Question

(a) Evaluate the given iterated integral, and (b) rewrite the integral using the other order of integration. $$ \int_{-\pi / 2}^{\pi / 2} \int_{0}^{\pi}(\sin x \cos y) d x d y $$

Short answer

(a) The integral evaluates to 4. (b) New order: \(\int_{0}^{\pi} \int_{-\pi/2}^{\pi/2} (\sin x \cos y) \, dy \, dx\).

Step-by-step solution

Evaluate Inner Integral

Consider the inner integral: \[\int_{0}^{\pi} \sin x \cos y \, dx\]The variable \(y\) is treated as a constant while integrating with respect to \(x\). Integrate \(\sin x\):\[\int \sin x \, dx = -\cos x + C\]Evaluate the definite integral:\[\int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = -\cos(\pi) + \cos(0) = 1 + 1 = 2\]Thus, the inner integral becomes:\[2 \cos y\]

Evaluate Outer Integral

Now, evaluate the outer integral using the result from the inner integral:\[\int_{-\pi / 2}^{\pi / 2} 2 \cos y \, dy\]Integrate \(\cos y\):\[\int \cos y \, dy = \sin y + C\]Evaluate the definite integral:\[2 \int_{-\pi / 2}^{\pi / 2} \cos y \, dy = 2 [\sin y]_{-\pi / 2}^{\pi / 2} = 2(\sin(\pi/2) - \sin(-\pi/2)) = 2(1 - (-1)) = 2 \times 2 = 4\]So, the value of the iterated integral is 4.

Change Order of Integration

Switch the order of integration. Initially, it was \(\int_{-\pi/2}^{\pi/2} \int_{0}^{\pi} ...\, dx \, dy\). Now change to \(\int_{0}^{\pi} \int_{-\pi/2}^{\pi/2} ... \ dy \, dx\).The new integral is:\[\int_{0}^{\pi} \int_{-\pi/2}^{\pi/2} (\sin x \cos y) \, dy \, dx\]Everything else remains the same, as the limits are constants and do not depend on other variables.

Key concepts

Order of Integration

In multivariable calculus, iterated integrals involve integrating a function over a two-dimensional region by integrating with respect to one variable first and then the other. The "order of integration" refers to the sequence or order in which these integrations are performed.
When changing the order of integration, especially in definite integrals, it often simplifies calculations or may be necessary to address the domain of integration more conveniently.

The original integral in our exercise, \( \int_{-\pi/2}^{\pi/2} \int_{0}^{\pi} \text{(some function)} \ dx \, dy \), involves first integrating with respect to \( x \), then \( y \). The task involves changing to \( \int_{0}^{\pi} \int_{-\pi/2}^{\pi/2} \text{(same function)} \ dy \, dx \). This switch doesn't change the integral's value because the limits are constants and independent of the other variable. Instead, it provides flexibility in computation.

• This skill is crucial when the region of integration depends on the function or when exploring alternative integrals.
• Choosing a convenient order can simplify the integration process significantly.

Definite Integral

Definite integrals represent a number that gives the net area under a curve within given boundary values. In the given problem, both inner and outer integrals are definite integrals, with specified limits of integration.

For instance, the inner integral \( \int_{0}^{\pi} \sin x \, dx \), evaluates the net area of \( \sin x \) from 0 to \( \pi \), resulting in a value of 2.
Similarly, the outer integration of \( 2\cos y \) from \(-\pi/2\) to \(\pi/2\) calculates the net area under \(2\cos y\), resulting in 4.

• It provides a concrete numerical answer to what would otherwise be an abstract concept.
• Helps quantify physical phenomena such as distance, area, or any quantity expressed as the integral of another function.

By using integration, we can find quantities that aren't easily described by algebraic formulas.

Trigonometric Integration

Trigonometric integration involves integrating functions that contain trigonometric expressions like \( \sin x \) or \( \cos y \). These functions are particularly significant when dealing with periodic phenomena or oscillations.

In our original exercise, we had to integrate \( \sin x \) over \( x \) and \( \cos y \) over \( y \). These integrations are straightforward thanks to standard trigonometric identities.
For example:

• Integrating \( \sin x \) gives \(-\cos x + C\), utilized to solve the inner integral.
• Integrating \( \cos y \) gives \( \sin y + C\), which helps evaluate the outer integral.

Trigonometric functions often repeat their values in consistent cycles (periodicity), making their integrals predictable and widely applicable in problems involving waves, vibrations, and similar situations.