Volumes - Slicing (1 Viewer)

[robbo_145]

It works out I think:

ax + x²/9 - (2√a/3)x^3/2 ... integrate

[(ax²)/2 + x³/27 - (4√a/15)x^5/2] from 0 to 9a.

[81a³/2 + 27a³ - 972a³/15]

The limits include an 'a' term so everything evens out to leave a³.

i have the same integral. yet get my answer :S

edit: goodo i can still add this close the hsc

 

[KFunk]

Haha, our answers agree now that I've added the coefficients up correctly .

 

[fantasy27]

And it looks like I can't add numbers together . My previously posted answer of 2547a²π/20 should probably be 27a³π/20 , that is, if I even added it up correctly this time.

yea, i was doing it the same method as you, but i kept getting a different answer and was wondering why.

so is the correct answer 27a³π/20 ?

 

[KFunk]

It looks good given that three of us have converged on it.

 

[haboozin]

yea, i was doing it the same method as you, but i kept getting a different answer and was wondering why.

so is the correct answer 27a³π/20 ?

yes thats the correct answer...

thnx KFunk with your reasoning.... why was i doing R^2 - r^2 instead of (R-r)^2 :s

btw its the knox trial (in the 8 trials thread)

should do it... some of the questions are pretty good.
Friend that goes to knox said 3 people got over 100.

 

[Vampire]

umm can someone post up the full working from start to finish please...thx

 

[KFunk]

umm can someone post up the full working from start to finish please...thx

Sticking together the working throughout the thread (plus some additions you get):

You have the curves y = 2√(ax) and y = 2x/3. The distance between them is (2√(ax) - 2x/3) so the radius of each semi-circle is r = (√(ax) - x/3). (because the distance between them is the diameter so you halve it).

δV = (1/2).πr²δx

V = lim(δx-->0) ∑ δV (from 0 to 9a)

= π/2 ∫ (√(ax) - x/3)² dx (from 0 to 9a)

= π/2 ∫ ax + x²/9 - (2√a/3)x^3/2 dx

= [(ax²)/2 + x³/27 - (4√a/15)x^5/2] from 0 to 9a.

= [81a³/2 + 27a³ - 972a³/15]

= 27a³π/20

 

[haboozin]

umm can someone post up the full working from start to finish please...thx

Legth of PQ (any point) = |y2 - y1| (diameter)
ie Radius of Semi-circle is |y2 - y1|/2

ie Area of Typical slice
(semi circle)
A= 1/2Pi((Y2-y1)/2)^2
A= 1/8pi (y2-y1)

but y^2 = 4ax
and y= 2/3x

so y2 = Sqrt(4ax)
y1=2/3x
so A is aprox Pi/8(2Sqrt(ax) 0 2/3x)^2
dv = Pi/8(2Sqrt(ax) 0 2/3x)^2 dx
lim
dx -->0

(9a SIGMA 0) Pi/8(2Sqrt(ax) 0 2/3x)^2 dx

so Pi/8 I(0 to 9a) Pi/8(2Sqrt(ax) 0 2/3x)^2 dx


and integrate..
= 27pi/20 a^3 units^3

 

[yrtherenonames]

If only we could do the actual exam in such a democratic manner...

 

[Vampire]

cool thx...just another thing, in hsc could you leave the answer in unsimplified form like KFunk's 2nd last line?

 

[haboozin]

cool thx...just another thing, in hsc could you leave the answer in unsimplified form like KFunk's 2nd last line?

they will prob give u the mark, in the standards package, the exempler on question 7 one of them isnt simplified (and they got 15/15 for that section) so i guess so.

 

[Stefano]

I have nfi... this is my weakest topic. Could someone tell me the general procedure for approaching slicing questions ?

I mean, is it: Int. [Area of once slice] {from x_1 to x_2} ?

 

[KFunk]

Slicing deals in areas of similar cross section. If you do a basic "rotate y = x² around the y axis" then you're summing infinitely thin circular discs. The cross sections you take are parallel to the axis along which you integrate. The key thing is to get the area of the cross section in terms of a single variable where that variable is 'x' if you're integrating along the x-axis or 'y' if you're integrating along the y-axis. The basic steps in doing the cross section thing would be:

Find the cross sectional area in terms of one variable. This may involve simultaneous equations, similar triangles or other tricks. You want to come out with a statement like:
Cross sectional area = A(x)

Your volume element is then δV = A(x)δx
( this gives you the area of the prisms that you're going to sum. They have the cross sectional area A(x) and a thickess δx)

You then go through the formalities of applying limits to get to an integral:

Volume of solid formed = lim (δx-->0) ∑ A(x)δx
= lim (δx-->0) ∑ δV where the limits of the sum (from x=a to x=b) are the limits between which you wish to integrate and sum the cross sectional strips.

This limit/sum is then equal to: ∫ A(x) dx
where the limits are the same as those used for the ∑ symbol. Then of course you integrate as you normally would.

This is a very general procedure for doing volumes by cross section but I hope it's of use.

 

[Stefano]

Slicing deals in areas of similar cross section. If you do a basic "rotate y = x² around the y axis" then you're summing infinitely thin circular discs. The cross sections you take are parallel to the axis along which you integrate. The key thing is to get the area of the cross section in terms of a single variable where that variable is 'x' if you're integrating along the x-axis or 'y' if you're integrating along the y-axis. The basic steps in doing the cross section thing would be:

Find the cross sectional area in terms of one variable. This may involve simultaneous equations, similar triangles or other tricks. You want to come out with a statement like:
Cross sectional area = A(x)

Your volume element is then δV = A(x)δx
( this gives you the area of the prisms that you're going to sum. They have the cross sectional area A(x) and a thickess δx)

You then go through the formalities of applying limits to get to an integral:

Volume of solid formed = lim (δx-->0) ∑ A(x)δx
= lim (δx-->0) ∑ δV where the limits of the sum (from x=a to x=b) are the limits between which you wish to integrate and sum the cross sectional strips.

This limit/sum is then equal to: ∫ A(x) dx
where the limits are the same as those used for the ∑ symbol. Then of course you integrate as you normally would.

This is a very general procedure for doing volumes by cross section but I hope it's of use.

Thank you VERY much Kfunk. You're the man!

 

[haboozin]

Thank you VERY much Kfunk. You're the man!

i swear your school taught u nothing :s - kids never go to his school

 

[Stefano]

i swear your school taught u nothing :s - kids never go to his school

I just didn't like my teachers teaching methods. I also hate this particular topic.