Posted: 3/10/2010 10:22:26 AM EDT
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a upside down circular cone (think traffic cone has a height of 3, radius 2. Whats is its volume. I came up with 4pi. solve this using integration.
I posed this question wrong. Find the volume of the area of the function y=3x/2 bounded by y=0 and y=3 after you rotate it around the Y-axis |
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Quoted:
A traffic cone is tapered. If your not using a tapered cone than you're trying to figure cylindrical volume. Your formula would be V= Pi X r squared X h. Could be a farggin trick question though if you don't specify ID or OD on r  A good turd is also tapered and will not allow one's bunghole to slam shut. |
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Since help has already been provided I thought I'd take this opportunity to pimp my Math / Science forum idea instead. http://www.ar15.com/forums/topic.html?b=1&f=2&t=999906 |
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Originally Posted By Bubbles: Using integration: V = ∫Adr A = 2πr r ranges from 0 to 2 V = ∫(2πr)dr from 0 to 2 V = πr^2 from 0 to 2 V = 4π I would have thought it more along the lines of this: V = Integral over a->b( pi * r^2 ) dx = Integral over a->b( pi * [f(x)]^2 ) dx = Integral over 0->3( pi * [3x/2]^2 ) dx = 9pi/4 Integral over 0->3( x^2 ) dx = 3pi/4 * x^3 evaluated over 0->3 = 48pi? GAH!!!! Sorry - I messed up - I rotated around the X-axis, which is incorrect. Here's the right way with INTEGRATION, as the OP asked... y = f(x) = 3x/2 x = f(y) = 2y/3 V = Integral over a->b( pi * r^2 ) dy = Integral over a->b( pi * [f(y)]^2 ) dy = Integral over 0->3( pi * [2y/3]^2 ) dy = 4pi/27 Integral over 0->3( y^2 ) dy = 4pi/27 * x^3 evaluated over 0->3 = 4pi/27 * 27 = 4pi |
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Quoted: Quoted: Using integration: V = ∫Adr A = 2πr r ranges from 0 to 2 V = ∫(2πr)dr from 0 to 2 V = πr^2 from 0 to 2 V = 4π I would have thought it more along the lines of this: V = Integral over a->b( pi * r^2 ) dx = Integral over a->b( pi * [f(x)]^2 ) dx = Integral over 0->3( pi * [3x/2]^2 ) dx = 9pi/4 Integral over 0->3( x^2 ) dx = 3pi/4 * x^3 evaluated over 0->3 = 48pi? GAH!!!! Sorry - I messed up - I rotated around the X-axis, which is incorrect. Here's the right way with INTEGRATION, as the OP asked... y = f(x) = 3x/2 x = f(y) = 2y/3 V = Integral over a->b( pi * r^2 ) dy = Integral over a->b( pi * [f(y)]^2 ) dy = Integral over 0->3( pi * [2y/3]^2 ) dy = 4pi/27 Integral over 0->3( y^2 ) dy = 4pi/27 * x^3 evaluated over 0->3 = 4pi/27 * 27 = 4pi Nevermind, I was rotating it around the x axis too, derp |
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Hey man. Lazy is how shit gets done. At least in my line of work. |
I keep getting 8pi. I'm not sure how you're getting your 27 Zhukov. I don't see a cube anywhere.  I'm using a bar napkin that has a ring of Guiness on it. ETA: Nevermind, I forgot the 1/3 in the formula.  I also wrote A instead of V. Whatever. It's the beer, I swear! Maybe an easier way is like this using dr & dy:  
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Quoted: I keep getting 8pi. I'm not sure how you're getting your 27 Zhukov. I don't see a cube anywhere. I'm using a bar napkin that has a ring of Guiness on it. ETA: Nevermind, I forgot the 1/3 in the formula. I also wrote A instead of V. Whatever. It's the beer, I swear! Maybe an easier way is like this using dr & dy: http://razoreye.net/forum_files/formula1.jpg http://razoreye.net/forum_files/formula2.jpg Yeah - the integral of y^2 is 1/3*y^3. That gets you 4pi/27 after you pull the constants outside of the integral. [ETA] What you did doesn't quite look right though. You started with the formula for the volume of a cylinder (1/3pi r^2 h), and integrated that? You got the right answer, but I don't see how that works. |