**Problem 6:****Objective:** Write the integral for the given volume (you don’t need to solve the integrals, just set them up).**Details:**Under the surface $ z = xy \, e^{-x^2-y^2} $ and above the region enclosed by $ y = 2 - x^2 $, $ y = 0 $, and $ x = 0 $. Sketch the region and explain what you did!**Tasks:**a. **Integrating with respect to $ x $ first:**1. Identify the limits for $ x $ by considering where the parabola $ y = 2 - x^2 $ meets the x-axis and y-axis.2. Setup the integral: The outer integral (with respect to $ y $) will run from $ y = 0 $ to $ y = 2 $. For each $ y $, $ x $ will range from $ x = 0 $ to $ x = \sqrt{2-y} $.3. Integrate $ z = xy \, e^{-x^2-y^2} $ with these limits.b. **Integrating with respect to $ y $ first:**1. Identify the limits for $ y $ which range from $ y = 0 $ to $ y = 2 - x^2 $.2. Setup the integral: The outer integral (with respect to $ x $) will run from $ x = 0 $ to $ x = \sqrt{2} $. For each $ x $, $ y $ will range from $ y = 0 $ to $ y = 2 - x^2 $.3. Integrate $ z = xy \, e^{-x^2-y^2} $ with these limits.**Diagram Explanation:**- The region of integration is bounded by the curve $ y = 2 - x^2 $ (a downward-facing parabola), the line $ y = 0 $, and the line $ x = 0 $.- Sketch the parabola on the xy-plane and highlight the area bounded by the axes and the curve. Use this setup to practice setting up double integrals for calculating volumes in multivariable calculus.

Answered: Image /qna-images/answer/28541355-f7ab-42a9-a814-f9bacb78ad7a.jpg

6. Write the integral for the given volume (you don't need to solve the integrals, just set them up) Under the surface z = xy e-**-y* and above the region enclosed by y = 2 – x² and y = 0 and x = 0. Sketch the region and explain what you did! a. Integrating with respect to x first b. Integrating with respect to y first

6. Write the integral for the given volume (you don't need to solve the integrals, just set them up) Under the surface z = xy e-**-y* and above the region enclosed by y = 2 – x² and y = 0 and x = 0. Sketch the region and explain what you did! a. Integrating with respect to x first b. Integrating with respect to y first

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Concept explainers

Optimization

Optimization comes from the same root as "optimal". "Optimal" means the highest. When you do the optimization process, that is when you are "making it best" to maximize everything and to achieve optimal results, a set of parameters is the base for the selection of the best element for a given system.

Integration

Integration means to sum the things. In mathematics, it is the branch of Calculus which is used to find the area under the curve. The operation subtraction is the inverse of addition, division is the inverse of multiplication. In the same way, integration and differentiation are inverse operators. Differential equations give a relation between a function and its derivative.

Application of Integration

In mathematics, the process of integration is used to compute complex area related problems. With the application of integration, solving area related problems, whether they are a curve, or a curve between lines, can be done easily.

Volume

In mathematics, we describe the term volume as a quantity that can express the total space that an object occupies at any point in time. Usually, volumes can only be calculated for 3-dimensional objects. By 3-dimensional or 3D objects, we mean objects that have length, breadth, and height (or depth).

Area

Area refers to the amount of space a figure encloses and the number of square units that cover a shape. It is two-dimensional and is measured in square units.

Question

**Problem 6:**

**Objective:** Write the integral for the given volume (you don’t need to solve the integrals, just set them up).

**Details:**
Under the surface $ z = xy \, e^{-x^2-y^2} $ and above the region enclosed by $ y = 2 - x^2 $, $ y = 0 $, and $ x = 0 $. Sketch the region and explain what you did!

**Tasks:**

a. **Integrating with respect to $ x $ first:**

1. Identify the limits for $ x $ by considering where the parabola $ y = 2 - x^2 $ meets the x-axis and y-axis.
2. Setup the integral: The outer integral (with respect to $ y $) will run from $ y = 0 $ to $ y = 2 $. For each $ y $, $ x $ will range from $ x = 0 $ to $ x = \sqrt{2-y} $.
3. Integrate $ z = xy \, e^{-x^2-y^2} $ with these limits.

b. **Integrating with respect to $ y $ first:**

1. Identify the limits for $ y $ which range from $ y = 0 $ to $ y = 2 - x^2 $.
2. Setup the integral: The outer integral (with respect to $ x $) will run from $ x = 0 $ to $ x = \sqrt{2} $. For each $ x $, $ y $ will range from $ y = 0 $ to $ y = 2 - x^2 $.
3. Integrate $ z = xy \, e^{-x^2-y^2} $ with these limits.

**Diagram Explanation:**

- The region of integration is bounded by the curve $ y = 2 - x^2 $ (a downward-facing parabola), the line $ y = 0 $, and the line $ x = 0 $.
- Sketch the parabola on the xy-plane and highlight the area bounded by the axes and the curve.

Use this setup to practice setting up double integrals for calculating volumes in multivariable calculus.

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