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Chapter 8: Problem 12

Find the double integral over the indicated region \(D\) in two ways. (a) Integrate first with respect to \(x\). (b) Integrate first with respect to \(y\). $$ \iint_{D} 6 x y^{2} d A, D=\\{(x, y): 1 \leq x \leq 2,2 \leq y \leq 3\\} $$

Short Answer

Expert verified
The double integral over the region \(D\) is 57, using both approaches.

Step by step solution

01

Set Up the Integral with Respect to x

First, integrate with respect to \(x\). Given the region \(D\), the limits for \(x\) are from 1 to 2, and for \(y\) from 2 to 3. Set up the double integral as: \[ \int_{2}^{3} \left( \int_{1}^{2} 6xy^2 \, dx \right) \, dy. \]
02

Integrate the Inner Integral with Respect to x

Evaluate the inner integral \( \int_{1}^{2} 6xy^2 \, dx \). Treat \(y^2\) as a constant, and integrate: \[ \int 6xy^2 \, dx = 6y^2 \cdot \left( \frac{x^2}{2} \right) = 3y^2x^2. \] Evaluate this from \(x=1\) to \(x=2\): \[ 3y^2(2^2) - 3y^2(1^2) = 12y^2 - 3y^2 = 9y^2. \]
03

Integrate the Outer Integral with Respect to y

Now evaluate \( \int_{2}^{3} 9y^2 \, dy \). Integrate: \[ \int 9y^2 \, dy = 9 \cdot \frac{y^3}{3} = 3y^3. \] Evaluate from \(y=2\) to \(y=3\): \[ 3(3^3) - 3(2^3) = 81 - 24 = 57. \]
04

Set Up the Integral with Respect to y

Now integrate first with respect to \(y\). The limits for \(y\) are from 2 to 3, and for \(x\) from 1 to 2. Set up the double integral as: \[ \int_{1}^{2} \left( \int_{2}^{3} 6xy^2 \, dy \right) \, dx. \]
05

Integrate the Inner Integral with Respect to y

Evaluate the inner integral \( \int_{2}^{3} 6xy^2 \, dy \). Treat \(x\) as a constant, and integrate: \[ \int 6xy^2 \, dy = 6x \cdot \frac{y^3}{3} = 2xy^3. \] Evaluate from \(y=2\) to \(y=3\): \[ 2x(3^3) - 2x(2^3) = 54x - 16x = 38x. \]
06

Integrate the Outer Integral with Respect to x

Now evaluate \( \int_{1}^{2} 38x \, dx \). Integrate: \[ \int 38x \, dx = 38 \cdot \frac{x^2}{2} = 19x^2. \] Evaluate from \(x=1\) to \(x=2\): \[ 19(2^2) - 19(1^2) = 76 - 19 = 57. \]

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Key Concepts

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

Calculus
Calculus is a branch of mathematics that defines changes and can describe a myriad of physical, natural, and economic phenomena. It consists of derivatives and integrals, both being fundamental concepts in the study of Calculus. Integrals, especially, are used to determine areas, volumes, and sums over continuous domains.
One key understanding in Calculus is the idea of limits, which helps make sense of functions as they approach particular points or infinity. Through limits, we establish continuity and define the derivative, a major concept for determining rates of change.
In the realm of Integrals, particularly double integrals, they allow us to integrate over two variables instead of one variable, which adds a layer of complexity and utility. This becomes particularly valuable in physics and engineering where we analyze systems in two or three dimensions.
Integration Techniques
Integration is a crucial part of calculus that helps find the total size of something by adding up parts. There are various techniques used to solve integrals, such as substitution, integration by parts, and partial fractions. Each technique simplifies complex integrals into simpler forms that can be evaluated more easily.
When working with double integrals, two main strategies are often employed:
  • Integrating first with respect to one variable.
  • Integrating with respect to the other variable.
This flexibility allows the solver to choose the most optimal path, often based on the region of integration or function behavior. In the current exercise, we solve using both orders to demonstrate that regardless of the order of integration, the outcome remains the same. This concept highlights the Fubini's Theorem, which validates switching the order of integration under certain conditions.
Multivariable Calculus
Multivariable Calculus extends single-variable calculus into higher dimensions. Instead of dealing with single-variable functions, it introduces functions that depend on two or more variables.
Double integrals are foundational in Multivariable Calculus. They allow us to compute volumes under surfaces described by functions of two variables. Given the exercise, the integral is conducted over a rectangular region, which simplifies the process by providing straightforward limits of integration.
For regions and functions that are more complex, it becomes critical to understand and possibly redefine the limits of integration or change coordinates for ease of integration. This entire practice enriches how areas, volumes, and other physical quantities can be computed for various applications, furthering the understanding and practical utility of calculus in multiple fields.

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Most popular questions from this chapter

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