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Calc 2 solids of revolution help

  • Thread starter kellyb1ll
  • Start date
  • #1
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A solid lies between planes perpendicular to the x-axis at x=-a and x=a for values of a>0 to be given below in parts (i) and (ii). In each case the cross-sections perpendicular to the x-axis between these planes run from the semicircle y=√(a^2-x^2) to the semicircle
y=-√(a^2-x^2).

If a=7 and the cross-sections are equilateral triangles with bases in the x-y plane, find a formula for the area A(x) of the cross-section at location x.


For the base i used 2*√(a^2-x^2), for the height i used √(a^2-x^2)/tan(30). so i get
(a^2-x^2)/tan(30) as an answer, which is not right. what am i doing wrong i cant seem to figure it out.

i uploaded a picture of what the problem gave me if it helps.
THANKS!!!
 

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Answers and Replies

  • #2
92
8
Why do you think that is not right?
 
  • #3
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its a problem on my online homework and when i type in that answer it says its wrong
 
  • #4
verty
Homework Helper
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You made a mistake finding the height of the triangle.
 

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