# Calc 2 solids of revolution help

1. Sep 2, 2013

### kellyb1ll

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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Last edited: Sep 2, 2013
2. Sep 2, 2013

### Jufro

Why do you think that is not right?

3. Sep 2, 2013

### kellyb1ll

its a problem on my online homework and when i type in that answer it says its wrong

4. Sep 3, 2013

### verty

You made a mistake finding the height of the triangle.