Post by: Andiio on June 23, 2011, 10:47:28 pm
Thanks!
Post by: taiga on June 23, 2011, 10:53:48 pm
Post by: Andiio on June 23, 2011, 11:02:08 pm
But how would you calculate the volume (rotated about y) for a question like this: y = √x with the lines x=1 and x=4?
I've been stumped on this question for the past half an hour, I had a looked at the worked sols but don't understand why they did some of the working :\
Post by: stonecold on June 23, 2011, 11:14:54 pm
Ahhh okay haha I get what you mean.
But how would you calculate the volume (rotated about y) for a question like this: y = √x with the lines x=1 and x=4?
I've been stumped on this question for the past half an hour, I had a looked at the worked sols but don't understand why they did some of the working :\
You just find the corresponding y values, then sub them in, so in this case, it would be between y=1 and y=2
Post by: Andiio on June 23, 2011, 11:17:47 pm
I just don't understand the second part and I tried visualising it but the lines x=1 & x=4 don't intersect at all.. and so I wasn't sure why they had (4)^2 - (1)^2? (i.e. x=4 curve - x=1 curve)
Post by: stonecold on June 23, 2011, 11:32:22 pm
V=pi x r^2 x thickness
where r=x=y^2
you are looking for the volume between x=1 and x=4, when x=1, y=1 and when x=4, y=2
One of us is wrong. :p
Post by: Andiio on June 23, 2011, 11:36:05 pm
You need to use this formula:
V=pi x r^2 x thickness
where r=x=y^2
you are looking for the volume between x=1 and x=4, when x=1, y=1 and when x=4, y=2
One of us is wrong. :p
I...don't think my brain is working.... :( :( :( :(
Post by: stonecold on June 23, 2011, 11:37:41 pm
You need to use this formula:
V=pi x r^2 x thickness
where r=x=y^2
you are looking for the volume between x=1 and x=4, when x=1, y=1 and when x=4, y=2
One of us is wrong. :p
I...don't think my brain is working.... :( :( :( :(
Yeah, well as always, I left out the pi haha.
Wait for the spesh heavyweights to come and confirm. :)
Post by: luken93 on June 23, 2011, 11:38:06 pm
You need to use this formula:Yeah, that's what I get as well...
V=pi x r^2 x thickness
where r=x=y^2
you are looking for the volume between x=1 and x=4, when x=1, y=1 and when x=4, y=2
One of us is wrong. :p
Which question is this anyway andiio?
Post by: Andiio on June 23, 2011, 11:40:21 pm
Urghhh does this mean the worked solutions are being retarded?
Q: For the regions bounded by the x-axis, the following curves, and the given lines, find the volume generated by rotating it about the y-axis.
y=√x; x=1, x=4
Post by: stonecold on June 23, 2011, 11:47:45 pm
You have to split it into two separate integrals, because between y=0 and y=1 you are integrating between straight lines, and between y=1 and y=2, you are integrating between a straight line and a curve.
Draw it out and you should see.
Post by: Andiio on June 23, 2011, 11:50:15 pm
Oh, yeah, it is a different question. The solutions are correct.
You have to split it into two separate integrals, because between y=0 and y=1 you are integrating between straight lines, and between y=1 and y=2, you are integrating between a straight line and a curve.
Draw it out and you should see.
Ahhh, I get that for when integrating between y=1 and y=2, it's a straight line and a curve, but don't you integrate between a line and a curve for between y=0 and y=1 as well? Or am I thinking completely wrong? I've shaded the areas that the curve itself makes WITH the y-axis :S
Post by: kamil9876 on June 23, 2011, 11:52:21 pm
The idea behind the solution you posted is to find the volume generated the pink part enclosed by y=0 and y=1 (that's the first integral) then the second integral finds the other pink part.
Post by: stonecold on June 23, 2011, 11:52:40 pm
Oh, yeah, it is a different question. The solutions are correct.
You have to split it into two separate integrals, because between y=0 and y=1 you are integrating between straight lines, and between y=1 and y=2, you are integrating between a straight line and a curve.
Draw it out and you should see.
Ahhh, I get that for when integrating between y=1 and y=2, it's a straight line and a curve, but don't you integrate between a line and a curve for between y=0 and y=1 as well? Or am I thinking completely wrong? I've shaded the areas that the curve itself makes WITH the y-axis :S
Yeah, you shaded the wrong bit. You should have the area under Y=SqRoot(x) shaded in between x=1 and x=4. this is what you rotate.
And it is actually an anulus, so r^2=r^2 outer - r^2 inner
Edit: The pink bit in kamil's diagram :p
Post by: Andiio on June 23, 2011, 11:55:56 pm
So basically for this question I just calculate the area bound between x=1 and x=4... wait is the green supposed to be pink in the above diagram?
Post by: stonecold on June 23, 2011, 11:56:52 pm
Ohhh... right.. I shaded in the blue part and half of the green part :| I got it confused with the ares with the y-axis then?
So basically for this question I just calculate the area bound between x=1 and x=4... wait is the green supposed to be pink in the above diagram?
No, it is right.
Have you done anulus' yet?
Post by: Andiio on June 24, 2011, 12:11:00 am
Ohhh... right.. I shaded in the blue part and half of the green part :| I got it confused with the ares with the y-axis then?
So basically for this question I just calculate the area bound between x=1 and x=4... wait is the green supposed to be pink in the above diagram?
No, it is right.
Have you done anulus' yet?
Am I right in saying that we just want the area the curve makes with the x-axis? i.e. the area below the curve?
Never heard of that anulus' :\ what are they?
Post by: stonecold on June 24, 2011, 12:15:22 am
Ohhh... right.. I shaded in the blue part and half of the green part :| I got it confused with the ares with the y-axis then?
So basically for this question I just calculate the area bound between x=1 and x=4... wait is the green supposed to be pink in the above diagram?
No, it is right.
Have you done anulus' yet?
Am I right in saying that we just want the area the curve makes with the x-axis? i.e. the area below the curve?
Never heard of that anulus' :\ what are they?
Firstly, I should spell it correctly lol. Annuli (Annulus sing.) are basically washers. They are a type of solid of revolution, but when you do your revolution around the axis, you get a hole in the middle, creating a washer type solid, for which the technical term is 'annulus.'
Can you see how if you revolve that pink shaded area around the y-axis, you create a solid with a hole in the middle?
To overcome this in your calculations, you still use
V=pi x r^2 x thickness
however now r^2= r^2 outer curve - r^2 inner curve.
I can post up a full solution tomorrow night if you can be bothered waiting. :)
Edit: And to answer your question, yes, you do want the area under between x=1, x=4 and the x-axis, but then you have to revolve this around the y-axis and find the volume.
Post by: Andiio on June 24, 2011, 12:18:59 am
Ohhh... right.. I shaded in the blue part and half of the green part :| I got it confused with the ares with the y-axis then?
So basically for this question I just calculate the area bound between x=1 and x=4... wait is the green supposed to be pink in the above diagram?
No, it is right.
Have you done anulus' yet?
Am I right in saying that we just want the area the curve makes with the x-axis? i.e. the area below the curve?
Never heard of that anulus' :\ what are they?
Firstly, I should spell it correctly lol. Annuli (Annulus sing.) are basically washers. They are a type of solid of revolution, but when you do your revolution around the axis, you get a hole in the middle, creating a washer type solid, for which the technical term is 'annulus.'
Can you see how if you revolve that pink shaded area around the y-axis, you create a solid with a hole in the middle?
To overcome this in your calculations, you still use
V=pi x r^2 x thickness
however now r^2= r^2 outer curve - r^2 inner curve.
I can post up a full solution tomorrow night if you can be bothered waiting. :)
Ahhh okay yeah I see how that works. Just wondering though, is it a specific topic in the VCE syllabus or are we meant to just deduce it logically?
Haha yes please, that'd be great. I think I'm still a tad sketchy over the areas and all.. :\
Post by: stonecold on June 24, 2011, 12:23:58 am
Post by: taiga on June 24, 2011, 12:33:19 am
Post by: Lols123 on June 24, 2011, 10:39:51 pm
Post by: Andiio on June 24, 2011, 10:44:26 pm
Is it just the comparison of the x-values between the certain bounds given?
E.g. y=x^2, y=-x from x=0 to x=1
Post by: luffy on June 25, 2011, 10:23:59 am
Just wondering, how exactly do you determine the curve that is 'the most to the right' and the curve that is 'the most to the left'?
Is it just the comparison of the x-values between the certain bounds given?
E.g. y=x^2, y=-x from x=0 to x=1
Depends, are you rotating it in the x-axis? or the y-axis?
If by "most to the right" and "most to the left", you are questioning which one to subtract from the other in the integral, I don't use a set rule. Instead, I simply visualise the 3D object produced and "subtract the unwanted area from the total area" and the rest becomes relatively simple.
Sorry - don't think I explained it properly. Hope I helped...
Post by: Andiio on June 26, 2011, 12:35:02 am
Edit: Is it accepted as formal working out?
Post by: vea on June 26, 2011, 02:54:07 am
Is the 'shell method' within the VCE syllabus? Also, in what situations do you use the 'shell' method?
Oh god, Dr He :P