A square of side length s lies in a plane perpendicular to a line L. One vertex of the square...

Question:

A square of side length s lies in a plane perpendicular to a line L. One vertex of the square lies on L. As this square moves a distance h along L, the square turns one revolution about L to generate a corkscrew-like column with square cross-sections.

a) Find the volume of the column.

b) What will the volume be if the square turns twice instead of once? Give reasons for your answer.

Volume:

A quantity that represents the space occupies by coordinate with the closed surface is known as volume. The area in which substance or fluid occupies space up to some height is volume. Its measurable unit is a cube of the meter.

Answer and Explanation:

Given Data:

  • The side of square of square is: {eq}s {/eq}

\*The distance move by square is: {eq}h {/eq}


(a)

The expression for area of square is

{eq}A = {s^2} {/eq}


The expression for volume of column for small distance {eq}dx {/eq} is

{eq}dV = Adx {/eq}


Integrate the above expression from {eq}0 {/eq} to {eq}h {/eq}

{eq}\begin{align*} \int {dV} &= \int_0^h {Adx} \\ V &= {s^2}\left[ x \right]_0^h\\ &= {s^2}\left[ {h - 0} \right]\\ &= {s^2}h \end{align*} {/eq}


Thus the volume of the column is {eq}{s^2}h {/eq}


(b)

The volume become half for each turn of volume of square when it turn twice instead of once

The expression for volume when square turn twice is

{eq}{V_o} = \dfrac{V}{2} + \dfrac{V}{2} {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} {V_o} &= \dfrac{{{s^2}h}}{2} + \dfrac{{{s^2}h}}{2}\\ &= {s^2}h \end{align*} {/eq}


Thus the volume of turn square remain same and it is {eq}{s^2}h {/eq}


Learn more about this topic:

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Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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