Volumes stuff (1 Viewer)

[CrashOveride]

i was not present when my class did work on volumes a while back....so i thought i'd catch up on it now and im looking at the chapter in the excel 4u extension maths book. It's outlined from the outset that when something is rotated around the X-axis, we take the strip perpendicular to the X-axis. When rotated around the Y-axis, we take the strip parallel to the Y-axis.

On the next page (70), the example (b) has the region bounded by the curve y² = x and the line x=4 revolved around the Y-axis. They take the strip perpendicular to Y-axis.....now do we just take each case on its own merits? coz what was the thing at the start talking about then???

Also umm i cant really visualise when these elementary strips make rings and stuff...does anyone know any sites where they do kinda volumes of revolution and maybe they illustrate via java applets or something?

 

[McLake]

You don't always take strips, there is another technique where you use infinitly thin, hollow cylinders with holes in the middle (or actually, think of egg rings).

So in this question yu end up with a badly shaped bowl. Think of the bowl as a series of rings, one inside the other. You then integrate with resect to x these cylinders, with height sqrt(x) and diameters from 0 to 8 (2 * x=4).

Does that make any sense?

 

[CrashOveride]

Not really coz im having a hard time visualing all this :S Do you have the excel4u book?

ACtually i can see what Patel is doing now. But i just wanted to qualify his earlier statements......also in the example 1b we are dealing with thickness of delta y....so when we want volume shudnt we be taking the limit as delta y appraoches 0 ? he has limit as delta x appraoches 0 ?

And also....i think i can see why its a ring now...its because delta y is infinitely small that (if u look at diagram) ...hard to explain but....the two lines of the strip parallel to the x-axis , even though they appear one longer than the other, they are approx. equal right because of delta y (which is infintely small) correct?

:-/

 

[McLake]

NB: I don't have Excel or Patel.

Originally posted by CrashOveride
ACtually i can see what Patel is doing now. But i just wanted to qualify his earlier statements......also in the example 1b we are dealing with thickness of delta y....so when we want volume shudnt we be taking the limit as delta y appraoches 0 ? he has limit as delta x appraoches 0 ?

No. If I unsetand correctly what you are saying, example 1b should be done with cylinders (not slices), and so is perpindicular to x, and so delta x --> 0

Originally posted by CrashOveride
And also....i think i can see why its a ring now...its because delta y is infinitely small that (if u look at diagram) ...hard to explain but....the two lines of the strip parallel to the x-axis , even though they appear one longer than the other, they are approx. equal right because of delta y (which is infintely small) correct?

:-/

I don't understand what you are saying. Sorry.

 

[CrashOveride]

But the strip taken was perpenicular to the Y-axis. And delta x is clearly labelled on the diagram. There is no mention of delta x until he starts taking the limit to get the volume. :S

 

[McLake]

Well that sounds like a typo ...

 

[CrashOveride]

Yeah my teacher said it should be delta y also....and it was a bit ambiguous because it was just the intro...yet Patel was delving into the shell and washer methods without clarifying it - shell method we take strip parallel to axis of rotation and wahser method we take strip perpendicular to axis of rotation.

 

[CrashOveride]

and also whats the mathematics behind saying that as n, the number of elements of volume, appraoches infinity, we can say the volume is equal to the integral, from b to a, of the element of volume w.r.t x (for eg)

 

[Harimau]

Originally posted by CrashOveride
and also whats the mathematics behind saying that as n, the number of elements of volume, appraoches infinity, we can say the volume is equal to the integral, from b to a, of the element of volume w.r.t x (for eg)

n is the number of partitations, slices or shells.

This is probably another way of thinking about it, but think of a set interval a <= x <= b for which you need to take the slices or shells of a shape. Because shells and "slices" are actually 3 dimensional objects, we can always just change them. Say if we had two "shells" or "slices" in this case, their "delta x" would be significantly large. (In other words, n=2) . Now in this case, you can see that the volume that is begotten as the sum of the two shells and slices are pretty inaccurate.

But say we increase n, to a very large number, we are going to get a very good approximation, as the "delta x" or the width of each slice or shell becomes very small. But as n goes to infinity, then the "delta x" goes infinitismally small. Because delta x, or the width is infinitismally small, the slices and shells are literally sheet thin, almost like an invisible line when you see it from the side... And if you sum them up, you will get an exact area. This is the same as the basic rectangles of the first time you saw intergration, right?

Sorry, i am just feeling a little off today, and can't explain things very well today.

 

[CrashOveride]

Summing them up won't given an *exact* volume, right ?

also what does it mean if a cross-sectional area is a continuous function of x ?

This older text i have says letting delta x ---> 0, if the cross-sectional area is a continuous function of x then the value of this limit<i>can be shown to be V*</i> ....it goes later on to say that defining volume isnt so str8 forward...says u nee dsophisticated limitintg procedures to develop satisfatory theory of volume, which is content of Hilbert's 3rd problem.

sorry i never lkearn about these "basic rectangles"...is there a direct mathematical way to say that as n--> inf...somehow an integral works out or was it shown by inspection or something

 

[McLake]

It is as exact as intergation ever is (ie: yes, but no)

Not sure about the rest ...

 

[CrashOveride]

ok as i do more, its making sense

just one thing...to make an integration easier i wanna show y^2 = 1/2 x^3 is an even function....as can be seen from Graphmatica
but when i arrange for y and do the standard test (shove in -x) ill get a negsative under the square root ....suggestions ?

 

[CrashOveride]

Also,

from an extremity A of the latus rectum AB of the parabola x^2 = 4ay, interval AK is drawn perpendicular to the x axis. Show volume formed by rotation of the region OAK about the line AK is 2/3pi(a^3)

i get 1/3 instead of 2/3 ?? :S

 

[wogboy]

i wanna show y^2 = 1/2 x^3 is an even function

It's not an even function as y is not defined for x < 0 over the real set. In fact y is not a function of x at all, because if you graph all the points satisfying that relation on an x,y plane, you'll find two values of y for every value of x>0 (just like graphing y^2 = x)

from an extremity A of the latus rectum AB of the parabola x^2 = 4ay, interval AK is drawn perpendicular to the x axis. Show volume formed by rotation of the region OAK about the line AK is 2/3pi(a^3)

Slicing the region perpendicular to the axis of rotation, the radius of each slice is r = 2a - x
pi*r^2 = pi*(2a - x)^2
= pi*(2a - 2*sqrt(a)*sqrt(y))^2
= pi*(4a^2 - 8a^(3/2)*y^(1/2) + 4ay)

integrating from y=0 to y=a,
V = pi * Integral{0->a} (4a^2 - 8a^(3/2)*y^(1/2) + 4ay)
= pi * (4a^3 - (16/3)*a^3 + 2a^2)
= (2/3)*pi*a^3

 

[CrashOveride]

Also the region in the cartesian plane satisfying 0<= x <= 2, 0<= y <= x^2 - 1/4x^4. the region is rotated about the y axis.....i used slices perpendicular to the y axis but then when i got up to the part of trying to express the x part in terms of y i get 4 roots so how do u do this one?

 

[CrashOveride]

Originally posted by wogboy
It's not an even function as y is not defined for x < 0 over the real set. In fact y is not a function of x at all, because if you graph all the points satisfying that relation on an x,y plane, you'll find two values of y for every value of x>0 (just like graphing y^2 = x)

ok maybe it was bad choice of words on my part....symmetric would have been a better word...is tehre any quick way to do show symmetry algebraically or i just have to draw the graph ?

thanks for answering that first question...i accidentally took the radius as a - x , doh

 

[CrashOveride]

Let "n" be a positive integer and "a" be a positive real number. Consider the graph of the function y=x^n whjich divides the rectangle formed by the lines x=a, y=a^n, x=0 and y = 0 into two regions bounded by two straight lines and a section of the curve itself. Show that the two solids obtained by rotating thee regions about the x axis is 1:2n

and the last one...i think if someone would be kind enough just to post up a diagram..i could handle the rest just not sure how to start (diagram, etc.)

Use the slicing technique to show that the volume of a spherical cap of heigh ''h'' and radius 'r' ( r > h) is V= 1/3 pi h²(3r - h)

 

[CrashOveride]

no one likes volumes

 

[wogboy]

Also the region in the cartesian plane satisfying 0<= x <= 2, 0<= y <= x^2 - 1/4x^4. the region is rotated about the y axis

This problem lends itself better to cylindrical shells. Draw up the region on an xy plane and if you take cylindrical shells you'll find:

- the radius of each shell:
r = x

- the height of each shell:
h = y = x^2 - 1/4x^4 (did you mean (1/4)x^4 or 1/(4x^4) ?)

So then you can use the normal method of integrating these cylindrical shells to find the volume, i.e. V = Integral(cuberoot(1/2) -> 2) 2*pi*r*h dx

Let "n" be a positive integer and "a" be a positive real number. Consider the graph of the function y=x^n whjich divides the rectangle formed by the lines x=a, y=a^n, x=0 and y = 0 into two regions bounded by two straight lines and a section of the curve itself. Show that the two solids obtained by rotating thee regions about the x axis is 1:2n

You mean find the ratio of the 2 volumes?

- region 1 (0 < x < a, 0 < y < x^n):

slicing perpendicular to the axis of rotation,
V1 = pi * Int {0->a} x^2n
= pi * (a^(2n+1))/(2n+1)

- region 2 (the other one):

You can see that if you add this volume to the first one, you'll get a cylinder of radius a^n, height a.

V2 = pi *(a^2n)*a - V1
so V2/V1 = {pi*(a^2n)*a / V1} - 1
= pi*a^(2n+1) / {pi*(a^(2n+1))/(2n+1)} - 1
= 2n + 1 - 1
= 2n
so V2:V1 = 2n:1

 

[Grey Council]

hoooiiieeeee!

Haven't seen you in a while, wogboy.
^_^

As usual, smooth solution
lol, we haven't started volumes yet, but I'm getting kinda scared.