VCE Specialist 3/4 Question Thread!

[zvezda]

how do i find the volume generated when the region is rotated about the y - axis

ans:126pi/5

This is really round-about, so someone contribute if there's a shorter way:
basically,
from x=1 to x=0, find the volume of the cylinder generated when rotating the square about the y-axis (as in, the square defined by 0,0 ; 0,1 ; 1,1 ; and 1,0).
Then transpose equation to make x the subject and find the volume generated when rotating the area enclosed the curve, the y-axis and the line y=1. Subtract this volume from the original volume, and you have the volume generated for the area of the graph enclosed by x=1 and x=0 when the graph is rotated about the y.

But, we want the area between x=4 and x=1. So repeat above process with area enclosed by x=4, x=0, the x-axis and the curve. Then subtract the volume found above^ from this one here, and there's your answer.

Apologies if this is hard to follow

EDIT: you sure its 126pi/5? I got 124pi/5 :/

[zvezda]

q8 multi choice in the essentials text. Chapter 8 review.
Help much appreciated

[brightsky]

D.

A and B are obviously true. A corresponds to the volume of the cylinder of radius f(b) and height (b-a) and B corresponds to the volume of the cylinder of radius f(a) and height (b-a). C is obviously true as well. D is dubious, since it seems to have been derived from incorrect calculus, but we shall suspend judgement for the time being. E is definitely true since the value is even greater than that in A (E implies the difference between two volumes, see if you can work out what it is physically speaking). So we can be sure that D is the required answer.

[Homer]

Then transpose equation to make x the subject and find the volume generated when rotating the area enclosed the curve, the y-axis and the line y=1. Subtract this volume from the original volume, and you have the volume generated for the area of the graph enclosed by x=1 and x=0 when the graph is rotated about the y.

EDIT: you sure its 126pi/5? I got 124pi/5 :/

hey thanks for that, but i dont understand why we have to subtracte the volume generated by rotating the area enclosed the curve the y-axis (31pi/5?)

also how would you find the volume when region rotated about the y-axis

[brightsky]

This is really round-about, so someone contribute if there's a shorter way:
basically,
from x=1 to x=0, find the volume of the cylinder generated when rotating the square about the y-axis (as in, the square defined by 0,0 ; 0,1 ; 1,1 ; and 1,0).
Then transpose equation to make x the subject and find the volume generated when rotating the area enclosed the curve, the y-axis and the line y=1. Subtract this volume from the original volume, and you have the volume generated for the area of the graph enclosed by x=1 and x=0 when the graph is rotated about the y.

But, we want the area between x=4 and x=1. So repeat above process with area enclosed by x=4, x=0, the x-axis and the curve. Then subtract the volume found above^ from this one here, and there's your answer.

Apologies if this is hard to follow

EDIT: you sure its 126pi/5? I got 124pi/5 :/

Another way:

Volume
= pi*4^2*sqrt(4) - pi*int^(2)_(1) y^4 dy - pi*(1)^2*sqrt(1)
= 32pi - pi*(2^5/5 - 1/5) - pi
= 31 pi - pi*(31/5)
=124 pi/5

[zvezda]

D.

A and B are obviously true. A corresponds to the volume of the cylinder of radius f(b) and height (b-a) and B corresponds to the volume of the cylinder of radius f(a) and height (b-a). C is obviously true as well. D is dubious, since it seems to have been derived from incorrect calculus, but we shall suspend judgement for the time being. E is definitely true since the value is even greater than that in A (E implies the difference between two volumes, see if you can work out what it is physically speaking). So we can be sure that D is the required answer.

cheers brightsky

[brightsky]

hey thanks for that, but i dont understand why we have to subtracte the volume generated by rotating the area enclosed the curve the y-axis (31pi/5?)

also how would you find the volume when region rotated about the y-axis

visualise the graph. first consider the top section first and then multiply by 2 later.

so y^2 = 2x + 1
x = (y^2 - 1)/2
volume of top section
= pi * int^(sqrt(7))_(1) ((y^2-1)/2)^2 dy
blah blah blah

then multiply result by 2 to get total area

[Homer]

thanks guys

[zvezda]

Hey,
If a•a = c•c, does a=c? Im referring to a question in essentials, q11 bi in short answers in fhe chapter 2 review.

[Alwin]

Hey,
If a•a = c•c, does a=c? Im referring to a question in essentials, q11 bi in short answers in fhe chapter 2 review.

a • a = |a|^2 and c • c = |c|^2
so, |a|^2= |c|^2 which does NOT mean a = c

This is because, |a| = |c| implies that only the lengths of a and c are similar, telling nothing about the direction.
Hence, they can be two completely different vectors.

An example is a=(1,0) and c=(0,1). clearly a • a = c • c but a ≠ c

btw, I didn't find that question anywhere in chapter 2 review.. or maybe I'm just didn't look hard enough

EDIT: Sorry guys, just had to go back and make all the a's and c's bold just a slight ocd tendency when it comes to vectors

[zvezda]

a • a = |a|^2 and c • c = |c|^2
so, |a|^2= |c|^2 which does NOT mean a = c

This is because, |a| = |c| implies that only the lengths of a and c are similar, telling nothing about the direction.
Hence, they can be two completely different vectors.

An example is a=(1,0) and c=(0,1). clearly a • a = c • c but a ≠ c

btw, I didn't find that question anywhere in chapter 2 review.. or maybe I'm just didn't look hard enough

EDIT: Sorry guys, just had to go back and make all the a's and c's bold just a slight ocd tendency when it comes to vectors

Ahh yeah if course....
Cheers for the help. You probably couldnt find it because it was one of the steps in my working out

[e^1]

Find the exact value of:


The answer is ; but I'm not sure how to go about this. If anyone could help, then that would be greatly appreciated!

[Alwin]

The answer is ; but I'm not sure how to go about this. If anyone could help, then that would be greatly appreciated!

This one is a bit long, but here:

so you have:

Complete the square of


Hence


Now rationalise






Which gives you
as required

[e^1]

Thanks for the help! but how do you complete the square in that case? :S

[lzxnl]

Let 2- sqrt 3 = (a+b sqrt 3)^2
We're hoping this works for rational a, b.
Expanding and equating rational and irrational parts, we get:
a^2+3b^2=2
-1=2ab
so from the second equation, b=-1/2a
Subbing into first equation
a^2+3/4a^2=2
a^4-2a^2+3/4=0
(a^2-1)^2=1/4
a^2=1+-1/2
It looks like we have a problem. But if we continue...
Let a^2=3/2
3b^2=1/2
b^2=1/6
b=sqrt(1/6)=sqrt(6)/6

Equation two requires that one of a or b be negative, so we have a=-sqrt(3/2)=-sqrt(6)/2

This should work. -sqrt(6/2)+sqrt(6)/6 should be a square root of 2-sqrt 3
You can then simplify everything down