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Calc 2 solids of revolution help
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[QUOTE="kellyb1ll, post: 4489887, member: 486736"] 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 can't seem to figure it out. i uploaded a picture of what the problem gave me if it helps. THANKS! [/QUOTE]
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