Question

Criticise the following 'proof' that \(\pi=0\) (a) Apply Green's theorem in a plane to the functions \(P(x, y)=\tan ^{-1}(y / x)\) and \(Q(x, y)=\tan ^{-1}(x / y)\), taking the region \(R\) to be the unit circle centred on the origin. (b) The RHS of the equality so produced is $$ \iint_{R} \frac{y-x}{x^{2}+y^{2}} d x d y $$ which, either by symmetry considerations or by changing to plane polar coordinates, can be shown to have zero value. (c) In the LHS of the equality set \(x=\cos \theta\) and \(y=\sin \theta\), yielding \(P(\theta)=\theta\) and \(Q(\theta)=\pi / 2-\theta .\) The line integral becomes $$ \int_{0}^{2} \pi\left[\left(\frac{\pi}{2}-\theta\right) \cos \theta-\theta \sin \theta\right] d \theta $$ which has value \(2 \pi\) (d) Thus \(2 \pi=0\) and the stated result follows.

Short answer

Errors in Green's theorem use and integral results mean the 'proof' is incorrect, \( \pi e 0 \).

Step-by-step solution

Introduction

Review the given 'proof' and identify each part. The goal is to find errors in logic or calculation.

Green's Theorem Application

Green's theorem states that for a region R bounded by a positively oriented simple closed curve C, \[\begin{equation} \int_{C} Pdx + Qdy = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx dy\end{equation}\]Given functions: P(x, y) = \tan^{-1}(y/x) and Q(x, y) = \tan^{-1}(x/y). Applying Green's theorem requires computing the partial derivatives.

Calculate Partial Derivatives

First, calculate \( \frac{\partial P}{\partial y}\) and \( \frac{\partial Q}{\partial x} \).For \( P(x, y) = \tan^{-1}(y/x) \), since \(\frac{d}{dx}\tan^{-1}(x) = \frac{1}{1+x^2} \), use the chain rule:\[ \frac{\partial P}{\partial y} = \frac{1}{1+(y/x)^2} \cdot \left(\frac{1}{x}\right) = \frac{x}{x^2+y^2} \]For \( Q(x, y) = \tan^{-1}(x/y)\), use a similar approach:\[ \frac{\partial Q}{\partial x} = \frac{1}{1+(x/y)^2} \cdot \left(\frac{1}{y}\right) = \frac{y}{x^2+y^2} \]

Substitute into Green's Theorem

Insert the computed partial derivatives into Green's theorem:\[ \int_{C} \tan^{-1}(y/x) dx + \tan^{-1}(x/y) dy = \iint_{R} \left( \frac{y}{x^2+y^2} - \frac{x}{x^2+y^2} \right) dx dy \]This simplifies to:\[ \iint_{R} \frac{y-x}{x^2+y^2} dx dy \]

Evaluate the Integral

Symmetry or polar coordinates can show that the double integral evaluates to zero. Under polar coordinates:\[ x = r \cos\theta, \, y = r \sin\theta \]So the integral becomes:\[ \iint_{R} \frac{r\sin\theta - r\cos\theta}{r^2} \cdot r \, dr \, d\theta = \iint_{R} (\sin\theta - \cos\theta) \, dr \, d\theta = 0 \] due to symmetry over the interval from \(0 \) to \( 2\pi \).

Check the Line Integral

For the line integral, substitute polar forms of x and y:\[ P(x, y) = \theta, \quad Q(x, y) = \frac{\pi}{2} - \theta \]Insert into the line integral:\[ \int_{0}^{2\pi} \left(\left(\frac{\pi}{2} - \theta\right) \cos \theta - \theta \sin \theta\right) d\theta \]

Evaluate the Line Integral and Error Identification

Evaluate the line integral and note discrepancies. Substituting and solving gives a value:\[ \int_{0}^{2\pi} 0 \, d\theta = 0 \]However, verifying calculation details evidently shows it would not provide \(2\pi \), discrediting the step followed. This confirms the conflict and error with the provided 'proof'.

Conclusion

The supposed 'proof' applying Green's theorem contains miscalculations and concept errors. Thus, concluding \( \pi = 0 \) is invalid.

Key concepts

Line integral

Line integrals are a type of integral where the function to be integrated is evaluated along a curve. In the context of Green's theorem, line integrals are used to relate the circulation of a vector field around a closed curve to the double integral over the region it encloses. Essentially, it links the behavior of a field along a boundary to its behavior inside the boundary.

For a curve \( C \) defined by parameter \( t \) from \( a \) to \( b \), the line integral of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} \) along \( C \) is given by:
\[ \int_{C} Pdx + Qdy = \int_{a}^{b} \left( P \frac{dx}{dt} + Q \frac{dy}{dt} \right) dt \]
In this exercise, we're dealing with the line integral over the unit circle, which simplifies our parameters and computations. By setting \( x = \cos \theta \) and \( y = \sin \theta \), we evaluate the line integral with respect to \( \theta \) from 0 to \( 2\pi \). This conversion often simplifies integrals, especially when the region has circular symmetry, like a unit circle.

Partial derivatives

Partial derivatives are derivatives of multivariable functions taken with respect to one variable while keeping others constant. They are essential in Green's theorem because the theorem involves the difference of partial derivatives.

Let's break down their use within Green's theorem:

• For a function \( P(x, y) \), the partial derivative \( \frac{\partial P}{\partial y} \) measures how \( P \) changes as \( y \) changes while holding \( x \) constant.
• Similarly, for a function \( Q(x, y) \), \( \frac{\partial Q}{\partial x} \) measures how \( Q \) changes as \( x \) changes while holding \( y \) constant.

In this exercise:

• For \( P(x, y) = \tan^{-1}(y/x) \), we have:
  \[ \frac{\partial P}{\partial y} = \frac{1}{1 + (y/x)^2} \cdot \left(\frac{1}{x}\right) = \frac{x}{x^2 + y^2} \]
• For \( Q(x, y) = \tan^{-1}(x/y) \), we have:
  \[ \frac{\partial Q}{\partial x} = \frac{1}{1 + (x/y)^2} \cdot \left(\frac{1}{y}\right) = \frac{y}{x^2 + y^2} \]

These partial derivatives are then plugged into Green's theorem formula to connect the line integral around the boundary with the double integral over the region.

Integration in polar coordinates

Polar coordinates are a two-dimensional coordinate system where each point is determined by a distance from a reference point (radius \( r \)) and an angle from a reference direction (angle \( \theta \)).

Using polar coordinates to solve integrals can greatly simplify the problem when dealing with circular or radial symmetry.
In polar coordinates:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)

For the given integral over the unit circle, converting to polar coordinates is advantageous because the limits of integration become more straightforward. The original integrals:\[ \iint_{R} \frac{y - x}{x^2 + y^2} \ dx \ dy \]
transform into:
\[ \iint_{R} \frac{r \sin \theta - r \cos \theta}{r^2} \ r \ dr \ d \theta \]
This simplifies to:\[ \iint_{R} ( \sin \theta - \cos \theta ) \ dr \ d \theta \]
The integrals often evaluate to simpler forms that can be easily computed, leveraging the symmetry of the problem. This shows that the integral over the unit circle evaluates to zero due to the symmetrical nature of sine and cosine functions around the circle.