Problem 11 - Thomas Calculus

Question

In Exercises \(1-14,\) evaluate the iterated integral. $$\int_{-1}^{2} \int_{0}^{\pi / 2} y \sin x d x d y$$

Step-by-Step Solution

Answer

The value of the iterated integral is 1.5.

Step 1: Identify the Integrals

We're given the iterated integral \[ \int_{-1}^{2} \int_{0}^{\pi / 2} y \sin x \, dx \, dy \]. Here, we notice that \( y \sin x \) is the integrand. The inner integral is with respect to \( x \) from \( 0 \) to \( \pi/2 \), and the outer integral is with respect to \( y \) from \( -1 \) to \( 2 \).

Step 2: Evaluate the Inner Integral

Evaluate the inner integral \( \int_{0}^{\pi / 2} y \sin x \, dx \). Since \( y \) is a constant with respect to \( x \), it can be factored out. Thus, the integral becomes \( y \int_{0}^{\pi / 2} \sin x \, dx \). The integral of \( \sin x \) is \( -\cos x \). Evaluating this from \( 0 \) to \( \pi/2 \) yields:\[ y \left[ -\cos x \right]_{0}^{\pi/2} = y \left( -\cos(\pi/2) + \cos(0) \right) = y(0 + 1) = y \].

Step 3: Simplify Result from Inner Integral

After evaluating the inner integral, we get \( y \). This simplifies our iterated integral to \( \int_{-1}^{2} y \, dy \).

Step 4: Evaluate the Outer Integral

Now evaluate the outer integral \( \int_{-1}^{2} y \, dy \). Use the formula for the integral of \( y \), which is \( \frac{1}{2}y^2 \). Evaluate this from \( -1 \) to \( 2 \):\[ \left[ \frac{1}{2}y^2 \right]_{-1}^{2} = \frac{1}{2} \left(2^2\right) - \frac{1}{2} \left((-1)^2\right) = \frac{1}{2} \times 4 - \frac{1}{2} \times 1 = 2 - 0.5 = 1.5 \].

Step 5: Conclude the Solution

The evaluation of the iterated integral \( \int_{-1}^{2} \int_{0}^{\pi / 2} y \sin x \, dx \, dy \) results in \( 1.5 \). This is the area under the surface described by the integral over the specified limits.

Key Concepts

Understanding Integrands

In the world of calculus, particularly while working with integrals, the term "integrand" is quite common. An integrand is essentially the function that you are integrating. It is the "meat" inside the sandwich of an integral.
For instance, in the iterated integral \( \int_{-1}^{2} \int_{0}^{\pi / 2} y \sin x \, dx \, dy \), the integrand is \( y \, \sin x \).
What's important here is recognizing that the integrand can be a combination of variables, like \( y \) and \( \sin x \) in this example. Its role is pivotal as it determines which mathematical operation - integration - needs to be applied.
The essence of understanding an integrand lies in realizing that it tells us what area or volume we are calculating when we perform integration operations. It's like identifying the shapes or curves we want to analyze.

Deciphering the Inner Integral

The concept of an "inner integral" is integral to understanding the process of evaluating double or iterated integrals. In our example, the inner integral is \( \int_{0}^{\pi / 2} y \sin x \, dx \).
Essentially, this means we hold one variable constant—in this case, \( y \)—and perform the integration with respect to the other variable: \( x \). This fixes our attention, first, on a slice of the region we are interested in.

• We treat \( y \) as a constant and integrate \( \sin x \) within the bounds from \( 0 \) to \( \pi/2 \).
• Since \( y \) acts as a constant, \( y \) comes out of the integral, simplifying our task significantly.

Once this step is completed, the result affects how the outer integral is solved, essentially simplifying the problem step-by-step.

Evaluating the Outer Integral

Once the inner integral is resolved, we transition smoothly into solving the "outer integral." In our scenario, this means looking at \( \int_{-1}^{2} y \, dy \).
This step focuses on the result we got from the inner integral, which now becomes the focus of the outer integral's process. The outer integral allows us to sweep through all the calculated slices to establish a complete picture of the area.
In this particular case, we use the antiderivative formula for \( y \), which simplifies our task tremendously. The integration of \( y \) results in \( \frac{1}{2}y^2 \), evaluated over the interval from \( -1 \) to \( 2 \).

• Evaluate \( \frac{1}{2}(2^2) - \frac{1}{2}((-1)^2) \)
• This simplification shows us how to get from complex iterated scenarios to an understandable, neat answer.

Interpreting Area Under the Surface

Understanding how an iterated integral relates to "area under the surface" greatly demystifies these calculations. Area, in terms of integral calculus, refers to the total proportion under a graph of a function over a specified range.
An iterated integral such as \( \int_{-1}^{2} \int_{0}^{\pi / 2} y \sin x \, dx \, dy \) signifies calculating an area, but across two dimensions - over \( x \) first, then \( y \).
This dual integration reflects the 'sweeping' action of moving from dealing with a line problem to a surface problem. The final result - \( 1.5 \) in this example - confirms that the double layers of integration accurately measure the total area described by our original integrand in its given limits.

• This value represents the accumulated 'weight' or quantity beneath the curve described in the multiple bounds.
• It provides a concrete numerical value showing the size of an abstract, complex shape.