Question

Evaluate \(\iint_{S} \sin \left(x y^{2}\right) d A\), where \(S\) is the annulus \(\left\\{(x, y): 1 \leq x^{2}+y^{2} \leq 4\right\\} .\) Hint: Done without thinking, this problem is hard; using symmetry, it is trivial.

Short answer

The integral evaluates to zero due to symmetry.

Step-by-step solution

Identify the Region

The region of integration, \(S\), is an annulus defined by \(1 \leq x^2 + y^2 \leq 4\). This is a ring-shaped area with an outer radius of 2 and an inner radius of 1.

Consider Symmetry of the Function

The function to be integrated is \(\sin(xy^2)\). Notice that for every point \((x, y)\), there is a symmetric point \((-x, y)\), such that \(\sin(xy^2) = -\sin(-xy^2)\). This indicates that the integral over a symmetric region centered around the origin will be zero.

Use Polar Coordinates

To further confirm, we could convert to polar coordinates. In polar coordinates, \(x = r\cos\theta\) and \(y = r\sin\theta\), so \(x^2 + y^2 = r^2\) and the annular region is converted to \(1 \leq r \leq 2\), with \(\theta\) ranging from \(0\) to \(2\pi\). Thus, \(dA = r \, dr \, d\theta\).

Evaluate the Integral

Using symmetry, we conclude that the integral of \(\sin(r^3\cos\theta\sin^2\theta)\) over the entire annular region must be zero, as the positive contributions from one half of the annulus cancel out with the negative contributions from the other half.

Key concepts

Polar Coordinates

When dealing with double integrals over circular or ring-shaped regions, switching to polar coordinates is incredibly beneficial.
In this case, our region of interest is an annulus, which simplifies our calculations when using polar coordinates.
In polar coordinates:

• The coordinate transformation is given by: \(x = r \cos \theta\) and \(y = r \sin \theta\).
• The expression \(x^2 + y^2 = r^2\) means the distance from the origin is represented by \(r\), making it straightforward to describe circular regions.

This transformation effectively changes a circular region in Cartesian coordinates into a rectangle in polar coordinates.
Consequently, calculating areas becomes much simpler, as integration can be performed wrt \(r\) and \(\theta\) individually. For the region in question, \(dA = r \, dr \, d\theta\) because the area element in polar coordinates incorporates the radial component, \(r\), into the differential area.
Thus, polar coordinates streamline the integration process significantly when working with circular regions.

Symmetry in Integration

Symmetry is a powerful tool in integration, as it often allows us to anticipate the behavior of an integral and simplify calculations.
In our exercise, we consider the symmetry of the integrand, which is \(\sin(xy^2)\).
Notice the function's property: for any point \((x, y)\), there's a corresponding symmetric point \((-x, y)\) such that \(\sin(xy^2) = -\sin(-xy^2)\).

• This points to the idea that the function exhibits odd symmetry.
• Integrating a function with this type of symmetry over a symmetric region around the origin results in a zero value.

Here's why: the contributions from the function value at each point and its symmetrically opposite point cancel each other out.
As the annular region is symmetric about the origin, the integral of our function over this region equals zero.
Often, identifying and utilizing symmetry can make a seemingly complex integration problem much simpler.

Annulus Region

An annulus is a ring-shaped region, defined by two concentric circles with different radii.
In our problem, the annulus is defined by \(1 \leq x^2 + y^2 \leq 4\), indicating the space between circles with radii 1 and 2.
This geometric structure is commonly encountered in exercises involving double integrals.

• In polar coordinates, such a region translates to \(1 \leq r \leq 2\), streamlining the representation.
• The angle \(\theta\) varies from \(0\) to \(2\pi\), covering the entire circle.

Visualizing an annulus involves imagining a doughnut-like shape, where the outer boundary is the larger circle and the inner boundary is the smaller circle within it.
Working within this ring-shaped region in polar coordinates efficiently allows for handling the symmetry and simplifying integration tasks.
As such, the concept allows the integral to be simplified significantly when combined with the symmetry property of the function, as seen previously.