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$r(x)$

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Calculus II

Definition

$r(x)$ represents the radius of a cross-section of a solid when determining volumes by slicing. This function is crucial as it defines the shape and size of the individual slices that make up the volume of a three-dimensional object. Understanding how to compute and manipulate $r(x)$ allows for accurate volume calculations using various methods, including the disk and washer methods.

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5 Must Know Facts For Your Next Test

  1. $r(x)$ is often expressed as a function that varies depending on the position along the axis of integration, allowing for dynamic cross-section sizes.
  2. When using the disk method, $r(x)$ is squared in the integral to calculate the area of each circular slice: $$A = \\pi [r(x)]^2$$.
  3. For shapes with holes, like tubes, $r(x)$ can represent both an outer radius and an inner radius when applying the washer method.
  4. Graphing $r(x)$ provides visual insights into how changes in the radius affect the overall volume calculation.
  5. $r(x)$ can be derived from equations of curves or surfaces that bound the solid, showcasing its role in connecting geometric shapes with algebraic functions.

Review Questions

  • How does the function $r(x)$ influence the choice between using the disk method and the washer method for calculating volume?
    • $r(x)$ plays a critical role in determining whether to use the disk or washer method by indicating whether there is an empty space within the solid. If there is no hole in the solid, $r(x)$ can directly represent the radius for a single solid cross-section, making the disk method appropriate. However, if there are inner and outer radii due to hollow sections, both outer and inner radii must be considered for accurate volume calculations using the washer method.
  • Explain how to set up an integral using $r(x)$ for calculating volume via slicing with both disk and washer methods.
    • To set up an integral using $r(x)$ for volume calculation with the disk method, you first identify $r(x)$ as the radius of your solid's cross-section and square it to find its area: $$A = \\pi [r(x)]^2$$. Then, integrate this area from a lower limit to an upper limit along your chosen axis. For the washer method, identify both outer radius $R(x)$ and inner radius $r(x)$. The area becomes $$A = \\pi [R(x)]^2 - \\pi [r(x)]^2$$, which you integrate over the same limits to find volume.
  • Evaluate how changes in $r(x)$ affect overall volume calculations and give examples of how this might impact real-world applications.
    • Changes in $r(x)$ can significantly impact volume calculations because they alter the size of each cross-section being integrated. For example, if $r(x)$ represents a tapering shape like a cone where $r(x)$ decreases linearly, this leads to a smaller volume than if $r(x)$ were constant. In real-world applications, such as designing tanks or vessels that change in diameter, accurately modeling $r(x)$ ensures that engineers can predict capacity and material requirements correctly. Understanding these relationships allows for more efficient designs in various fields such as manufacturing and architecture.

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