more efficiently. This technique is not only useful for mathematical rigor but also for practical applications in fields such as physics, engineering, and data science.

For a more accurate and specific answer, one would need to calculate the exact value of the integral (A) in the second example. However, the general approach remains the same: transform, simplify, and integrate.

Keywords: polar coordinates, two-dimensional integral, transformation technique