Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(V=\iint_{S} \sin \sqrt{x^{2}+y^{2}} d A\) and \(W=\) \(\iint_{S}\left|\sin \sqrt{x^{2}+y^{2}}\right| d A\), where \(S\) is the region inside the circle \(x^{2}+y^{2}=4 \pi^{2} .\) (a) Without calculation, determine the sign of \(V\). (b) Evaluate \(V\). (c) Evaluate \(W\).

Step-by-Step Solution

Answer

(a) 0, (b) 0, (c) 8π.

Step 1: Understanding the Region

The region \(S\) is the interior of the circle \(x^2 + y^2 = 4\pi^2\). This means that \(x\) and \(y\) can take any values such that they lie within a circle of radius \(2\pi\) centered at the origin.

Step 2: Sign of V (Conceptual)

Since \(\sin\sqrt{x^2 + y^2}\) is integrated over a symmetric region around the origin, and \(\sin\) is positive and negative over this range, the positive and negative contributions will cancel out, therefore \(V\) will be zero.

Step 3: Polar Coordinates Transformation

Convert the double integral into polar coordinates. Here, \(x = r\cos\theta\) and \(y = r\sin\theta\), so the formula \(\iint_S \sin\sqrt{x^2 + y^2} \, dA\) becomes \(\int_0^{2\pi} \int_0^{2\pi} \sin(r) \, r \, dr \, d\theta\).

Step 4: Solving for V

The integral simplifies to \(\int_0^{2\pi} \left[\int_0^{2\pi} r\sin r \, dr\right] \, d\theta\). The inner integral evaluates to zero because \(\int_a^b f'(x)g(x)\, dx = (f(b)g(b) - f(a)g(a)) - \int_a^b f(x)g'(x) \, dx\), applying partial integration to \(\sin(r)\). Hence, \(V = 0\).

Step 5: Setting Up W

To evaluate \(W\), where the function is \(|\sin \sqrt{x^2 + y^2}|\), symmetry is no longer a factor to apply cancellation. Convert \(W\) into polar coordinates as before: \(\int_0^{2\pi} \int_0^{2\pi} |\sin(r)| \, r \, dr \, d\theta\).

Step 6: Evaluating W

The integral \(\int_0^{2\pi} \int_0^{2\pi} |\sin(r)| \, r \, dr \, d\theta\) results in \(2\int_0^{2\pi} \left[-\cos(r)\right]_0^{\pi} \, dr \cdot \int_0^{2\pi} d\theta\), giving \(W = (2[-(-1)] \cdot 2) \cdot 2\pi = 8\pi\).

Key Concepts

Polar Coordinates

In the context of double integrals, polar coordinates can prove to be a valuable tool for simplifying integration over circular regions. In our original exercise, the region of integration is a circle defined by the equation \(x^2 + y^2 = 4\pi^2\), which is perfectly suited for polar coordinates. Polar coordinates transform every point in the plane (\(x, y\)) into a radius \(r\) and an angle \(\theta\), so \(x = r\cos\theta\) and \(y = r\sin\theta\). This transformation makes it easier to handle circular regions because we work directly in terms of \(r\), the radius, and \(\theta\), the angle.
When converting a double integral to polar coordinates, the differential area element \(dA\) is represented as \(r \, dr \, d\theta\). This change accounts for the fact that each infinitesimal section of the region increases in size as you move further from the origin. For our case, the integrals for both \(V\) and \(W\) were transformed using these rules to simplify the work required when integrating over a circle.

Symmetry in Calculus

Symmetry can be a powerful ally in calculus, often allowing us to simplify complex integrals or predict results without full computation. In the original problem, symmetry played a crucial role in establishing the sign of \(V\). The integrand \(\sin\sqrt{x^2 + y^2}\) changes its sign over certain symmetric regions relative to the origin, much like the sine function oscillates between positive and negative values.
The region of integration is symmetric around the origin since it is a circle, meaning that it involves areas that are perfectly balanced across axes. Because of this symmetry, the positive and negative values of \(\sin\sqrt{x^2 + y^2}\) cancel out over the entire region, resulting in the integral \(V\) equating to zero. Such symmetric properties in integrals can save a considerable amount of computation by allowing conclusions to be drawn based on shape and behavior rather than numeric intervention.

Integration Techniques

Mastering integration techniques is essential for handling a variety of calculus problems, particularly those involving double integrals. The exercise involved two important methods:

• Polar Coordinate Integration
• Partial Integration

For \(V\), transforming into polar coordinates simplified the function considerably, especially the radial part. This allowed us to easily evaluate the integral as it highlighted a natural symmetry in the \(\sin\) expression.
The method of partial integration was used to deal with the inner integral \(\int_0^{2\pi} r\sin r \, dr\), providing a straightforward technique to solve for a variable that doesn’t simplify easily, such as \(\sin(r)\). Like product rule in differentiation, partial integration (or integration by parts) is a derivative in reverse; it helps in reworking an integral into a potentially simplifiable form.
For \(W\), after converting to polar coordinates, evaluating the absolute value integral required recognizing that the absolute \(\sin(r)\) changed the traditional \(-\cos(r)\) formula into one where values are strictly non-negative, which affects integral boundaries. These techniques in unison help in navigating and simplifying tasks involving double and more complex integrals.