ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: ben4386 on August 14, 2008, 05:24:40 pm

Title: 2004 Exam 1 MC 16
Post by: ben4386 on August 14, 2008, 05:24:40 pm

Hi guys, Im doing found this question in an old exam and I got the wrong answer, I know that my answer is finding the volume of the inner part rotated around the y axis but im having a little trouble finding the area in terms of an intergral from the y axis,

here is the q

(http://i184.photobucket.com/albums/x214/ben4386/untitled.jpg)

any help would be greatly appreciated

Title: Re: 2004 Exam 1 MC 16
Post by: cara.mel on August 14, 2008, 05:45:51 pm

Rearrange to get into form x = f(y)
y = sqrt(x^2-9)
y^2 = x^2 - 9
y^2 + 9 = x^2

Formula for solid of revolution is pi * int x^2 dy, and we want the outside part: ie, the area between the line y = 5 and the curve y^2 + 9 = x^2
therefore pi * int (5^2 - (y^2 + 9) dy)
= pi * int(16 - y^2)
Now I hope that is right because I'm too lazy to check vcaa :P

Title: Re: 2004 Exam 1 MC 16
Post by: ben4386 on August 14, 2008, 07:23:54 pm

thanks heaps!