Maths question [MTH1020]

[alondouek]

Haha yeah, I forgot to mention the y-axis is also a bound

[lzxnl]

In that case, you could do it using VCE methods as well; large cylinder formed by rotating y=1 about the x axis minus the solid formed by rotating the curve bound by x=0, x=1 and y=x^3 around the x axis.
Which from my quick calculations seems to yield pi-pi/7=6pi/7

While the other way requires you to find x in terms of y; the integral is 2pi*integral of xy dy = 2pi*integral of y^4/3 dy from 0 to 1 = 6pi/7 as well.

Doesn't really matter which method you use.

[alondouek]

hmm, I got something a little different; for the question, I need to take the area enclosed by y=x³, y=1 and the y-axis, then revolve that area around the x-axis.

So, using the shell method, I got:

[lzxnl]

hmm, I got something a little different; for the question, I need to take the area enclosed by y=x³, y=1 and the y-axis, then revolve that area around the x-axis.

So, using the shell method, I got:

http://en.wikipedia.org/wiki/Shell_integration

Says that rotating around the x axis means you integrate with respect to y.
Which makes sense. The formula looks to me like integrating 2pi*r*h.
When revolving around the x axis, if you cut up your volume into tiny slices:
Each slice can be seen as a thin, tiny cylinder for a given y value, with radius y, length x and width dy. The area of one of these slices is 2pi*y*x, so integrating yields 2pi*y*x dy.

Draw up a diagram and it'll make sense.

[alondouek]

Okay, that makes sense Thanks!

[alondouek]

Problem though, even when I try to use the formula from wiki:

I still end up with my prior (presumably wrong) solution;









What am I doing wrong?

[lzxnl]

Problem though, even when I try to use the formula from wiki:

I still end up with my prior (presumably wrong) solution;









What am I doing wrong?

y=y
But f(y)=x=y^1/3

[alondouek]

y=y
But f(y)=x=y^1/3

So to be clear, I'm taking the inverse of f(x), rather than just subbing y in as x?

[lzxnl]

According to the article I showed you, if you have something of the form y=f(x) and you revolve that, y=a and y=b, you integrate 2pi*xy dy from y=a to y=b. Remember it in that form; it's probably easier IMO to remember. Now y=f(x), so x=f^-1(y). Sub that in.

In short, just remember the formulas in terms of x and y, or even better, remember where they came from to minimise chances of error.

[alondouek]

Brilliant, thanks for the elaboration

[BigAl]

The way I remembered this formula in spesh was that I formed little differential element, dx or dy, depending on what axis the function was being rotated. Youre essentially adding up little circles separated by dx or dy. by making x or y the subject youre actually changing dependent and independent value

[lzxnl]

This formula actually isn't in spesh, which is sort of stupid. Although the reasoning is quite simple, yeah.

[BigAl]

You mean  it isn't on the formula sheet?

[lzxnl]

It's not in the course. Ask a student how to do that and they'll use the wrong formula. Or give you a blank stare.

[alondouek]

It's not in the course. Ask a student how to do that and they'll use the wrong formula. Or give you a blank stare.

Hah! I didn't even do spesh and I can do that