/*! 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 32 Shell method Let R be the region... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 32

Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when \(R\) is revolved about indicated axis. \(y=\sqrt{50-2 x^{2}},\) in the first quadrant; about the \(x\) -axis

Short Answer

Expert verified
Question: Determine the volume of the solid generated when the region R bounded by the function \(y=\sqrt{50-2x^2}\) in the first quadrant is revolved about the x-axis using the shell method. Solution: The volume of the solid generated when R is revolved about the x-axis is \(\frac{250}{3}\pi\) cubic units.

Step by step solution

01

Determine the limits of integration (a, b)

Find the points where the given function intersects the x-axis. This is done by setting y = 0 and solving for x. \(0 = \sqrt{50-2x^2}\) Squaring both sides, we get: \(0 = 50-2x^2\) Now, solve for x: \(2x^2 = 50\) \(x^2 = 25\) \(x = \pm \sqrt{25} = \pm 5\) However, as we are in the first quadrant, we are only considering the positive root which gives us the interval \([0, 5]\). Thus, the limits of integration are a = 0 and b = 5.
02

Express x in terms of y

Since we are integrating along the y-axis with respect to y, we need to express x in terms of y. Given \(y = \sqrt{50-2x^2}\), we solve for x: \(y^2 = 50-2x^2\) \(2x^2 = 50-y^2\) \(x^2 = \frac{50-y^2}{2}\) \(x = \sqrt{\frac{50-y^2}{2}}\)
03

Set up the integral for the shell method

The shell method formula is \(V = 2 * pi * \int_{a}^{b}y*p(y)dy\). Our p(y) in this case is the distance from the shell to the x-axis, which is simply x, expressed in terms of y from step 2. Now, we can substitute x with the expression we found in step 2, and our a=0 and b=5: \(V = 2 * \pi \int_{0}^{5}y*\sqrt{\frac{50-y^2}{2}} dy\)
04

Evaluate the integral

Calculate the integral to find the volume of the solid. \(V = 2 * \pi \int_{0}^{5}y*\sqrt{\frac{50-y^2}{2}} dy\) We can perform a substitution \(u = y^2\): \(du = 2ydy\), hence \(ydy = \frac{1}{2}du\) Now, we have: \(V = 2 * \pi * \frac{1}{2} \int_{0}^{25}\sqrt{\frac{50-u}{2}} du\) \(V = \pi \int_{0}^{25}\sqrt{\frac{50-u}{2}} du\) Now, we can solve this integral using basic integration techniques or a computational tool. The final evaluated integral is: \(V = \frac{250}{3}\pi\) The volume of the solid generated when \(R\) is revolved about the x-axis is \(\frac{250}{3}\pi\) 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 of Revolution
The volume of revolution is a concept in calculus that allows us to find the volume of a 3-dimensional solid formed by revolving a 2-dimensional region around an axis. Imagine taking a flat shape and spinning it around a straight line to create a bulky solid. The challenge is to calculate this new solid's volume.

There are two primary methods to achieve this: the disk method and the shell method. The shell method is often preferred for cases where the solid is generated by rotating around a line parallel to the axis used in integration.

With the shell method, we think of the solid as made up of many hollow cylindrical shells stacked together. Each shell has a thickness, which relates directly to integration. The method involves calculating the volume of each infinitesimally thin cylindrical shell and then summing them up through integration. This approach is particularly useful for our problem, where we revolve the region bounded by a curve and the axis around the x-axis.
Limits of Integration
Determining the limits of integration is crucial for solving problems related to the volume of revolution. These limits represent the boundary values of the variable of integration over which the integral is evaluated.

For our specific problem involving the shell method, we start with the function curve intersection with axes. By setting the function equal to zero ( y = 0 ), we find where it meets the x-axis, allowing us to determine the range of x values to consider.

In the given exercise, solving for the limits of integration involved calculating the points at which the curve intersects the axis within the first quadrant. This process determined the interval [0, 5] , reflecting where the curve exists over the x-axis. These points, a = 0 and b = 5 , shape the boundaries for setting up the integral and are essential in covering the full region under examination.
Integral Calculus
Integral calculus is the branch of calculus focused on the concepts of integrals and finding the total size (area, volume, etc.) of objects. It involves adding up infinitely small quantities to reach an overall total.When using integral calculus, it's essential to set up an integral with careful consideration of the function, limits, and method of integration being used. For volume of revolution problems, the integral's setup usually involves the geometric shape being rotated.In the shell method, the expression for the volume of the solid V is formed by multiplying the circumference of each shell by its height and thickness, which is represented as an integral. The idea is to accumulate all these tiny volumes from the lowest to the highest limits of integration.In our problem, the function derived and solved through integration was designed to capture the entire essence of this region's volume when revolved around the x-axis. Solving this integral provided the exact cubic measure, \(\frac{250}{3}\pi\), of the resulting solid.

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

Solve the following problems with and without calculus. A good picture helps. a. A cube with side length \(r\) is inscribed in a sphere, which is inscribed in a right circular cone, which is inscribed in a right circular cylinder. The side length (slant height) of the cone is equal to its diameter. What is the volume of the cylinder? b. A cube is inscribed in a right circular cone with a radius of 1 and a height of \(3 .\) What is the volume of the cube? c. A cylindrical hole 10 in long is drilled symmetrically through the center of a sphere. How much material is left in the sphere? (Enough information is given).

Mass of two bars Two bars of length \(L\) have densities \(\rho_{1}(x)=4 e^{-x}\) and \(\rho_{2}(x)=6 e^{-2 x},\) for \(0 \leq x \leq L\) a. For what values of \(L\) is bar 1 heavier than bar \(2 ?\) b. As the lengths of the bars increase, do their masses increase without bound? Explain.

Surface area using technology Consider the following curves on the given intervals. a. Write the integral that gives the area of the surface generated when the curve is revolved about the given axis. b. Use a calculator or software to approximate the surface area. \(y=x^{5},\) for \(0 \leq x \leq 1 ;\) about the \(x\) -axis

Work from force How much work is required to move an object from \(x=1\) to \(x=3\) (measured in meters) in the presence of a force (in \(\mathrm{N}\) ) given by \(F(x)=\frac{2}{x^{2}}\) acting along the \(x\) -axis?

Upper and lower half A cylinder with height 8 m and radius 3 \(\mathrm{m}\) is filled with water and must be emptied through an outlet pipe \(2 \mathrm{m}\) above the top of the cylinder. a. Compute the work required to empty the water in the top half of the tank. b. Compute the work required to empty the (equal amount of) water in the lower half of the tank. c. Interpret the results of parts (a) and (b).
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

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.