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 November 5th, 2012, 06:07 AM #1 Newbie   Joined: Oct 2011 Posts: 27 Thanks: 0 Rotation of a circle Hello People, I have to solve a question, most of it is done. It is about the function f(x)=Sqr(r^2-x^2) d) for the restriction -r<=x<=r , calculate the volume and compare it to common equations in your equation booklet. i have done that, it gives the normal equation for the volume of a sphere. e) now its restricting the x-axis with -r<=x<=a. For the volume i get this answer: ((r^2)a-(1/3)(a^3))-(2/3)(r^3) BUT, it asks me for common equations again, from my booklet. but neither in the internet nor in my booklet im able to find anything similar. Additionally, it does not make any sense to me, what would this equation calculate? Is there anything or is this just a trick question? Thanks
 November 5th, 2012, 07:03 AM #2 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,767 Thanks: 5 Re: Rotation of a circle How are you computing the volume of a curve?? Are you in calculus, perhaps studying surfaces of revolution???
 November 5th, 2012, 09:34 AM #3 Newbie   Joined: Oct 2011 Posts: 27 Thanks: 0 Re: Rotation of a circle using pi times the defined integral of the squared function. gives me the volume of the rotated circle.
 November 5th, 2012, 11:45 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Rotation of a circle We should get: $V=\pi\int_{-r}^a r^2-x^2\,dx$ where $-r\le a\le r$ $V=\pi$r^2x-\frac{1}{3}x^3$_{-r}^a=\pi$$ar^2-\frac{1}{3}a^3+r^3-\frac{1}{3}r^3$$=\pi$$ar^2-\frac{1}{3}a^3+\frac{2}{3}r^3$$$ Now if we define: $h=a+r\,\therefore\,a=h-r$ we have: $V=\pi$$(h-r)r^2-\frac{1}{3}(h-r)^3+\frac{2}{3}r^3$$=\pi$$hr^2-r^3-\frac{1}{3}\(h^3-3h^2r+3hr^2-r^3$$+\frac{2}{3}r^3\)=$ $V=\frac{\pi}{3}$$3hr^2-3r^3-h^3+3h^2r-3hr^2+r^3+2r^3$$=\frac{\pi h^2}{3}$$3r-h$$$

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