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Evaluate the following double integrals over the region \(R\) $$\iint_{R} \sqrt{\frac{x}{y}} d A ; R=\\{(x, y): 0 \leq x \leq 1,1 \leq y \leq 4\\}$$

Short Answer

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Question: Evaluate the double integral $\int\limits_1^4\int\limits_0^1 \sqrt{\frac{x}{y}} dx dy$ over the region R defined by $0 \leq x \leq 1$ and $1 \leq y \leq 4$. Answer: The value of the given double integral over the region R is 5.

Step by step solution

01

Set up the double integral

According to the bounds of the region R, we have the following double integral: $$\int\limits_1^4\int\limits_0^1 \sqrt{\frac{x}{y}} dx dy$$
02

Find the integral with respect to x

First, we find the integral inner with respect to x: $$\int\limits_0^1 \sqrt{\frac{x}{y}} dx$$ Let's substitute \(u=\sqrt{\frac{x}{y}}\). Then, \(u^2 = \frac{x}{y}\) and \(x=yu^2\), and also \(dx=2yu\,du\). Now, the integral becomes: $$\int \sqrt{\frac{yu^2}{y}} (2yu) du = 2y \int u^2 du = 2y\left(\frac{u^3}{3}\right)$$ Using the substitution \(u=\sqrt{\frac{x}{y}}\), we get: $$\frac{2y}{3}\left(\frac{x}{y}\right)^\frac{3}{2}$$ Now, we plug in the limits of integration: $$\left.\frac{2y}{3}\left(\frac{x}{y}\right)^\frac{3}{2}\right|_0^1 = \frac{2y}{3} \left(1 - 0\right) = \frac{2y}{3}$$
03

Find the integral with respect to y

Now, we find the outer integral with respect to y: $$\int\limits_1^4 \frac{2y}{3} dy = \frac{2}{3}\int\limits_1^4 y dy = \frac{2}{3}\left(\frac{y^2}{2}\right)\Big|_1^4$$
04

Evaluate the integral and find the result

Finally, let's evaluate the integral by plugging in the limits of integration: $$\frac{2}{3}\left(\frac{y^2}{2}\right)\Big|_1^4 = \frac{2}{3}\left(\frac{16}{2} - \frac{1}{2}\right) = \frac{2}{3}(8-\frac{1}{2}) = \frac{2}{3}\cdot\frac{15}{2} = 5$$ Thus, the value of the given double integral over the region R is 5.

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Key Concepts

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

Iterated Integrals
Double integrals are often evaluated by transforming them into an iterated integral, which involves integrating a function of two variables over a region step-by-step. For the given exercise, we have a double integral over the region \( R \), given by
  • \( 0 \leq x \leq 1 \)
  • \( 1 \leq y \leq 4 \)
The double integral is written as \( \int_1^4 \int_0^1 \sqrt{\frac{x}{y}} \, dx \, dy \), demonstrating the process of iterated integration. This is structured to first integrate the function with respect to \( x \) while \( y \) is treated as a constant, and then integrate the result with respect to \( y \). Changing the order of integration may sometimes simplify the problem, depending on the region boundaries and the function. However, always ensure that the order used respects the limits defined by the region of integration.
Substitution in Integrals
Substitution is a valuable technique in calculus for simplifying integrals, making complex functions easier to integrate. In our exercise, a substitution was performed during the inner integral
  • Let \( u=\sqrt{\frac{x}{y}} \)
  • Then \( u^2 = \frac{x}{y} \) and thus \( x = yu^2 \)
  • The differential changes as \( dx = 2yu \, du \)
These transformations replaced \( \sqrt{\frac{x}{y}} \) with a more manageable expression in terms of \( u \). Integrating in terms of \( u \) simplifies the calculations because \( \int u^2 \, du \) is straightforward to compute. Once integrated, the result is reverted to the original variables to find the desired solution. Substitution is crucial for reducing complex variable dependencies and simplifying integration bounds.
Region of Integration
Understanding the region of integration is essential for setting up a double integral properly. This region, denoted as \( R \) here, dictates the bounds for the integrals.
  • It is often defined by inequalities: \( 0 \leq x \leq 1 \) and \( 1 \leq y \leq 4 \).
These boundaries help us determine the limits for the iterated integrals. The specified region may represent geometric shapes such as rectangles, circles, or other bounded areas across the coordinate plane. Visualizing this region can often make it clearer what limits need to be considered for integration, especially if we change the order of integration. Proper identification of the region ensures the integration covers all the necessary areas for the problem.
Calculus Techniques
Solving double integrals involves several calculus techniques, including integration by substitution, understanding the geometric implications of the integral, and manipulating integral bounds. In our original exercise, we used:
  • Iterated Integration: Breaking down the double integral into iterated single integrals over specified limits.
  • Substitution: Transforming the integrand into a simpler form to facilitate easier integration.
  • Careful evaluation of integration limits: Ensuring that integrals are calculated properly by substituting and computing at bounds.
These techniques exemplify how calculus can transform complex problems into manageable sequences of steps, ultimately yielding a solution that is both accurate and derived from fundamental principles. Keeping these techniques in mind and practicing their application becomes invaluable as more complex integrals are encountered.

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