shuning
Member
- Joined
- Aug 23, 2008
- Messages
- 654
- Gender
- Male
- HSC
- 2009
-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
moriah college 2001 paper (1 Viewer)
- Thread starter shuning
- Start date
untouchablecuz
Active Member
- Joined
- Mar 25, 2008
- Messages
- 1,693
- Gender
- Male
- HSC
- 2009
to make it simpler, you can consider y>0 and y<0 separately
y>0
dV = pi[(x+dx)^2-x^2]y = pi(x+dx+x)(x+dx-x)[4-(x-3)^2] = 2pix[4-(x-3)^2]dx (after simplifying and disregarding terms in dx^2)
similarly, y<0
dV = pi[(x+dx)^2-x^2]y = pi(x+dx+x)(x+dx-x){sqrt[4-(x-3)^2]} = 2pix{sqrt[4-(x-3)^2]}dx (the sign of y here doesnt matter)
.'. overall dv = 2pix[4-(x-3)^2]dx + 2pix{sqrt[4-(x-3)^2]}dx
solving 0 = 4-(x-3)^2, we get the limits of integration 1--->5
hence,
V = S [2pix[4-(x-3)^2]]dx + S [2pixsqrt[4-(x-3)^2]}dx, from 1 to 5
from a) i) S xsqrt[4-(x-3)^2]}dx = 12(pi)^2
using that and expanding the first part of the integral, you get V
(sorry about the typeface, cbf for latex)
y>0
dV = pi[(x+dx)^2-x^2]y = pi(x+dx+x)(x+dx-x)[4-(x-3)^2] = 2pix[4-(x-3)^2]dx (after simplifying and disregarding terms in dx^2)
similarly, y<0
dV = pi[(x+dx)^2-x^2]y = pi(x+dx+x)(x+dx-x){sqrt[4-(x-3)^2]} = 2pix{sqrt[4-(x-3)^2]}dx (the sign of y here doesnt matter)
.'. overall dv = 2pix[4-(x-3)^2]dx + 2pix{sqrt[4-(x-3)^2]}dx
solving 0 = 4-(x-3)^2, we get the limits of integration 1--->5
hence,
V = S [2pix[4-(x-3)^2]]dx + S [2pixsqrt[4-(x-3)^2]}dx, from 1 to 5
from a) i) S xsqrt[4-(x-3)^2]}dx = 12(pi)^2
using that and expanding the first part of the integral, you get V
(sorry about the typeface, cbf for latex)
kwabon
Banned
yep, very true, just consider them as seperate things and find the volume individually and add them together, and wallah bob is your uncle
and btw shuning, where do u get these papers from, coz i am in desperate need of them, please dont tell me from the resource site, coz i have most of them already
thang ye
you can reduce the amount of calculations by a substitution u = x-3
^2) + \sqrt{4 - (x-3)^2} \right) x \, dx\\ =2\pi\int_{-2}^2 \left( (4-u^2) + \sqrt{4 - u^2} \right) (u+3) \, du \\ = 2\pi\left( \int_{-2}^2 u\left( (4-u^2) + \sqrt{4 - u^2} \right) \, du + 6\pi\left( \int_{-2}^2 \left( (4-u^2) + \sqrt{4 - u^2} \right) \, du)
And the great thing is the first integrand is an odd function.. so the first integral is 0. the 2nd integrand is just the shaded area
And the great thing is the first integrand is an odd function.. so the first integral is 0. the 2nd integrand is just the shaded area
Last edited: