# Math Help - Adding complicated quotients

Again, the book has confused me. What on Earth have they done?
I got a bit further on, and I've got the same answer they reached doing this, except I don't know where their 1/2 went. Theirs doesn't have it and mine does.

2. Hi !

Ok, we have :

$S=\frac{\pi r^3}{3 (100-r^2)^{\frac 12}}+\frac{2 \pi r (100-r^2)^{\frac 12}}{3}$

Let $\boxed{k=(100-r^2)^{\frac 12}}$, so we have :

$S=\frac{\pi r^3}{3k}+\frac{2 \pi r \cdot k}{3}$

What is the common denominator ?
In one side, we have 3k and in the other side, 3.

So the common denominator is $\boxed{3k}$

To make such a thing, we will have to multiply the fraction on the right side by $\tfrac kk$ so that we will have 3k at the bottom.

$S=\frac{\pi r^3}{3k}+\frac{2 \pi r \cdot k^{\color{red}2}}{3k}=\boxed{\frac{\pi r^3+2 \pi r k^2}{3k}}$

But what is $k^2$ ?
A rule of exponents tells us : $(a^b)^c=a^{bc}$

Therefore $k^2=\left((100-r^2)^{\frac 12}\right)^2=(100-r^2)^{\frac 12 \cdot 2}=(100-r^2)^1=100-r^2$

---> $S=\frac{\pi r^3+2 \pi r (100-r^2)}{3(100-r^2)^{\frac 12}}$

Is it clear enough ? :x

3. :O It makes sense! That's phenomenal.
It's like if you were to do:
2/10 + 1/5
You would end up with:
2/10 + 2/10, to get a common denominator...And then you get 4/10. It's just a coincidence that the 1/2 power cancels out and all, and it caused it to slip my mind :P
BUT, there is one thing left unexplained. The pi symbol that becomes negative. Do you know why that is?

Forgive me, I don't quite know how to use the math symbols.

4. :O It makes sense! That's phenomenal.
It's like if you were to do:
2/10 + 1/5
You would end up with:
2/10 + 2/10, to get a common denominator...And then you get 4/10. It's just a coincidence that the 1/2 power cancels out and all, and it caused it to slip my mind :P

BUT, there is one thing left unexplained. The pi symbol that becomes negative. Do you know why that is?
Nope
Mistake ?

Forgive me, I don't quite know how to use the math symbols.
Well, it's still understandable ^^

But you can try learning how to use the latex in this section : http://www.mathhelpforum.com/math-help/latex-help/

5. Oh, I think a few lines before it was a negative, and then when it wasn't it was a typo.
So it's negative. :P
Thank you very much. You've been very helpful.