Author Topic: Recreational Problems (SM level) (Read 92729 times) Tweet Share

Glockmeister · « **Reply #180 on:** June 26, 2008, 09:59:34 pm »

$\mbox{The area enclosed by }y=x^2,\; x=1,\; x=2\mbox{ and the x axis is spun around the y axis}$

$\mbox{Find the volume of this solid.}$

Examination of the curve $y = x^2$ shows that the shape produced will be a cylinder with a hole in the middle. The hole will bound at x = 1 and x = 0

Therefore:

$V = \int_{0}^{4} \pi x^{2}dy - \int_{0}^{1} \pi x^{2}dy$

$V = \pi (\int_{0}^{4} y dy - \int_{0}^{1} y dy)$

$V = \pi (\frac{16}{2} - \frac{1}{2})$

$V = \frac{15 \pi}{2} cubic units$

EDITED: Added pie in

ell · « **Reply #181 on:** June 26, 2008, 11:19:55 pm »

Quote from: Glockmeister on June 26, 2008, 09:59:34 pm

$\mbox{The area enclosed by }y=x^2,\; x=1,\; x=2\mbox{ and the x axis is spun around the y axis}$

$\mbox{Find the volume of this solid.}$

Examination of the curve $y = x^2$ shows that the shape produced will be a cylinder with a hole in the middle. The hole will bound at x = 1 and x = 0

Therefore:

$V = \int_{0}^{4} x^{2}dy - \int_{0}^{1} x^{2}dy$

$V = \int_{0}^{4} y dy - \int_{0}^{1} y dy$

$V = \frac{16}{2} - \frac{1}{2}$

$V = \frac{15}{2} cubic units$

$\pi$ ?

dcc · « **Reply #182 on:** June 26, 2008, 11:54:39 pm »

Yes, Pi.

Glockmeister · « **Reply #183 on:** June 27, 2008, 12:30:07 am »

oh fuck...

Mao · « **Reply #184 on:** June 27, 2008, 10:04:05 am »

whilst the answer is correct, can you give an argument for why $\int_a^b \pi x^2\; dy$ will give the appropriate volume?

When i tried with a few other functions, the answers were surprisingly similar but not the same:

$y=x^3 \; \mbox{from x=1 to x=2}$
your method yields $\frac{3(31)\pi}{5}$ whereas mine yielded $\frac{2(31)\pi}{5}$

$y=x^4 \; \mbox{from x=1 to x=2}$
your method yields $\frac{4096\pi}{3}$ whereas mine yielded $21\pi$

Glockmeister · « **Reply #185 on:** June 27, 2008, 12:47:49 pm »

Because the area is being rotated around the y-axis,

dcc · « **Reply #186 on:** June 27, 2008, 03:17:09 pm »

Quote from: Glockmeister on June 26, 2008, 09:59:34 pm

$\mbox{The area enclosed by }y=x^2,\; x=1,\; x=2\mbox{ and the x axis is spun around the y axis}$

$\mbox{Find the volume of this solid.}$

Consider the following picture (with coloured regions!)

Clearly, the question is asking us to find the volume of the solid formed by rotating the red & blue sections of the graph around the y-axis.

First, find the volume of the green section:

$V_{green} = \pi \int_1^4 x^2\; dy = \pi \int_1^4 y\; dy = \pi \left[\dfrac{y^2}{2}\right]_1^4 = \dfrac{15 \pi}{2} \mbox{ cubic units.}$

Now, if we consider the big chunk bound the lines $x = 0, x = 2, y = 1, y = 4$ , we determine that by rotating this solid around the y-axis, we attain a volume of:

$V_{gb} = \pi (2)^2 (3) = 12\pi$ . Also consider the fact that $V_{gb} = V_{green} + V_{blue}$ which gives:

$V_{blue} = 12\pi - \dfrac{15 \pi}{2} = \dfrac{9 \pi}{2} \mbox{ cubic units.}$

Now consider the red, which is just a cylinder, we find that the volume of the red is given by:

$V_{red} = \pi (2)^2 (1) - \pi (1)^2 (1) = 3\pi \mbox{ cubic units.}$

$\mbox{Hence the volume of the solid is: } \dfrac{15 \pi}{2} \mbox{ cubic units.}$

Mao · « **Reply #187 on:** June 27, 2008, 03:21:29 pm »

Quote from: Glockmeister on June 27, 2008, 12:47:49 pm

Because the area is being rotated around the y-axis,

as dcc's workings above show, you have found $V_{green}$ , it is by fluke that the two of them are of equal volumes

Glockmeister · « **Reply #188 on:** June 27, 2008, 05:26:42 pm »

umm so i have... thats pretty interesting. I think ii could see what went wrong.

sb3700 · « **Reply #189 on:** July 14, 2008, 07:56:28 pm »

Quote from: Ahmad on October 20, 2007, 11:43:27 am

$\mbox{6. Describe a method to solve }6x^4+25x^3+x^2+25x+6=0\mbox{ (This one is out to enwiabe)}$

16. Solve the equation $6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0$ .

17. Solve $6x^4 - 24x^3 + 12x^2 + 24x + 6 = 0$ .

Reviving thread with old problems and a better method.

Solving 16 but applicable to all three problems.

$6x^4 - 24x^3 + 12x^2 + 24x + 6 = 0$
Clearly x = 0 is not a solution, so x <> 0 (not equal to)
so $6x^2 + 6/x^2 - 24x + 24/x + 12 = 0$

As $(1/x - x)^2 = x^2 + 1/x^2 - 2$

$6((1/x - x)^2 + 2) + 24(1/x - x) + 12 = 0$
$6k^2 + 24k + 24 = 0$ with $k = 1/x - x$
$6(k+2)^2 = 0$
$k = 1/x - x = -2$
$x^2 - 2x - 1 = 0$
$(x-1)^2 - 2 = 0$
$x = 1 + \sqrt{2}, 1 - \sqrt{2}$

/0 · « **Reply #190 on:** July 14, 2008, 08:04:54 pm »

Quote from: sb3700 on July 14, 2008, 07:56:28 pm

Quote from: Ahmad on October 20, 2007, 11:43:27 am
$\mbox{6. Describe a method to solve }6x^4+25x^3+x^2+25x+6=0\mbox{ (This one is out to enwiabe)}$

16. Solve the equation $6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0$ .

17. Solve $6x^4 - 24x^3 + 12x^2 + 24x + 6 = 0$ .

Reviving thread with old problems and a better method.

Solving 16 but applicable to all three problems.

$6x^4 - 24x^3 + 12x^2 + 24x + 6 = 0$
Clearly x = 0 is not a solution, so x <> 0 (not equal to)
so $6x^2 + 6/x^2 - 24x + 24/x + 12 = 0$

As $(1/x - x)^2 = x^2 + 1/x^2 - 2$

$6((1/x - x)^2 + 2) + 24(1/x - x) + 12 = 0$
$6k^2 + 24k + 24 = 0$ with $k = 1/x - x$
$6(k+2)^2 = 0$
$k = 1/x - x = -2$
$x^2 - 2x - 1 = 0$
$(x-1)^2 - 2 = 0$
$x = 1 + \sqrt{2}, 1 - \sqrt{2}$

holy crap... that was nice

bigtick · « **Reply #191 on:** July 19, 2008, 10:56:20 pm »

Show $cos\frac{\pi }{5}+cos\frac{3\pi }{5}=\frac{1}{2}$ .

bigtick · « **Reply #192 on:** July 25, 2008, 11:55:36 pm »

Show that $cos\frac{\pi }{11}-cos\frac{2\pi }{11}+cos\frac{3\pi }{11}-cos\frac{4\pi }{11}+cos\frac{5\pi }{11}=\frac{1}{2}$ .

hard · « **Reply #193 on:** July 28, 2008, 06:34:40 pm »

how about this, my teacher asked this but he says that that he was able to prove it and told us it's a hard question.
Can you solve it? I'm looking for a proof that arg z1 + arg z2 = arg(z1z2)

dcc · « **Reply #194 on:** July 28, 2008, 07:32:15 pm »

Quote from: hard on July 28, 2008, 06:34:40 pm

how about this, my teacher asked this but he says that that he was able to prove it and told us it's a hard question.
Can you solve it? I'm looking for a proof that arg z1 + arg z2 = arg(z1z2)

I assume by this you do not mean the principal argument, as the relation shown above is incorrect if the principal argument is considered.

Let $z_{1} = \cos \theta_{1} + i \sin \theta_{1},\ z_{2} = \cos \theta_{2} + i \sin \theta_{2}$

$\begin{align*} z_{1} \cdot z_{2} &= \left( \cos \theta_{1} + i \sin \theta_{1} \right) \cdot \left( \cos \theta_{2} + i \sin \theta_{2} \right)\\ z_{1} \cdot z_{2} &= \left ( \cos \theta_{1} \cos \theta_{2} - \sin \theta_{1} \sin \theta_{2} \right) + i \left(\sin \theta_{1} \cos \theta_{2} + \cos \theta_{1} \sin \theta_{2} \right)\\ z_{1} \cdot z_{2} &= \cos \left(\theta_{1} + \theta_{2}\right) + i \sin \left(\theta_{1} + \theta_{2}\right)\\ \end{align*} $

$\arg(z_{1}) = \theta_{1},\ \arg(z_{2}) = \theta_{2},\ \arg(z_{1} \cdot z_{2}) = \theta_{1} + \theta_{2}$

As required.

Search the forums now!

We have moved!

Author Topic: Recreational Problems (SM level) (Read 92729 times) Tweet Share

Glockmeister

Re: Recreational Problems (SM level)

ell

Re: Recreational Problems (SM level)

dcc

Re: Recreational Problems (SM level)

Glockmeister

Re: Recreational Problems (SM level)

Mao

Re: Recreational Problems (SM level)

Glockmeister

Re: Recreational Problems (SM level)

dcc

Re: Recreational Problems (SM level)

Mao

Re: Recreational Problems (SM level)

Glockmeister

Re: Recreational Problems (SM level)

sb3700

Re: Recreational Problems (SM level)

/0

Re: Recreational Problems (SM level)

bigtick

Re: Recreational Problems (SM level)

bigtick

Re: Recreational Problems (SM level)

hard

Re: Recreational Problems (SM level)

dcc

Re: Recreational Problems (SM level)

Recent Posts

User:
Password: