VCE Specialist 3/4 Question Thread!

[keltingmeith]

Because we were checking for area bound by the line x = 4. If you tried to integrate it from 0 to 1/3, you're not accounting for the fact that you've bounded that area there. In fact, if you do try it, you'll get an infinite volume.

[Bestie]

thank you everyone !!!

[Bestie]

another question?
please see attachment
is there a way to work this out algebraically, without the use of a CAS?
can someone please enlighten me

thank you

[Bestie]

i'm sorry.. but i also have another question... sorry... but if someone could help me here too i would really appreciate it
question 8 please
i have no idea where option A, B and E came from?
please see attachment
thank you

[keltingmeith]

Not at your level. While it is certainly possible to integrate (ln(x))^2, you can't do it with the techniques you're taught in specialist. So in this case, you'll just have to use the calculator.

Also - that typo is beautiful.

[EspoirTron]

I am pretty sure that the question about y=ln(x) is beyond the scope of VCE Specialist. The solution by hand is by integration by parts if I am not mistaken, which to my current knowledge, is beyond the scope of the course.

[keltingmeith]

The solution by hand is by integration by parts if I am not mistaken, which to my current knowledge, is beyond the scope of the course.

Yep - that's the technique. I actually had to solve (ln(x))^3 the other day, which required having to integrate (ln(x))^2, hahah - breaking into parts everywhere!

Funnily enough, it's not EXACTLY beyond the scope of the course - integration by parts is a natural consequence of "integration by recognition", which is something methods likes to do a lot. However, you certainly can't be asked to integrate something like this just off the go, they'd have to lead you into it.

[EspoirTron]

Yep - that's the technique. I actually had to solve (ln(x))^3 the other day, which required having to integrate (ln(x))^2, hahah - breaking into parts everywhere!

Funnily enough, it's not EXACTLY beyond the scope of the course - integration by parts is a natural consequence of "integration by recognition", which is something methods likes to do a lot. However, you certainly can't be asked to integrate something like this just off the go, they'd have to lead you into it.

Yes, I do understand what you mean there haha!

I agree, there are some very cheeky examples that can pop up actually. Most of the times I have found a good u-substitution can do the trick, but again that just depends on the problem at hand. It has been so long since I have done methods and now I remember 'Integration by recognition', probably one of the best bits or the Integral calculus syllabus.

[Conic]

i'm sorry.. but i also have another question... sorry... but if someone could help me here too i would really appreciate it
question 8 please
i have no idea where option A, B and E came from?
please see attachment
thank you

In option A we have a cylinder with radius f(b), and height b-a. Because f(b)>f(a), this will have greater volume than the original.


In option B we have a cylinder with radius f(a), and height b-a. Because f(a)<f(b), this will have a smaller volume than the original.


In option D we have a cylinder of radius f(b) and height b with a cylinder of radius f(a) and height a removed. This is a larger volume than in option A, so it is also larger than the original.

[Eugenet17]

Could someone explain to me how to do this question? The worked solutions doesn't make any sense to me :|

http://gyazo.com/215276a04748ee1027603a8d592ec405

[keltingmeith]

This question is a little cheeky - you see, you actually have to do the top half MINUS the bottom half to make it so you're only getting the area contained in the circle before you revolve it around. So, let's first find the equation of each half:



Now, we just plug and play with the formula:

[lzxnl]

Could someone explain to me how to do this question? The worked solutions doesn't make any sense to me :|

http://gyazo.com/215276a04748ee1027603a8d592ec405

Actually, there is another cheeky way of doing this.
Can you see that if you unravelled your donut, you get a cylinder? Well, the height of the cylinder is just the distance the centre of the circle has travelled, or 4*2pi = 8pi. Then, the radius is 2, so V = pi*r^2*h = pi*2^2*8pi = 32pi^2

But please don't do this in a SAC; use this only as a check

[alchemy]

Actually, there is another cheeky way of doing this.
Can you see that if you unravelled your donut, you get a cylinder? Well, the height of the cylinder is just the distance the centre of the circle has travelled, or 4*2pi = 8pi. Then, the radius is 2, so V = pi*r^2*h = pi*2^2*8pi = 32pi^2

nice!

[Eugenet17]

Thanks alot guys, just another question out of curiosity, how would I go about a question if it was something like rotating a circle around the y-axis to get a torus?

[keltingmeith]

Exact same approach as I used, but solve for x instead of y. Then, you'll need to do , where is more positive in the x-axis, and is more negative in the x-axis.

Think of it like flipping your axis along the line y=x, so that the y-axis becomes the x-axis and vice-versa. Then, you should hopefully see it behaving very similarly to how you know integrals behave.

Also note: if you tried to rotate the circle in your question around the y axis, you'll get a sphere, not a torus.