/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Write an iterated integral that ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 1

Write an iterated integral that gives the volume of the solid bounded by the surface \(f(x, y)=x y\) over the square \(R=\\{(x, y): 0 \leq x \leq 2,1 \leq y \leq 3\\}\)

Short Answer

Expert verified
Answer: The volume of the solid is 8.

Step by step solution

01

Write down the given information

From the exercise, we are given the surface \(f(x, y) = xy\), and the region \(R = \{(x, y): 0 \leq x \leq 2, 1 \leq y \leq 3\}\).
02

Set up the iterated integral

Based on the given region \(R\), we will integrate with respect to \(x\) first, and then with respect to \(y\). The bounds for \(x\) are \(0 \leq x \leq 2\), and the bounds for \(y\) are \(1 \leq y \leq 3\). Therefore, our iterated integral will look like this: \[V = \int_{1}^3 \int_{0}^2 f(x, y) dxdy = \int_{1}^3 \int_{0}^2 xy\, dxdy\]
03

Integrate with respect to x

Let's now integrate with respect to \(x\): \[\int_{0}^2 xy\, dx = \frac{1}{2}x^2y\Big|_0^2 = (2^2y)\frac{1}{2} - (0^2y)\frac{1}{2} = 2y\]
04

Integrate with respect to y

Now we will integrate the result from step 3 with respect to \(y\): \[V = \int_{1}^3 2y\, dy = y^2\Big|_1^3 = (3^2) - (1^2) = 9 - 1 = 8\]
05

Write down the final answer

After integrating with respect to both \(x\) and \(y\), we found that the volume of the solid bounded by the surface \(f(x, y)=xy\) and the square \(R = \{(x, y): 0 \leq x \leq 2, 1 \leq y \leq 3\}\) is \(V = 8\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the following two-and three-dimensional regions. Specify the surfaces and curves that bound the region, choose a convenient coordinate system, and compute the center of mass assuming constant density. All parameters are positive real numbers. A tetrahedron is bounded by the coordinate planes and the plane \(x / a+y / a+z / a=1 .\) What are the coordinates of the center of mass?

Mass from density Find the mass of the following solids with the given density functions. Note that density is described by the function \(f\) to avoid confusion with the radial spherical coordinate \(\rho\). The solid cone \(\\{(r, \theta, z): 0 \leq z \leq 4,0 \leq r \leq \sqrt{3} z\) \(0 \leq \theta \leq 2 \pi\\}\) with a density \(f(r, \theta, z)=5-z\)

Evaluate the following integrals in spherical coordinates. $$\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{1}^{2 \sec \varphi}\left(\rho^{-3}\right) \rho^{2} \sin \varphi d \rho d \varphi d \theta$$

Use a change of variables to evaluate the following integrals. \(\iiint_{D} d V ; D\) is bounded by the planes \(y-2 x=0, y-2 x=1\) \(z-3 y=0, z-3 y=1, z-4 x=0,\) and \(z-4 x=3\)

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.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.