Calculate The Double Integral Xy 2 X 2 1

This guide explains how to calculate the double integral of xy² over the region defined by x² + y² ≤ 1 (a unit circle). We'll cover the mathematical approach, provide a step-by-step calculator, and discuss practical applications.

What is a double integral?

A double integral extends the concept of single integration to two dimensions. It calculates the volume under a surface z = f(x,y) over a region R in the xy-plane. For this calculation, we're integrating xy² over the unit circle.

The double integral is fundamental in physics, engineering, and statistics for calculating quantities like mass, charge, or probability over a two-dimensional region.

How to calculate the double integral xy² x² + y² ≤ 1

The integral ∫∫ xy² dA over the region x² + y² ≤ 1 can be evaluated using polar coordinates, which simplifies the circular boundary.

Key formulas

1. Polar coordinate transformation: x = r cosθ, y = r sinθ

2. Area element: dA = r dr dθ

3. Integral becomes: ∫₀²π ∫₀¹ (r cosθ)(r sinθ)² r dr dθ

Step-by-step calculation

Convert to polar coordinates: xy² = r² cosθ sin²θ
Multiply by area element: r² cosθ sin²θ * r dr dθ = r³ cosθ sin²θ dr dθ
Integrate with respect to r: ∫₀¹ r³ dr = [r⁴/4]₀¹ = 1/4
Integrate with respect to θ: ∫₀²π cosθ sin²θ dθ
Use trigonometric identities to simplify the θ integral
Final result: The integral evaluates to 0

The result is zero because the integrand xy² is an odd function, and the region of integration is symmetric about the origin. The positive and negative contributions cancel each other out.

Example calculation

Let's verify the result with specific limits:

Step	Calculation	Result
1	Convert to polar coordinates	xy² = r³ cosθ sin²θ
2	Integrate r from 0 to 1	1/4
3	Integrate θ from 0 to 2π	0

The final result confirms that the double integral is indeed zero, demonstrating the symmetry of the integrand over the circular region.

FAQ

Why does this integral evaluate to zero?

The integrand xy² is an odd function, and the region of integration is symmetric about the origin. The positive and negative contributions cancel each other out.

Can I use Cartesian coordinates for this integral?

Yes, but polar coordinates simplify the calculation due to the circular boundary. The result remains the same in either coordinate system.

What if the region wasn't symmetric?

For asymmetric regions, the integral would not necessarily be zero. The symmetry of the unit circle and the odd function property are key to this result.