The base of S is a circular disk with radius 4r. Parallel cross-sections perpendicular to the base are squares. Find the volume V of this solid. I'd write out my work on this problem, but I don't know how to use the thing that lets you do that.
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Originally Posted by Hellacious D The base of S is a circular disk with radius 4r. Parallel cross-sections perpendicular to the base are squares. Find the volume V of this solid. I'd write out my work on this problem, but I don't know how to use the thing that lets you do that. side of a square cross-section perpendicular to the x-axis = area of a square cross-section =
I'm getting (-512r^3)/3 which is what I've gotten, but webassign says no deal. I've tried (-512x^3)/3 and -512/3 with no success.
Originally Posted by Hellacious D I'm getting (-512r^3)/3 which is what I've gotten, but webassign says no deal. I've tried (-512x^3)/3 and -512/3 with no success. show how you arrived at this "no deal" solution.
How do I write out my work? I can do it in text, but what's the program you guys use on here to make it look nice?
Originally Posted by Hellacious D How do I write out my work? I can do it in text, but what's the program you guys use on here to make it look nice? http://www.mathhelpforum.com/math-help/latex-help/
eh, no time to learn that today. 2*(the integral from 0 to 4r) of 4(16(r^2)-(x^2))dx 2*(4(16(r^2)-((x^3)/3)) evaluated from 0 to 4r 8(16(r^2)-(64r^3)/3) - 8(16r^2) (128r^2)-((512r^3)/3)-128r^2 (-512r^3)/3
Originally Posted by Hellacious D eh, no time to learn that today. 2*(the integral from 0 to 4r) of 4(16(r^2)-(x^2))dx 2*(4(16(r^2)x - ((x^3)/3)) evaluated from 0 to 4r is a constant ... antiderivative is
Thank you much, I don't know why I kept missing that.
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