/*! 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 10 Evaluate the following iterated ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 10

Evaluate the following iterated integrals. $$\int_{0}^{3} \int_{-2}^{1}(2 x+3 y) d x d y$$

Short Answer

Expert verified
Question: Evaluate the iterated integral $\int_{0}^{3} \int_{-2}^{1}(2 x+3 y) d x d y$. Answer: 36

Step by step solution

01

Evaluate the inner integral with respect to x

First, we'll evaluate the inner integral, considering y as a constant: $$\int_{-2}^{1}(2x + 3y) dx$$ To do this, integrate the function with respect to x and apply the limits of integration: $$\left[ x^2 + 3xy \right]_{-2}^{1} = (1 + 3y) - ((-2)^2 + 3(-2)y) = 1 - 4 + 3y + 6y = -3 + 9y$$ Now the inner integral has been evaluated and can be rewritten as a function of y: $$-3 + 9y$$
02

Evaluate the outer integral with respect to y

Now, we'll evaluate the outer integral, using the result from the inner integral: $$\int_{0}^{3} (-3 + 9y) dy$$ Integrate the function with respect to y and apply the limits of integration: $$\left[-3y + \frac{9}{2}y^2 \right]_{0}^{3} = (-9 + \frac{9}{2}(3)^2) - (0) = -9 + \frac{9}{2}(9)$$ Finally, simplify the result: $$-9 + \frac{9}{2}(9) = -9 + \frac{81}{2} = \frac{72}{2} = 36$$ The result of evaluating the iterated integral is 36.

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

Use polar coordinates to find the centroid of the following constant-density plane regions. The region bounded by the cardioid \(r=3-3 \cos \theta\)

Use a double integral to find the area of the following regions. The region bounded by all leaves of the rose \(r=2 \cos 3 \theta\)

Intersecting spheres One sphere is centered at the origin and has a radius of \(R\). Another sphere is centered at \((0.0, r)\) and has a radius of \(r,\) where \(r>R / 2 .\) What is the volume of the region common to the two spheres?

Density distribution A right circular cylinder with height \(8 \mathrm{cm}\) and radius \(2 \mathrm{cm}\) is filled with water. A heated filament running along its axis produces a variable density in the water given by \(\rho(r)=1-0.05 e^{-0.01 r^{2}} \mathrm{g} / \mathrm{cm}^{3}(\rho\) stands for density here, not for the radial spherical coordinate). Find the mass of the water in the cylinder. Neglect the volume of the filament.

A cake is shaped like a hemisphere of radius 4 with its base on the \(x y\)-plane. A wedge of the cake is removed by making two slices from the center of the cake outward, perpendicular to the \(x y\)-plane and separated by an angle of \(\varphi\). a. Use a double integral to find the volume of the slice for \(\varphi=\pi / 4 .\) Use geometry to check your answer. b. Now suppose the cake is sliced horizontally at \(z=a>0\) and let \(D\) be the piece of cake above the plane \(z=a\). For what approximate value of \(a\) is the volume of \(D\) equal to the volume in part (a)?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

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.