2004 Exam 1 MC 16

[ben4386]

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



any help would be greatly appreciated

[cara.mel]

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

[ben4386]

thanks heaps!