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Chapter 6: Problem 1

Find the volumes of the solids. The solid lies between planes perpendicular to the \(x\) -axis at \(x=0\) and \(x=4 .\) The cross-sections perpendicular to the axis on the interval \(0 \leq x \leq 4\) are squares whose diagonals run from the parabola \(y=-\sqrt{x}\) to the parabola \(y=\sqrt{x}\).

Short Answer

Expert verified
The volume of the solid is 16 cubic units.

Step by step solution

01

Analyze the Problem

We are given a region where the cross-sections perpendicular to the x-axis are squares. The diagonals of these squares extend from the parabola \(y = -\sqrt{x}\) to \(y = \sqrt{x}\), situated between \(x = 0\) and \(x = 4\). The key task is to find the volume of the solid formed by these squares.
02

Calculate the Diagonal of the Square

The distance from the parabola \(y = -\sqrt{x}\) to \(y = \sqrt{x}\) is the diagonal of the square. This distance is given by calculating \(\sqrt{x} - (-\sqrt{x}) = 2\sqrt{x}\).
03

Determine the Side Length of the Square

The relationship between the diagonal \(d\) and the side \(s\) of the square is given by \(d = s\sqrt{2}\). Thus, we solve for \(s\): \(s = \frac{d}{\sqrt{2}} = \frac{2\sqrt{x}}{\sqrt{2}} = \sqrt{2x}\).
04

Calculate the Area of the Cross-sectional Square

The area \(A\) of a square with side \(s\) is \(s^2\). Thus, \(A = (\sqrt{2x})^2 = 2x\).
05

Integrate to Find the Volume

To find the volume \(V\), integrate the area of the square from \(x = 0\) to \(x = 4\): \[ V = \int_{0}^{4} A \ dx = \int_{0}^{4} 2x \ dx \]Calculate this integral: \[ V = 2 \int_{0}^{4} x \ dx = 2 \left[ \frac{x^2}{2} \right]_{0}^{4} = \left[ x^2 \right]_{0}^{4} = 16 \ - 0 = 16 \].
06

State the Final Volume

The volume of the solid is derived from the integration calculation, resulting in a value of 16 cubic units.

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Key Concepts

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

Cross-sectional Area
Understanding the cross-sectional area is crucial when finding the volume of solids with specific shapes. In this problem, the cross-sections are squares. These squares are formed perpendicularly on the interval from \(x = 0\) to \(x = 4\). The squares' diagonals stretch between the parabolas \(y = -\sqrt{x}\) and \(y = \sqrt{x}\). This gives the diagonal length as \(2\sqrt{x}\). Once you know the diagonal, you can find the side length using the relationship between a square's diagonal and side: \(d = s\sqrt{2}\). Solving this gives the side \(s = \sqrt{2x}\). The area of the square, which is the cross-sectional area, is then \(A = s^2 = (\sqrt{2x})^2 = 2x\). This area equation \(2x\) becomes integral as it's needed to find the volume of the solid.
Integration in Calculus
Integration is a powerful concept in calculus used to calculate various properties such as area and volume. In this exercise, integration is necessary to find the total volume of the solid created from the square cross-sections.The solid鈥檚 volume \(V\) can be determined by integrating the cross-sectional area over the given interval from \(x = 0\) to \(x = 4\). The integral becomes \[ V = \int_{0}^{4} 2x \, dx. \]This integral represents adding up all the infinitesimal cross-sectional areas \(2x\) along the \(x\)-axis. Solving the integral:
  • Find the antiderivative of \(2x\), which is \(x^2\).
  • Evaluate the definite integral from \(0\) to \(4\).
  • Compute the difference \( \left[ x^2 \right]_{0}^{4} = 16 - 0 = 16 \).
Consequently, the final volume is confirmed to be 16 cubic units, thanks to successfully performing this integration.
Parabola
Parabolas are curves that can be represented by quadratic equations, and in this problem, they serve as the boundaries between which the squares' diagonals extend. Specifically, we have the parabolas described by \(y = -\sqrt{x}\) and \(y = \sqrt{x}\).These are reflections of each other and form symmetric curves about the \(x\)-axis. As \(x\) increases, the \(y\)-values represent half the length of the diagonal across the middle of these squares. At each \(x\) value, the distance between these parabolas is the full diagonal \(2\sqrt{x}\).Understanding how the parabola affects the formation and size of the square cross-sections is crucial. As \(x\) changes, this diagonal dictates the size of the square and, consequently, the area that gets integrated to find the total volume.

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Most popular questions from this chapter

Find the volume of the solid generated by revolving each region about the \(y\) -axis. The region enclosed by the triangle with vertices (0,1),(1,0) and ( 1,1 )

Use the shell method to find the volumes of the solids generated by revolving the regions bounded by the given curves about the given lines. \(y=x^{3}, \quad y=8, \quad x=0\) a. The \(y\) -axis b. The line \(x=3\) c. The line \(x=-2\) d. The \(x\) -axis e. The line \(y=8\) f. The line \(y=-1\)

Find the arc length function for the graph of \(f(x)=2 x^{3 / 2}\) using (0, 0) as the starting point. What is the length of the curve from (0,0) to (1,2)\(?\)

The strength of Earth's gravitational field varies with the distance \(r\) from Earth's center, and the magnitude of the gravitational force experienced by a satellite of mass \(m\) during and after launch is $$F(r)=\frac{m M G}{r^{2}}.$$ Here, \(M=5.975 \times 10^{24} \mathrm{kg}\) is Earth's mass, \(G=6.6720 \times\) \(10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} \mathrm{kg}^{-2}\) is the universal gravitational constant, and \(r\) is measured in meters. The work it takes to lift a 1000 -kg satellite from Earth's surface to a circular orbit \(35,780 \mathrm{km}\) above Earth's center is therefore given by the integral $$\text { Work }=\int_{6,370,000}^{35,780,000} \frac{1000 M G}{r^{2}} d r \text { joules. }$$ Evaluate the integral. The lower limit of integration is Earth's radius in meters at the launch site. (This calculation does not take into account energy spent lifting the launch vehicle or energy spent bringing the satellite to orbit velocity.)

The viewing portion of the rectangular glass window in a typical fish tank at the New England Aquarium in Boston is \(1.6 \mathrm{m}\) wide and runs from \(0.01 \mathrm{m}\) below the water's surface to \(0.85 \mathrm{m}\) below the surface. Find the fluid force against this portion of the window. The weight-density of seawater is \(10,050 \mathrm{N} / \mathrm{m}^{3} .\) (In case you were wondering, the glass is \(2 \mathrm{cm}\) thick and the tank walls extend \(10 \mathrm{cm}\) above the water to keep the fish from jumping out.)
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