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Chapter 13: Problem 24

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

Short Answer

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

Step by step solution

01

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: $$\int_{0}^{1} \int_{-1}^{2} (x^{2}-y^{2})^{2} dx dy$$
02

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. $$\int_{-1}^{2}(x^4 - 2x^2y^2 + y^4)dx = \left[\frac{1}{5}x^5 - \frac{2}{3}x^3y^2 + xy^4\right]_{-1}^{2}$$ $$=\left(\frac{1}{5}(2)^5 - \frac{2}{3}(2)^3y^2 + (2)y^4\right) -\left(\frac{1}{5}(-1)^5 - \frac{2}{3}(-1)^3y^2 + (-1)y^4\right)$$ $$=\left(\frac{32}{5} - \frac{16}{3}y^2 + 2y^4\right) - \left(-\frac{1}{5} + \frac{2}{3}y^2 - y^4\right)$$
03

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 \(\frac{61}{45}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Changing order of integration If possible, write iterated integrals in spherical coordinates for the following regions in the specified orders. Sketch the region of integration. Assume that \(f\) is continuous on the region. $$\begin{aligned}&\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{4 \sec \varphi} f(\rho, \varphi, \theta) \rho^{2} \sin \varphi d \rho d \varphi d \theta \text { in the orders }\\\&d \rho d \theta d \varphi \text { and } d \theta d \rho d \varphi\end{aligned}$$

Let \(R\) be the region bounded by the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1,\) where \(a>0\) and \(b>0\) are real numbers. Let \(T\) be the transformation \(x=a u, y=b v\) Find the average square of the distance between points of \(R\) and the origin.

Spherical to rectangular Convert the equation \(\rho^{2}=\sec 2 \varphi\) where \(0 \leq \varphi<\pi / 4,\) to rectangular coordinates and identify the surface.

Evaluate the following integrals in spherical coordinates. $$\int_{0}^{2 \pi} \int_{0}^{\pi / 3} \int_{0}^{4 \sec \varphi} \rho^{2} \sin \varphi d \rho d \varphi d \theta$$

Open and closed boxes Consider the region \(R\) bounded by three pairs of parallel planes: \(a x+b y=0, a x+b y=1\) \(c x+d z=0, c x+d z=1, e y+f z=0,\) and \(e y+f z=1\) where \(a, b, c, d, e,\) and \(f\) are real numbers. For the purposes of evaluating triple integrals, when do these six planes bound a finite region? Carry out the following steps. a. Find three vectors \(\mathbf{n}_{1}, \mathbf{n}_{2},\) and \(\mathbf{n}_{3}\) each of which is normal to one of the three pairs of planes. b. Show that the three normal vectors lie in a plane if their triple scalar product \(\mathbf{n}_{1} \cdot\left(\mathbf{n}_{2} \times \mathbf{n}_{3}\right)\) is zero. c. Show that the three normal vectors lie in a plane if ade \(+b c f=0\) d. Assuming \(\mathbf{n}_{1}, \mathbf{n}_{2},\) and \(\mathbf{n}_{3}\) lie in a plane \(P,\) find a vector \(\mathbf{N}\) that is normal to \(P .\) Explain why a line in the direction of \(\mathbf{N}\) does not intersect any of the six planes and therefore the six planes do not form a bounded region. e. Consider the change of variables \(u=a x+b y, v=c x+d z\) \(w=e y+f z .\) Show that $$J(x, y, z)=\frac{\partial(u, v, w)}{\partial(x, y, z)}=-a d e-b c f$$ What is the value of the Jacobian if \(R\) is unbounded?
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