/*! 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 30 Use a double integral to find th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 30

Use a double integral to find the volume. The volume under the surface \(z=3 x^{3}+3 x^{2} y\) and over the rectangle \(R=\\{(x, y): 1 \leq x \leq 3,0 \leq y \leq 2\\}\)

Short Answer

Expert verified
The volume is 81 cubic units.

Step by step solution

01

Understand the Problem

We need to find the volume under the surface given by the equation \(z = 3x^3 + 3x^2y\) and above the rectangle \(R\) which is defined in the xy-plane as \(1 \leq x \leq 3\) and \(0 \leq y \leq 2\). To find this volume, we will use a double integral over the region \(R\).
02

Set Up the Double Integral

The expression for the volume \(V\) under the surface and over the rectangle can be written as a double integral: \[ V = \int_{1}^{3} \int_{0}^{2} (3x^3 + 3x^2y) \, dy \, dx \] Here, the bounds reflect the rectangle \(R\), and the function being integrated is the given surface equation \(z\).
03

Integrate with Respect to y

First, integrate the function \(3x^3 + 3x^2y\) with respect to \(y\), treating \(x\) as a constant: \[ \int_{0}^{2} (3x^3 + 3x^2y) \, dy = \left[ 3x^3y + \frac{3x^2y^2}{2} \right]_{0}^{2} \] When evaluated, this becomes: \[ \left( 3x^3 \cdot 2 + \frac{3x^2 \cdot 4}{2} \right) - \left( 3x^3 \cdot 0 + \frac{3x^2 \cdot 0}{2} \right) = 6x^3 + 6x^2 \]
04

Integrate with Respect to x

Next, integrate \(6x^3 + 6x^2\) with respect to \(x\): \[ \int_{1}^{3} (6x^3 + 6x^2) \, dx \] Calculate the integral: \[ = \left[ \frac{6x^4}{4} + \frac{6x^3}{3} \right]_{1}^{3} \] This simplifies to: \[ = \left[ \frac{3x^4}{2} + 2x^3 \right]_{1}^{3} \]
05

Evaluate the Final Expression

Substitute the upper and lower bounds into \(\left[ \frac{3x^4}{2} + 2x^3 \right]_{1}^{3}\): \[ \left( \frac{3 \cdot 3^4}{2} + 2 \cdot 3^3 \right) - \left( \frac{3 \cdot 1^4}{2} + 2 \cdot 1^3 \right) \] Calculate each part: \(\frac{243}{2} + 54 - \frac{3}{2} - 2 = 81\).
06

Conclusion

The volume under the given surface \(z = 3x^3 + 3x^2y\) and over the rectangle \(R\) is 81 cubic units.

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!

Key Concepts

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

Volume Under Surface
When working with double integrals to find the volume under a surface, we have a powerful tool at our disposal. Imagine a surface described by a function, such as \( z = 3x^3 + 3x^2y \), hovering above a section of the xy-plane. The double integral helps us calculate the exact volume of space between this surface and the plane underneath it.

Think of it like this: you are slicing this region into infinite tiny slices, calculating the volume of each slice, and then adding up all these individual volumes. The result is the total volume under the given surface. This method is especially useful when dealing with surfaces that form complex shapes or curves, where straightforward geometrical formulas do not apply.
Rectangular Region
In double integrals, the region over which we integrate plays a crucial role. For this problem, our region is a rectangle defined by the inequalities \(1 \leq x \leq 3\) and \(0 \leq y \leq 2\). This rectangle, residing in the plane defined by the x and y axes, serves as the base over which we calculate the volume under the surface.

The significance of defining a specific region, like a rectangle, is to provide limits for the integration. These bounds, in turn, dictate the range of x and y values over which the function is evaluated. By knowing exactly where the function lies in relation to the coordinate plane, we ensure accuracy in our calculations.
Integration with Respect to Variables
Performing a double integral involves integrating with respect to one variable first, followed by the other. In our example, we integrate the surface function \(3x^3 + 3x^2y\) first with respect to \(y\), treating \(x\) as a constant. Once completed, we then integrate the result with respect to \(x\).

This process, known as repeated integration or iterated integration, makes the problem manageable by breaking it into simpler, step-by-step integrals.
  • Start with the innermost integral and handle one variable at a time.
  • Integrate with respect to \(y\) over its defined interval, which is 0 to 2.
  • Next, take this result and integrate it with respect to \(x\), over its interval from 1 to 3.
By doing so, we systematically find the cumulative volume as influenced by changes in each variable within the specified bounds.

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

The tendency of a solid to resist a change in rotational motion about an axis is measured by its moment of inertia about that axis. If the solid occupies a region \(G\) in an \(x y z\) -coordinate system, and if its density function \(\delta(x, y, z)\) is continuous on \(G,\) then the moments of inertia about the \(x\) -axis, the \(y\) -axis, and the \(z\) -axis are denoted by \(I_{x}, I_{y},\) and \(I_{z},\) respectively, and are defined by $$\begin{aligned} I_{x} &=\iiint_{G}\left(y^{2}+z^{2}\right) \delta(x, y, z) d V \\ I_{y} &=\iiint_{G}\left(x^{2}+z^{2}\right) \delta(x, y, z) d V \\ I_{z} &=\iiint_{G}\left(x^{2}+y^{2}\right) \delta(x, y, z) d V \end{aligned}$$ In these exercises, find the indicated moments of inertia of the solid, assuming that it has constant density \(\delta .\) \(I_{z}\) for the solid cylinder \(x^{2}+y^{2} \leq a^{2}, 0 \leq z \leq h\)

Express the volume of the solid described as a double integral in polar coordinates. $$ \begin{array}{l}{\text { Below } z=\left(x^{2}+y^{2}\right)^{-1 / 2}} \\\ {\text { Outside of } x^{2}+y^{2}=1} \\ {\text { Inside of } x^{2}+y^{2}=9} \\\ {\text { Above } z=0}\end{array} $$

Find the mass and center of gravity of the solid. The solid that has density \(\delta(x, y, z)=x z\) and is enclosed by \(\left.y=9-x^{2} \text { (for } x \geq 0\right), x=0, y=0, z=0,\) and \(z=1\)

Evaluate the iterated integral. $$ \int_{0}^{\pi} \int_{0}^{1-\sin \theta} r^{2} \cos \theta d r d \theta $$

Use spherical coordinates to find the centroid of the solid. The solid in the first octant bounded by the coordinate planes and the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

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.