Problem 19 - Calculus Early Transcendentals

Question

Evaluate each double integral over the region \(R\) by converting it to an iterated integral. $$\iint_{R} 4 x^{3} \cos y d A ; R=\\{(x, y): 1 \leq x \leq 2,0 \leq y \leq \pi / 2\\}$$

Step-by-Step Solution

Answer

Question: Calculate the double integral of the function $4x^3\cos y$ over the rectangular region $1 \leq x \leq 2$ and $0 \leq y \leq \pi / 2$. Answer: The value of the double integral over the given region is equal to $15$.

Step 1: Write the double integral as an iterated integral

The double integral can be written as an iterated integral by integrating first with respect to \(x\) and then with respect to \(y\). The given region is rectangular, so we can simply write down the limits for both variables. The iterated integral will look like this: $$ \int_{0}^{\pi/2} \int_{1}^{2} 4x^3\cos y\, dx\, dy $$

Step 2: Evaluate the inner integral with respect to \(x\)

Now we need to evaluate the integral with respect to \(x\). We will integrate the function \(4x^3\cos y\) with respect to \(x\) and then use the limits for \(x\): \begin{align*} \int_{1}^{2} 4x^3\cos y\, dx = \frac{4\cos y}{4}\bigg|_{1}^2x^4 \\ &=\cos y(x^4\big|_1^2)\\ &= \cos y(16 - 1)= 15\cos y \end{align*}

Step 3: Evaluate the outer integral with respect to \(y\)

Now we need to evaluate the integral of the result from the previous step with respect to \(y\). We will integrate the function \(15\cos y\) with respect to \(y\) and then use the limits for \(y\): \begin{align*} \int_{0}^{\pi/2} 15\cos y\, dy = \int_{0}^{\pi/2} 15(\cos y)\, dy = 15\sin y\bigg|_0^{\pi \over 2}\\ &=15(\sin (\pi \over 2)-\sin(0))\\ &=15(1 - 0)=15 \end{align*}

Step 4: Interpret the final result

The value of the double integral \(\iint_{R} 4 x^{3} \cos y\, dA\) over the given region is equal to \(15\). This is our final answer.

Key Concepts

Iterated Integral

Let's start by understanding what an iterated integral is. A double integral allows us to calculate the accumulation of quantities over a two-dimensional region. However, evaluating a double integral directly can be complex, so we often convert it into two simpler, iterated, one-dimensional integrals.

In the given exercise, the double integral over region \( R \) is transformed into an iterated integral. This means we first integrate with respect to \( x \), and then with respect to \( y \). This process simplifies the calculation:

• Inner Integral: Integrate with respect to \( x \).
• Outer Integral: Integrate the result with respect to \( y \).

Converting to an iterated integral often involves using the boundaries of the given region. In this exercise, our region \( R = \{ (x, y) : 1 \leq x \leq 2, 0 \leq y \leq \frac{\pi}{2} \} \) is a rectangle, making it straightforward to set limits for \( x \) and \( y \). For more complex regions, these limits might vary or require deeper analysis.

Integration Limits

Integration limits determine the range over which we perform each integration step. Correctly identifying and applying these limits is key to solving a double integral problem.

In this exercise, our limits are uniform across a rectangular region. That's why they are constant:

• For \( x \), the limits are from 1 to 2.
• For \( y \), the limits are from 0 to \( \frac{\pi}{2} \).

These boundaries define how far we "slice" along each axis, which in turn controls how much of the surface over region \( R \) we account for as we integrate.

It's essential to ensure these limits align with the problem's geometry. Changing the order of integration (if allowed by the problem) would still require careful attention to how those limits translate between the two variables.

Trigonometric Integration

Trigonometric integration involves functions like sine, cosine, and others, which frequently appear in calculus problems and solutions. These functions have distinctive integrals that are often straightforward but need careful handling.

In our step-by-step solution, we see trigonometric integration in action during the outer integral, where we integrate \( 15\cos(y) \) with respect to \( y \). The rules for integrating basic trigonometric functions let us directly find:

• The integral of \( \cos(y) \) is \( \sin(y) \).
• Adding a constant multiplier, like 15, applies directly to the result.

The calculation here becomes, after bounds integration, \( 15(\sin(\frac{\pi}{2}) - \sin(0)) = 15 \). Remember that certain values, like \( \sin(0) = 0 \) and \( \sin(\frac{\pi}{2}) = 1 \), are key to quickly solving such problems.

Recognizing these fundamental integrals vastly simplifies evaluating trigonometric expressions and supports solving more complex integration challenges effectively.