Volumes Problem (1 Viewer)

[George W. Bush]

The region bounded by y=sinx (0<= x <= 2pi) and the x-axis is rotated about the line y=c. Find c such that the volume is minimised.

From the diagram I feel the answer is c=0 but proving it...

 

[wogboy]

If we slice the volume vertically the radius^2 of each slice is:

r^2 = (sinx - c)^2
pi*r^2 = pi * (sinx - c)^2
= pi*(sin^2(x) - 2csin(x) + c^2)

so V = Int{0->2*Pi} pi*r^2
= pi*Int{0->2*Pi} sin^2(x) - 2csin(x) + c^2 dx
= pi^2 + 2*Pi^2*c^2

So if you plot V against c, it's a parabola which has a min pt at c=0.

 

[wogboy]

Just something interesting I want to add...

What your doing here is integrating wrt to one variable and differentiating wrt to another. There's a theorem that goes like this:

d/dt Int f dx = Int (df/dt) * dx

so in our case

dV/dc
= d/dc Pi * Int{0->2*Pi} sin^2(x) - 2csin(x) + c^2 dx
= Pi * Int{0->2*Pi} -2sin(x) + 2c dx
= 4*Pi^2*c

so dV/dc = 0 when c = 0, which is consistent with our previous result. This theorem is called Leibniz's Rule.

 

[George W. Bush]

Originally posted by wogboy
If we slice the volume vertically the radius^2 of each slice is:

r^2 = (sinx - c)^2

This is the bit I don't understand. Say we take a sample line of y=1/2 just to illustrate my point - the area above y=1/2 between pi/6 and 5pi/6 does have a radius of (sinx - c)^2, but if we only consider this part, aren't we neglecting the volume beneath y=1/2 from pi/6 -> 5pi/6? Some of this will cancel out, but I'm pretty sure you'll still have some extra volume.

 

[wogboy]

Hmm you have a point there, I didn't read the question completely at first (I thought it was just after the area between y=c and y=sinx, not between y=sinx and the x-axis.

 

[JJ]

Hello everyone.
WARNING: the following is probably bodgy late-night maths, but here goes
oh, and S stands for integral sign

Vi=|([c-sin(xi)]^2-[c-0]^2)|πdx
V=πS([c-sin(x)]^2-c^2)dx
V=πS(sin^2(x)-2csin(x))dx
V=πS(1/2[1-cos(2x)]-2csin(x))dx
V=π[x/2-1/4sin(2x)+2ccos(x)]
dV/dc= π[2sin(x)]
dV/dc=0 when
2sin(x)=0
therefore turning points at x=0, π, 2π
minimum at x=0
at x=0
0= 2cπ (volume between x=0 and x=0 is zero)
therefore c=0


and one question. I am the only 4-unit person at my school, and I am having trouble with resisted motion. Is it mkv^n or kv^n? my textbook can't seem to make up it's mind

 

[gman03]

Hello everyone.
WARNING: the following is probably bodgy late-night maths, but here goes
oh, and S stands for integral sign

Vi=|([c-sin(xi)]^2-[c-0]^2)|πdx
V=πS([c-sin(x)]^2-c^2)dx
V=πS(sin^2(x)-2csin(x))dx
V=πS(1/2[1-cos(2x)]-2csin(x))dx
V=π[x/2-1/4sin(2x)+2ccos(x)]
dV/dc= π[2sin(x)]
dV/dc=0 when
2sin(x)=0
therefore turning points at x=0, π, 2π
minimum at x=0
at x=0
0= 2cπ (volume between x=0 and x=0 is zero)
therefore c=0


and one question. I am the only 4-unit person at my school, and I am having trouble with resisted motion. Is it mkv^n or kv^n? my textbook can't seem to make up it's mind

We seriously need LaTeX... can't read a line

From memory it is F = -kvⁿ

Edit: In a sense it doesn't matter as <i>m</i> itself is a constant anyway

 

[Grey Council]

ma = mg +/- mkv^n
more or less
point is:
the m's cancel out, so basically, no m's.


just remember to add in the With m line though, markers might be dodgy.

 

[CM_Tutor]

and one question. I am the only 4-unit person at my school, and I am having trouble with resisted motion. Is it mkv^n or kv^n? my textbook can't seem to make up it's mind

Actually, it can be either - look at the question in detail, as each must be done in a way appropriate to its definitions. For example, a question I recently answered in this forum had a pushing force P and a resistance of mkv. Thus, the equation was:

&Sigma;F = P - mkv = mx''

and the m's did not cancel.

 

[Iota]

It's interesting. I'm currently reading a book on what could be called extra-curricular physics. It uses L^aT_eX.

I would be happy if this subforum used L^aT_eX. It is used on the AOPS forums very well.