What is the reason for the double integral ∫∫ydσ=0 in polar coordinates?

Clever proof of double integral ∬ydσ=0 under polar coordinates and common error analysis

This article analyzes a question about double integrals under polar coordinates and explains why the integral result is zero and common errors in the calculation process. In the question, the integral region is the heart-shaped region described by the polar coordinate equation r = ½ (⅓)sinθ, and the integrated function is f(x,y) = y.

Quick solution using symmetry:

The key is to observe the symmetry of the integrated function and the integral region. The integraved function y is an odd function about the y-axis, i.e. f(x,-y) = -f(x,y). Meanwhile, the heart-shaped region is symmetric about the x-axis. This means that for any point (x,y) in the area, the point (x,-y) is also in the area. Therefore, the integral values ​​on the upper and lower parts of the x-axis are equal in size and opposite in the symbols, and the final results are canceled out with each other, and the double integral result is zero:

∬ _σ y dσ = 0

More stringent mathematical expression:

∬ σ f(x,y)dxdy = ∫dx ∫ y 0 -y 0 f(x,y)dy = 0 (because ∫ y 0 -y 0 f(x,y)dy = 0)

Common errors and correct solutions:

Many students try to use polar coordinate transformation to calculate directly, but errors are prone to occur. The correct polar coordinate integration steps are as follows:

∬ _σ y dσ = ∫ ₀ ^2π ∫ ₀ ^{½ (⅓)sinθ} (r sinθ) * r dr dθ = ∫ ₀ ^2π ∫ ₀ ^{½ (⅓)sinθ} r² sinθ dr dθ

An error usually occurs after the integral of r, and the integral of θ is calculated. Some students will mistakenly calculate ∫ ₀ ^2π (½ (⅓)sinθ) dθ, ignore the integral of the constant term ½, or mistakenly believe that ∫ ₀ ^2π sinθ dθ is not equal to zero.

Correct calculation steps:

1. Inner layer integral (for r integral):

∫ ₀ ^{½ (⅓)sinθ} r² sinθ dr = [⅓r³ sinθ] ₀ ^{½ (⅓)sinθ} = ⅓(½ (⅓)sinθ)³ sinθ

2. Outer integration (integration for θ):

∫ ₀ ^2π ⅓(½ (⅓)sinθ)³ sinθ dθ

Although this integral seems complicated, since the integral result of sinθ on [0, 2π] is 0, and (½ (⅓)sinθ)³ is an odd function about sinθ, the final result is still 0.

Summary: Using symmetry, we can quickly determine that the result of the double integral is zero, avoiding complex calculations. If it is necessary to calculate through polar integral, the inner and outer layer integrals need to be carefully calculated to avoid common calculation errors. Remember, symmetry analysis is a powerful tool to solve the integral problem.

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