# Thread: Volumes of a solid of revolutions

1. Originally Posted by VonNemo19
Yes, I'm sure.

$\displaystyle [x^{1/3}]^2=x^{2/3}$
Can you show me how you solved that I am sort of confused with that thanks. DO you add the powers... explain thanks

2. Originally Posted by Awsom Guy
Can you show me how you solved that I am sort of confused with that thanks. DO you add the powers... explain thanks
$\displaystyle (x^m)^n=(x^m)(x^m)...ntimes$ Therefore, $\displaystyle (x^m)^n=x^{nm}$. so you multiply powers.

3. Oh man... I have forgotton my basics thanks heaps for that explanation I have got the correct answer thank you.

4. Originally Posted by Awsom Guy
Can you show me how you solved that I am sort of confused with that thanks. DO you add the powers... explain thanks
What he's doing is squaring the expression, which is what is required in the volumes of revolution formulas. It's an application of laws of indices if you look at it in that kind of way.

For instance, your value for y might also be something like $\displaystyle 3sin^2 \theta$ - here, you would get $\displaystyle (3sin^2 \theta)^2$ which leads to $\displaystyle 9sin^4 \theta$.

I think what you might need to familiarise yourself with though is the whole concept of $\displaystyle x^a . x^b = x^{ab}$. Here, what you've just basically witnessed was the a, which in your case was 1/3, multiplied by 2. And two thirds is...2/3. Which is what you get - and then when you integrate x to that power, you're adding one and getting a coefficient of 3/5.

5. yes thank you I understand.

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