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Vi Nguyen · May 15, 2009, 10:49 AM

How would I find the volume enclosed by these two curves if I were to rotate it about the x-axis?

curves: y=x and y=(x^2)-4x+4

I know the curves intersect at x=1 and 4, I think that the area enclosed by the graph is 4.5 by taking the integral between the two limits, but volume when rotated about the x-axis?

galactus · May 15, 2009, 11:11 AM

Since we are rotating about the x-axis, we can use washers.

Note that $x^{2}-4x+4=(x-2)^{2}$

${\pi}\int_{1}^{4}[x^{2}-[(x-2)^{2}]^{2}]dx={\pi}\int_{1}^{4}[x^{4}-8x^{3}+23x^{2}-32x+16]dx$

Can you integrate that?

Notice what this looks like? It is related to the area of a circle formula, ${\pi}r^{2}$ .

What the two functions generate is a solid having washer-shaped cross sections. The cross sections at x has a inner radius of g(x) and the outer radius f(x).

It's area is ${\pi}\left[[f(x)]^{2}-[g(x)]^{2}\right]$ .

Therefore, the volume is ${\pi}\int_{a}^{b}\left[[f(x)]^{2}-[g(x)]^{2}\right]dx$

Where f(x) and g(x) are the functions given.

Just Google it and you'll find lots about the washer and shell methods.

IN this case, the shell method is more complicated, so we won't bother with it since washers is the easier way to go.

Here is an animated graph of your region. Can you see the washers? See what it is about now a little better?

I always liked solids of revolution. It is one of the coolest and most applicable parts of calculus.

You can use it to find the volume or surface area of any odd-shaped region if you apply it correctly.

If you want to show off, you can give the Riemann sum of it:

$\lim_{n\to \infty}\left[\frac{3{\pi}\sum_{k=0}^{n-1}\left(\left(1+\frac{3(k+\frac{1}{2})}{n}\right)^ {2}-\left(\frac{3(k+\frac{1}{2})}{n}-1\right)^{4}\right)}{n}\right]$

I'm sorry, I reckon I got carried away and didn't have much to do today.

Unknown008 · May 16, 2009, 10:37 AM

Lol galactus! What program did you use? It's the first one I've seen that projects in three dimensions!

galactus · May 16, 2009, 10:57 AM

I used Maple 10's animation. And then, exported it to a gif and posted:)

Unknown008 · May 16, 2009, 11:00 AM

Great!! :)

Vi Nguyen · May 16, 2009, 10:02 PM

Your getting carried away is extremely welcomed!! Your graph is crazy.. lol!! But when I integrated I ended up with a -ve no so I went back and expanded g(x), I think you left out the -ve, after expanding it from scratch I got the same as you but just different sign, integrated this and got V=72pi/5. But thanks heaps for explaining and the visuals, really helped.

P.S still haven't figured out how to rate the answer, my computer or whatever won't allow me to when I click "rate this answer".

Originally Posted by galactus

Since we are rotating about the x-axis, we can use washers.

Note that $x^{2}-4x+4=(x-2)^{2}$

${\pi}\int_{1}^{4}[x^{2}-[(x-2)^{2}]^{2}]dx={\pi}\int_{1}^{4}[x^{4}-8x^{3}+23x^{2}-32x+16]dx$

Can you integrate that?.

Notice what this looks like?. It is related to the area of a circle formula, ${\pi}r^{2}$ .

What the two functions generate is a solid having washer-shaped cross sections. The cross sections at x has a inner radius of g(x) and the outer radius f(x).

It's area is ${\pi}\left[[f(x)]^{2}-[g(x)]^{2}\right]$ .

Therefore, the volume is ${\pi}\int_{a}^{b}\left[[f(x)]^{2}-[g(x)]^{2}\right]dx$

Where f(x) and g(x) are the functions given.

Just google it and you'll find lots about the washer and shell methods.

IN this case, the shell method is more complicated, so we won't bother with it since washers is the easier way to go.

Here is an animated graph of your region. Can you see the washers?. See what it is about now a little better?.

I always liked solids of revolution. It is one of the coolest and most applicable parts of calculus.

You can use it to find the volume or surface area of any odd-shaped region if you apply it correctly.

If you want to show off, you can give the Riemann sum of it:

$\lim_{n\to \infty}\left[\frac{3{\pi}\sum_{k=0}^{n-1}\left(\left(1+\frac{3(k+\frac{1}{2})}{n}\right)^ {2}-\left(\frac{3(k+\frac{1}{2})}{n}-1\right)^{4}\right)}{n}\right]$

I'm sorry, I reckon I got carried away and didn't have much to do today.

Unknown008 · May 16, 2009, 11:40 PM

Originally Posted by Vi Nguyen

P.S still haven't figured out how to rate the answer, my computer or whatever won't allow me to when I click "rate this answer".

Have you rated one of galactus' answers lately? If so, that explains why. You are not allowed to give good reps to someone in the same topic successively. You have to spread the reputation, by giving others good ratings.

If that's not the case, then, I don't know.

galactus · May 17, 2009, 04:49 AM

What does -ve mean? You got me there. Negative volume? Yes, you have the correct solution and what I gave is correct. Nothing forgotten. I just reversed the negative sign after expanding. Technically, there should be an absolute value sign around it.

Now, try it with shells. That is tougher.

galactus · May 17, 2009, 06:14 AM

You are not allowed to give good reps to someone in the same topic successively. You have to spread the reputation, by giving others good ratings.

That is probably the case. I have tried and got the same message. Personally, I think that should be changed because sometimes you can not give someone else a rate first.

Unknown008 · May 17, 2009, 07:58 AM

I agree with you only if that is applied to some topics. That rule was introduced because of a small group giving each other numerous good reps even though their response was not so good.

Or allowing giving reps successively only to some privileged members.

galactus · May 17, 2009, 08:01 AM

Oh, I see. Makes sense. There's always someone to abuse and ruin it for the rest of us.

Vi Nguyen · May 17, 2009, 06:08 PM

Yes, thanks, that's probably the problem, there should be a better explanation than "you must spread the reputaion".

Originally Posted by Unknown008

Have you rated one of galactus' answers lately? If so, that explains why. You are not allowed to give good reps to someone in the same topic successively. You have to spread the reputation, by giving others good ratings.

If that's not the case, then, I don't know.

Vi Nguyen · May 17, 2009, 06:15 PM

wouldn't know were to begin with shells, I thought that may be the case that you probably forgot to include the absolute symbol. But I don't know the technique you used to expand ((x-2)^2)^2 I know you must have a method of expanding without having to do it from scratch (multiply it out then calculate). I showed some of my friend your graph and everyone's fairly impressed. :)

Originally Posted by galactus

What does -ve mean?. You got me there. Negative volume?. Yes, you have the correct solution and what I gave is correct. Nothing forgotten. I just reversed the negative sign after expanding. Technically, there should be an absolute value sign around it.

Now, try it with shells. That is tougher.

Unknown008 · May 18, 2009, 08:00 AM

That, I get it this way:

$((x-2)^2)^2 = (x-2)^4$

OOOOOOO $= x^4 - ^4C_1(x)^3(2)^1 + ^4C_2(x)^2(2)^2 - ^4C_3(x)^1(2)^3 +(2)^4$

OOOOOOO $= x^4 - 8x^3 + 24x^2 - 32x + 16$