Question

The following integrals can be evaluated only by reversing the order of integration. Sketch the region of integration, reverse the order of integration, and evaluate the integral. $$\int_{0}^{\sqrt[3]{\pi}} \int_{y}^{\sqrt[3]{\pi}} x^{4} \cos \left(x^{2} y\right) d x d y$$

Short answer

Answer: The reversed order of integration for the double integral is: $$\int_{0}^{\sqrt[3]{\pi}} \int_{0}^x x^{4} \cos \left(x^{2} y\right) d y d x$$

Step-by-step solution

Sketch the region of integration

The given limits of integration are 0 to \(\sqrt[3]{\pi}\) for the outer integral and y to \(\sqrt[3]{\pi}\) for the inner integral. To sketch the region of integration, we'll determine the boundaries between which y and x lie. For the outer integral: \(0 \le y \le \sqrt[3]{\pi}\) For the inner integral: \(y \le x \le \sqrt[3]{\pi}\) The boundaries for our region of integration are: 1) The x-axis: \(y = 0\) 2) The line \(x = y\) 3) The line \(x = \sqrt[3]{\pi}\) Sketch the region enclosed by these lines.

Find new limits of integration

To reverse the order of integration, we need to convert the given limits in terms of x and y. In this case, we would have: For the outer integral (x): \(0 \le x \le \sqrt[3]{\pi}\) For the inner integral (y): \(0 \le y \le x\) Now we can write the double integral using the new limits of integration: $$\int_{0}^{\sqrt[3]{\pi}} \int_{0}^x x^{4} \cos \left(x^{2} y\right) d y d x$$

Evaluate the double integral

First, we'll evaluate the inner integral. $$\int_{0}^{x} x^{4} \cos \left(x^{2} y\right) d y$$ To do this, we will use a substitution: \(u = x^{2} y\). Thus, \(\frac{du}{dy} = x^2\) and \(dy = \frac{1}{x^2} du\). Our limits of integration for u will be when y = 0 and y = x: When \(y = 0\): \(u = x^2 \cdot 0 = 0\) When \(y = x\): \(u = x^2 \cdot x = x^3\) So our inner integral becomes: $$\int_{0}^{x^3} x^{4} \cos \left(u\right) \frac{1}{x^2} d u$$ Which simplifies to: $$x^{2} \int_{0}^{x^3} \cos \left(u\right) d u$$ Now, we can integrate with respect to u: $$x^{2}\left[\sin \left(u\right)\Big|_{0}^{x^3}\right]$$ This gives us: $$x^{2} (\sin (x^{3}) - \sin(0)) = x^{2} \sin(x^{3})$$ Now, we'll use this result to evaluate the outer integral: $$\int_{0}^{\sqrt[3]{\pi}} x^{2} \sin(x^{3}) d x$$ To evaluate this integral, we will use another substitution: \(v = x^3\). So, \(\frac{dv}{dx} = 3x^2\) and \(dx = \frac{1}{3x^2} dv\). Our limits of integration for v will be when x = 0 and x = \(\sqrt[3]{\pi}\): When \(x = 0\): \(v = 0^{3} = 0\) When \(x = \sqrt[3]{\pi}\): \(v = (\sqrt[3]{\pi})^{3} = \pi\) So our outer integral becomes: $$\int_{0}^{\pi} \sin \left(v\right) \frac{1}{3} d v$$ Integrating with respect to v: $$\frac{1}{3}\left[-\cos \left(v\right)\Big|_{0}^{\pi}\right]$$ This gives us the final answer: $$\frac{1}{3}(-\cos(\pi) + \cos(0)) = \frac{1}{3}(2) = \boxed{\frac{2}{3}}$$

Key concepts

Order of Integration

The order of integration refers to the sequence in which multiple integrations are performed in a double integral. It's an important concept in calculus, particularly when dealing with complex regions of integration. When given a double integral, you might need to reverse this order to simplify calculations. Let’s explore how that works.

• In the original exercise, the limits first presented were for an outer integral with respect to y, and an inner integral with respect to x.
• The given order \( \int_{0}^{\sqrt[3]{\pi}} \int_{y}^{\sqrt[3]{\pi}} f(x, y) \, dx \, dy \) was reversed, allowing integration with respect to y first, then x.
• The new order became \( \int_{0}^{\sqrt[3]{\pi}} \int_{0}^{x} f(x, y) \, dy \, dx \).

Reversing the order of integration is beneficial when the re-evaluated integral covers a simpler, more manageable region of the coordinate plane. Often, it relies on the symmetry or constraints of the region to facilitate computations.

Integration Techniques

When evaluating double integrals, a variety of integration techniques might come into play. Each offers unique solutions for handling different types of mathematical challenges.

• U-substitution is a common technique. It involves substituting a part of the integrand with a new variable to simplify the integral. This was used in the exercise with \( u = x^{2} y \).
• Integration by parts can also be employed when dealing with products of functions, though it was not necessary in this example.
• Carefully reversing limits can simplify the integration process, making it easier to tackle complex expressions.

Choosing the right technique is crucial for simplifying and correctly evaluating integrals, as complex integrals often require multi-step processes and thoughtful variable substitutions.

Trigonometric Substitution

Trigonometric substitution is a powerful method in calculus, often used when dealing with integrals that involve square roots or quadratic expressions. While the exercise at hand primarily required substitutions based on variable products like \( x^2 y \), trigonometric substitution remains an essential tool.

• Typically employed when an integral contains expressions like \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), or similar forms.
• Substitution leverages trigonometric identities to transform the integral into a simpler algebraic form.
• Though not used directly in the presented exercise, this technique showcases the utility of substitution as a broader strategy in calculus.

Understanding trigonometric substitution broadens one's toolkit, preparing for efficiently handling more diverse and complex integrals.

Calculus

Calculus is the mathematics of change and motion, and an essential tool in the domain of integral calculus is the double integral. It allows us to calculate the volume under surfaces, area of regions, and more.

• Double integrals are an extension of single-variable calculus, requiring integration over two variables instead of one.
• Applications include physics, engineering, and economics, where they help in finding volumes and averages across multi-dimensional systems.
• The fundamental theorem of calculus can extend to higher dimensions, facilitating the understanding of double integrals.

In mastering calculus and integrals, constant practice through problem-solving builds a strong foundation. It's key to understanding the behavior of complex functions and the spaces they inhabit.