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Chapter 6: Problem 9

Find the volume of solids The solid lies between planes perpendicular to the \(y\) -axis at \(y=0\) and \(y=2 .\) The cross-sections perpendicular to the \(y\) -axis are circular disks with diameters running from the \(y\) -axis to the parabola \(x=\sqrt{5} y^{2}\)

Short Answer

Expert verified
The volume of the solid is \( 8\pi \).

Step by step solution

01

Determine the Diameter of the Cross-Sectional Disks

The diameter of each circular cross-section as a function of \( y \) runs from the \( y \)-axis to the parabola \( x = \sqrt{5} y^2 \). Therefore, the diameter is \( x = \sqrt{5} y^2 \).
02

Calculate the Radius of the Disks

The radius \( r(y) \) of each disk is half of the diameter. Thus, \( r(y) = \frac{1}{2} \sqrt{5} y^2 \).
03

Set Up the Formula for the Volume of a Disk

The volume \( V \) of a thin disk of thickness \( dy \) is given by \( V = \pi [r(y)]^2 dy \). Substituting our expression for \( r(y) \), we have \( V = \pi \left( \frac{1}{2} \sqrt{5} y^2 \right)^2 dy \).
04

Simplify the Expression for the Volume of Each Disk

Simplifying, \( V = \pi \left( \frac{1}{2} \sqrt{5} y^2 \right)^2 dy = \pi \cdot \frac{5}{4} y^4 dy = \frac{5\pi}{4} y^4 dy \).
05

Integrate to Find the Total Volume

To find the total volume, integrate the expression from \( y=0 \) to \( y=2 \): \[ V = \int_0^2 \frac{5\pi}{4} y^4 dy \].
06

Evaluate the Integral

Calculate the integral: \[ V = \frac{5\pi}{4} \int_0^2 y^4 dy = \frac{5\pi}{4} \left[ \frac{y^5}{5} \right]_0^2 \]. When evaluated, this becomes \[ \frac{5\pi}{4} \times \frac{32}{5} = 8\pi \].

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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 key to visualizing how a solid is formed by slicing it into shapes. Here, we're dealing with circular disks because the cross-sections are perpendicular to the y-axis and form circles. The diameter of these disks depends on the distance from the y-axis to the parabola, as given by the function \( x=\sqrt{5}y^2 \). To find the radius, which is essential for calculating area, we take half of the diameter length. So, the expression for the radius \( r(y) \) of each cross-section is \( \frac{1}{2}\sqrt{5}y^2 \).
  • The cross-sectional area of a circle is computed using the formula \( \pi r^2 \).
  • For this problem, each disk cross-section has an area of \( \pi \left( \frac{1}{2}\sqrt{5}y^2 \right)^2 \).
This enables us to precisely compute the volume by stacking these infinitesimally thin disks together.
Integration
Integration is a mathematical technique used to calculate the entirety of something, like areas under curves or the total volume occupied by a solid. In this scenario, we use integration to sum up the volumes of countless disks from \( y=0 \) to \( y=2 \).We start by setting up the integral for the given expression: \( V = \int_0^2 \frac{5\pi}{4} y^4 dy \). By integrating, we are effectively adding up the volume of each small disk to find the total volume of the entire solid.
  • Integration requires calculating the antiderivative of the function within the integral.
  • Here, the antiderivative of \( y^4 \) is \( \frac{y^5}{5} \).
  • By applying the limits from \( y=0 \) to \( y=2 \), the computation involves substituting these values into the antiderivative.
This process will give us the complete volume of the solid.
Disk Method
The disk method is a strategy used in calculus to find volumes of solids of revolution. In this method, the solid is visualized as being made up of circular disks stacked against each other, similar to pancakes.For this problem, since the cross-sectional areas are circles (diski) with varying radii as we move along the y-axis, it is apt to employ the disk method.
  • The volume \( V \) of one disk is calculated using \( \pi [r(y)]^2 \) multiplied by the disk's height, \( dy \).
  • The height \( dy \) represents an infinitesimally small thickness.
  • All disks from the first slice at \( y=0 \) to the last at \( y=2 \) are then integrated to find the solid鈥檚 total volume.
This method is particularly useful when dealing with solids of rotation around an axis, as in this scenario, where the diameter of the disks is modeled by the equation of a parabola.

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Most popular questions from this chapter

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