Problem 24 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Evaluate the double integral ∬(x^2 - y^2)^2 dA over the region R, where R is defined by -1 ≤ x ≤ 2 and 0 ≤ y ≤ 1. Solution: Evaluating the double integral over the region R gives us a result of (61/45).

Step 1: Convert the double integral to an iterated integral

To convert the given double integral to an iterated integral, we need to split the double integral into two single interpolates. The first will be with respect to x, and the second with respect to y. Using the bounds for the region R, we can write the iterated integral in the form: $$\int_{0}^{1} \int_{-1}^{2} (x^{2}-y^{2})^{2} dx dy$$

Step 2: Evaluate the inner integral

Now, we need to evaluate the inner integral with respect to x. The integral is as follows: $$\int_{-1}^{2} (x^{2}-y^{2})^{2} dx$$ To compute this integral, we need to expand the expression \((x^{2}-y^{2})^{2}\). $$(x^{2}-y^{2})^{2} = (x^4 - 2x^2y^2 + y^4)$$ Now, integrate this expression over the interval \([-1, 2]\) with respect to x. $$\int_{-1}^{2}(x^4 - 2x^2y^2 + y^4)dx = \left[\frac{1}{5}x^5 - \frac{2}{3}x^3y^2 + xy^4\right]_{-1}^{2}$$ $$=\left(\frac{1}{5}(2)^5 - \frac{2}{3}(2)^3y^2 + (2)y^4\right) -\left(\frac{1}{5}(-1)^5 - \frac{2}{3}(-1)^3y^2 + (-1)y^4\right)$$ $$=\left(\frac{32}{5} - \frac{16}{3}y^2 + 2y^4\right) - \left(-\frac{1}{5} + \frac{2}{3}y^2 - y^4\right)$$

Step 3: Evaluate the outer integral

Now, it's time to evaluate the outer integral with respect to y using the resulting expression from Step 2: $$\int_{0}^{1}\left(\frac{32}{5} - \frac{16}{3}y^2 + 2y^4 + \frac{1}{5} - \frac{2}{3}y^2 + y^4\right)dy$$ Simplify the expression inside the integral: $$\int_{0}^{1}\left(\frac{33}{5} - \frac{22}{3}y^2 + 3y^4\right)dy$$ Now, integrate the expression over the interval \([0, 1]\) with respect to y. $$\left[\frac{33}{5}y - \frac{22}{9}y^3 + \frac{3}{5}y^5\right]_{0}^{1}$$ Finally, substituting the bounds, we get the value of the integral: $$\left(\frac{33}{5}(1) - \frac{22}{9}(1)^3 + \frac{3}{5}(1)^5\right) - \left(\frac{33}{5}(0) - \frac{22}{9}(0)^3 + \frac{3}{5}(0)^5\right)$$ $$=\frac{33}{5} - \frac{22}{9} + \frac{3}{5} = \boxed{\frac{61}{45}}$$ So, the double integral over the region R evaluates to \(\frac{61}{45}\).

Key Concepts

Iterated Integrals

Iterated integrals are a way to compute double integrals by breaking them into two separate single integrals, one inside the other. The idea is to first integrate with respect to one variable and then with respect to another. This technique simplifies the calculation of double integrals and allows us to handle each variable independently. In our original exercise, this meant converting

• the double integral \(\iint_{R}\left(x^{2}-y^{2}\right)^{2} dA\) into two consecutive computations
• an inner integral evaluated with respect to \(x\), followed by
• an outer integral evaluated with respect to \(y\).

The bounds for these integrals come from the problem's specified region \(R\), giving clear limits of integration for each integral involved. This method maintains the order of integration, which is crucial as it determines how we approach the region. By setting up an iterated integral, it becomes possible to compute complex areas or volumes defined by the integrand over the specified region.

Integral Calculus

Integral calculus is a branch of calculus focused on the concept of integration. It helps in finding the accumulation of quantities, like areas under curves or total values over a specific region. In the context of our double integral problem, integral calculus allows us to calculate the area under a surface expressed by \( (x^{2} - y^{2})^{2} \) over a specified region \( R \). By breaking the problem into iterated integrals, we leverage the idea of calculating the area in sections, making the comprehensive task more manageable. Integral calculus offers tools to manage complex expressions, like breaking down \( (x^{2} - y^{2})^{2} \) into simpler parts, allowing integrations to be

• expanded, evaluated separately, and then
• combined to find the complete integral result.

This strategic manipulation of the integral is a cornerstone of integral calculus, utilizing both algebraic skills and knowledge of calculus.

Integration Techniques

Integration techniques are methods and strategies that help us evaluate integrals effectively. They are highly valuable in addressing both single and double integrals. In our case study, several such techniques are employed:First, the method involved expanding the expression

• \((x^{2}-y^{2})^{2}\)
• Before integrating, so it simplified the function into separate terms: \(x^4 - 2x^2y^2 + y^4\).

This makes the integral easier to handle compared to the original composite expression. Such algebraic manipulation is key to solving complex integrals by reducing them into simpler, manageable forms. Afterwards, the process of individually integrating each term separately with respect to \(x\), before recombination and focusing on \(y\) in the outer integral, demonstrates linear integration's utility and prowess in handling variable-specific challenges in multiple integral problems. Such techniques empower us to obtain clean solutions while managing intricate mathematical forms effectively.