Problem 25 - Calculus Early Transcendentals

Question

Evaluate each double integral over the region \(R\) by converting it to an iterated integral. $$\iint_{R}\left(x^{5}-y^{5}\right)^{2} d A ; R=\\{(x, y): 0 \leq x \leq 1,-1 \leq y \leq 1\\}$$

Step-by-Step Solution

Answer

Question: Write down an iterated integral for the double integral of $(x^{5}-y^{5})^{2}$ over the rectangle defined by \(0 \leq x \leq 1, -1 \leq y \leq 1\), and calculate its value. Answer: The iterated integral of $(x^{5}-y^{5})^{2}$ over the given rectangle is: $$\int_{-1}^1 \int_0^1 (x^{5}-y^{5})^2 dx dy$$. Its value is $\frac{1}{6}$.

Step 1: Convert the double integral to an iterated integral

First, we need to notice that since the region \(R\) is rectangular, we can easily convert the double integral into an iterated integral by writing the bounds of integration for \(x\) and \(y\) separately. This gives us: $$\iint_{R}\left(x^{5}-y^{5}\right)^{2} d A= \int_{-1}^1 \int_0^1 (x^{5}-y^{5})^2 dx dy$$

Step 2: Evaluate the inner integral

Now we want to integrate the inner integral with respect to \(x\): $$\int_0^1 (x^{5}-y^{5})^2 dx$$ We can expand the square, which gives us \(x^{10}-2x^5y^5+y^{10}\). To integrate each term, we get: $$\int_0^1 x^{10} dx - 2y^5 \int_0^1 x^5 dx + y^{10} \int_0^1 dx = \left[\frac{1}{11}x^{11} - 2y^5\frac{1}{6}x^6 + y^{10}x\right]_0^1 = \frac{1}{11} - \frac{1}{3}y^5 + y^{10}$$

Step 3: Evaluate the outer integral

Now we want to integrate the outer integral with respect to \(y\): $$\int_{-1}^1 \left(\frac{1}{11} - \frac{1}{3}y^5 + y^{10}\right) dy$$ Each term can be integrated separately: $$\left[\frac{1}{11}y - \frac{1}{18}y^6 + \frac{1}{11}y^{11}\right]_{-1}^1 = \left(\frac{1}{11} - \frac{1}{18} + \frac{1}{11}\right) - \left(-\frac{1}{11} + \frac{1}{18} + \frac{1}{11}\right) = \boxed{\frac{1}{6}}$$ So the result of evaluating the iterated integral is \(\frac{1}{6}\).

Key Concepts

Double Integral

When you come across the term double integral, it refers to a process in calculus that is used to integrate functions of two variables over a two-dimensional region. Imagine a surface that represents a function in three dimensions. The double integral helps you find the volume of the shape that forms between this surface and a region on the plane. It's analogous to finding the area under a curve, but this time in three dimensions.

For example, in the exercise, the function \( (x^5-y^5)^2 \) is being integrated over a specific region, \((R)\), which is a rectangle in this case. The double integral \( \iint_{R}(x^{5}-y^{5})^{2} dA \) represents the volume under the surface created by \( (x^5-y^5)^2 \) and above the rectangular region \((R)\). Integrating such an expression requires breaking it down into simpler parts called iterated integrals, which is precisely what the step-by-step solution shows us.

Integration Bounds

The terms integration bounds or limits of integration specify the range over which you are integrating. They define the dimensions of the region that you're interested in. For a double integral, you'll have two sets of bounds—a pair for each variable. In the given exercise, the bounds for \(x\) are from 0 to 1, while the bounds for \(y\) are from -1 to 1.

Setting the correct bounds is crucial. For the integral to represent the volume over region \((R)\), you must match the bounds with the correct variable. In our example, these bounds tell us that for each slice along the \(y\)-axis, \(x\) varies from 0 to 1. Similarly, as we move along the \(x\)-axis, the \(y\)-value can range from -1 to 1. This information allows us to construct the appropriate iterated integral to correctly find the desired volume.

Integral Evaluation

The integral evaluation step is where the actual computation happens. Once you have defined your iterated integral, as with our double integral from the region \((R)\), you need to perform the integration step by step. Often, starting with the innermost integral, you'll find the antiderivative with respect to one variable while treating the other as a constant.

In our exercise, we first integrate with respect to \(x\), treating \(y\) as a constant. It's like slicing the volume into vertical slices and stacking them side by side. After completing the inner integral, we get a function in terms of \(y\), which we then integrate over the specified \(y\)-bounds. This is akin to taking the sum of all the vertical slices to get the total volume. The final result of this integration process gives us a number that represents the combined area of all these slices, which, in the context of our exercise, was found to be \(\frac{1}{6}\).

By carefully following these steps and correctly evaluating the integrals, you ensure that the calculated volume is accurate for the given region. This process is at the heart of many applications in physics, engineering, and probability theory, where understanding the cumulative effect over a region is essential.