How do I find the area if the base is bounded by y = x^5, y = 1, and x = 0. whil

Question

How do I find the area if the base is bounded by y = x^5, y = 1, and x = 0. while using equilateral triangles perpendicular to the y axis. I got as far as solving for x (x = y^(1/5) )

Explanation / Answer

Draw two vertical lines at x = x and x = x+dx Volume of that thin square slice = (1 - x^2)^2 dx y = 1 => x = 1 Integrating between limits x = 0 to x = 1 ?(x=0 to 1) (1 - 2x^2 + x^4) dx = [x - (2/3)x^3 + (1/5)x^5] (x=0 to 1) = 1 - 2/3 + 1/5 (Edited after Mathsman pointed out the error.) = 8/15. Mathsman : I don't know what has happened to me. This is the second time I had to accept my mistake within half an hour. I think I need to take a few days rest from Yahoo Answers. Thanks for pointing out the error which I am going to correct. Tabula Rasa has already solved your semicircle part. For the equilateral triangle, dV = (v3/4) (1 - x^2)^2 * dx shall give you the result. V = (v3/4) (x - (2/3)x^3 + (1/5)x^5) (x=0 to 1) = (v3/4) (1 - 2/3 + 1/5) = (2v3/15) Perhaps one more way of seeing the equilateral triangle is to consider ( 1 - x^2) as the altitude. In that cas, base will be 2(1 - x^2) cot60° and area = (1 - x^2)^2 * (1/v3) and volume will work out as 8/(15v3). please rate me....

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