/*! 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 4 Find \(\iiint_{W}(x z+y) d V\) i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 4

Find \(\iiint_{W}(x z+y) d V\) if \(W\) is the solid region in \(\mathbb{R}^{3}\) above the \(x y\) -plane bounded by the surface \(y=x^{2}\) and the planes \(z=y, y=1,\) and \(z=0\).

Short Answer

Expert verified
The integral evaluates to \( \frac{4}{7} \).

Step by step solution

01

Identify Region Boundaries

The solid region \( W \) is bounded by several geometrical features: \( y = x^2 \) (a parabolic surface), the planes \( z = y \) and \( z = 0 \) (horizontal planes), and \( y = 1 \) (a vertical plane). We start by expressing the limits of integration in terms of these conditions.
02

Set up the Triple Integral

For the given solid \( W \), the order of integration is \( dz \, dy \, dx \). Thus, for each fixed \( x \):1. \( z \) varies from \( 0 \) to \( y \).2. \( y \) varies from \( x^2 \) to \( 1 \).3. \( x \) varies from \(-1 \) to \( 1 \) (since \( y = x^2 \) is bounded at \( y = 1 \), solving \( x^2 = 1 \), gives \( x = \pm 1 \)).The integral setup is:\[\iiint_{W}(x z+y) \, dV = \int_{-1}^{1} \int_{x^2}^{1} \int_{0}^{y} (x z + y) \, dz \, dy \, dx\]
03

Integrate with Respect to \( z \)

Begin with the innermost integral:\[\int_{0}^{y} (x z + y) \, dz = \int_{0}^{y} x z \, dz + \int_{0}^{y} y \, dz\]The first part is:\[\int_{0}^{y} x z \, dz = x \left[ \frac{z^2}{2} \right]_0^y = x \frac{y^2}{2}\]The second part is:\[\int_{0}^{y} y \, dz = y [z]_0^y = y^2\]Combine these results:\[x \frac{y^2}{2} + y^2 = y^2 \left( \frac{x}{2} + 1 \right)\]
04

Integrate with Respect to \( y \)

Substitute the result of the \( z \) integration into the \( y \) integral:\[\int_{x^2}^{1} y^2 \left( \frac{x}{2} + 1 \right) \, dy\]This becomes:\[\left( \frac{x}{2} + 1 \right) \int_{x^2}^{1} y^2 \, dy = \left( \frac{x}{2} + 1 \right) \left[ \frac{y^3}{3} \right]_{x^2}^{1}\]Evaluate the bounds:\[\left( \frac{x}{2} + 1 \right) \left( \frac{1^3}{3} - \frac{(x^2)^3}{3} \right)= \left( \frac{x}{2} + 1 \right) \left( \frac{1}{3} - \frac{x^6}{3} \right)\]
05

Integrate with Respect to \( x \)

Compute the outermost integral:\[\int_{-1}^{1} \left( \frac{x}{2} + 1 \right) \left( \frac{1}{3} - \frac{x^6}{3} \right) \, dx\]Expanding, we solve:\[\int_{-1}^{1} \left( \frac{x}{6} + \frac{1}{3} - \frac{x^7}{6} - \frac{x^6}{3} \right) \, dx\]Each term:- \(\int_{-1}^{1} \frac{x}{6} \, dx = 0\) because it's an odd function over symmetric limits.- \(\int_{-1}^{1} \frac{1}{3} \, dx = \left[ \frac{x}{3} \right]_{-1}^{1} = \frac{2}{3}\)- \(\int_{-1}^{1} \frac{x^7}{6} \, dx = 0\) because it's an odd function over symmetric limits.- \(\int_{-1}^{1} \frac{x^6}{3} \, dx = \left[ \frac{x^7}{21} \right]_{-1}^{1} = \frac{2}{21}\)Combine these to find:\[ \frac{2}{3} - \frac{2}{21} = \frac{14}{21} - \frac{2}{21} = \frac{12}{21} = \frac{4}{7}\]
06

Final Result

The final result of the integral over the solid region \( W \) is \( \frac{4}{7} \).

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.

Solid Region
When tackling problems involving triple integrals, it's crucial to first understand the solid region, or the volume in space defined by the problem. In this exercise, the solid region \( W \) is bounded in \( \mathbb{R}^3 \). This means we are working in a three-dimensional space, and the region is encapsulated by various surfaces and planes.
  • The parabolic surface \( y = x^2 \) creates a shape that limits how far the region extends in the \( y \) direction relative to \( x \).
  • The plane \( y = 1 \) acts as a vertical boundary. This tells us that for \( y \), values are only considered from where the parabolic boundary starts, up to 1.
  • Horizontally, the boundaries are created by the planes \( z = 0 \) and \( z = y \). \( z = y \) means the solid can rise vertically up to the value of \( y \), and \( z = 0 \) keeps it above the \( xy \)-plane.
By recognizing these boundaries, we define the limits over which we will integrate our function. It's like understanding the walls of a room that we're going to fill with paint, making sure not to exceed them.
Limits of Integration
Once the solid region is identified, establishing the correct limits of integration is the next step. Integrating a function over a region means summing an infinite number of small values over an area, or in this case, a volume.
Here, the problem specifies to use the order \( dz \, dy \, dx \):
  • The innermost integral limits \( z \) between 0 and \( y \). This is dictated by the planes \( z = 0 \) and \( z = y \), which constrict \( z \) to evolve vertically from the base up to the height at \( y \).
  • The second integral, \( y \), spans from \( x^2 \) to 1. This respects the curve \( y = x^2 \) as the lower boundary and the plane \( y = 1 \) as the upper limit. This establishes how \( y \) can spread laterally depending on \( x \).
  • Finally, \( x \) varies from -1 to 1. These are the solutions for \( y = x^2 \) when \( y = 1 \), showing the lateral reach of the parabolic surface.
The sequence \( dz \, dy \, dx \) is important. It specifies the order for evaluating the integrals, and each step builds upon the preceding integral. Comprehending these boundaries allows solving the triple integral accurately.
Parabolic Surface
In this type of problem, the parabolic surface \( y = x^2 \) plays a huge part. Let's explore what this means in the real world. The surface is vital for defining how the solid region extends in the coordinate plane.
  • The equation \( y = x^2 \) describes a parabola in the \( xy \)-plane that opens upwards. Its vertex, or the turning point, is at the origin \( (0, 0) \).
  • This parabola essentially curtails the region from expanding arbitrarily in the \( x \) direction, forming a symmetric boundary that restricts \( y \) from being less than \( x^2 \) at any point between \(-1\) and \(1\).
  • A crucial observation is how it relates directly to the chosen limits for \( x \) in the integral. Solving \( y = x^2 \) when \( y = 1 \) gives us the endpoints for \( x \) as \( \pm1 \).
Understanding this parabolic boundary allows us to visualize how the three-dimensional integral behaves, emphasizing the coordinated effort between all limits of integration to define the precise volume of integration.

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

A small cookie has the shape of the region in the first quadrant bounded by the curves \(y=x^{2}\) and \(x=y^{2},\) where \(x\) and \(y\) are measured in inches. Chocolate is poured unevenly on top of the cookie in such a way that the density of chocolate at the point \((x, y)\) is given by \(f(x, y)=100(x+y)\) grams per square inch. Find the total mass of chocolate on the cookie.

Let \(R=[0,1] \times[0,1],\) and let \(f: R \rightarrow \mathbb{R}\) be the function given by: $$ f(x, y)=\left\\{\begin{array}{ll} 1 & \text { if }(x, y)=\left(\frac{1}{2}, \frac{1}{2}\right), \\ 0 & \text { otherwise } \end{array}\right. $$ (a) Let \(R\) be subdivided into a 3 by 3 grid of subrectangles of equal size (Figure 5.36 ). What are the possible values of Riemann sums based on this subdivision? (Different choices of sample points may give different values.) (b) Repeat for a 4 by 4 grid of subrectangles of equal size. (c) How about for an \(n\) by \(n\) grid of subrectangles of equal size? (d) Let \(\delta\) be a positive real number. If \(R\) is subdivided into a grid of subrectangles of dimensions \(\triangle x_{i}\) by \(\triangle y_{j},\) where \(\triangle x_{i}<\delta\) and \(\triangle y_{j}<\delta\) for all \(i, j,\) show that any Riemann sum based on the subdivision lies in the range \(0 \leq \sum_{i, j} f\left(\mathbf{p}_{i j}\right) \Delta x_{i} \Delta y_{j}<4 \delta^{2}\). (e) Show that \(\iint_{R} f(x, y) d A=0,\) i.e., that \(\lim _{\triangle x_{i} \rightarrow 0 \atop \triangle y_{j} \rightarrow 0} \sum_{i, j} f\left(\mathbf{p}_{i j}\right) \Delta x_{i} \Delta y_{j}=0 .\) Technically, this means you need to show that, given any \(\epsilon>0,\) there exists a \(\delta>0\) such that, for any subdivision of \(R\) into subrectangles with \(\triangle x_{i}<\delta\) and \(\Delta y_{j}<\delta\) for all \(i, j\), every Riemann sum based on the subdivision satisfies \(\left|\sum_{i, j} f\left(\mathbf{p}_{i j}\right) \Delta x_{i} \Delta y_{j}-0\right|<\epsilon\).

Find the volume of the pyramid-shaped solid in the first octant bounded by the three coordinate planes and the planes \(x+z=1\) and \(y+2 z=2\).

The sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) of radius \(a\) sits inscribed in the circular cylinder \(x^{2}+y^{2}=a^{2}\) Given real numbers \(c\) and \(d\), where \(-a \leq c

Here is a result that will come in handy increasingly as we calculate more integrals. Let \(R=[a, b] \times[c, d]\) be a rectangle, and let \(F: R \rightarrow \mathbb{R}\) be a real-valued function such that the variables \(x\) and \(y\) separate into two continuous factors, that is, \(F(x, y)=f(x) g(y)\), where \(f:[a, b] \rightarrow \mathbb{R}\) and \(g:[c, d] \rightarrow \mathbb{R}\) are continuous. Show that: $$ \iint_{R} f(x) g(y) d x d y=\left(\int_{a}^{b} f(x) d x\right)\left(\int_{c}^{d} g(y) d y\right). $$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

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.