# Calculate the volume with a rational funtion

1. Jan 24, 2012

### yaakob7

1. The problem statement, all variables and given/known data

Calculate the volume (V) generated by the given function:
$y^{2}=\frac{x^3}{2a-x}$ about the line $x=2a$

I suppose you have to use cylindrical shells [because it's difficult to fin $y$], so:

$V=2\pi(2a-x)\sqrt{\frac{x^3}{2a-x}}Δx$

now:

$V=\int^{2a}_{0}2\pi(2a-x)\sqrt{\frac{x^3}{2a-x}}dx$

but I tried a million things but nothing works:

First, simplifying the integrand to this:

$V=2\pi\int^{2a}_{0}\sqrt{2a-x}\sqrt{x^3}dx$

Next, I've been searching for any substitution function, but nothing works. If you have some idea that could help me I'd be grateful.

Thanks

2. Jan 24, 2012

### SammyS

Staff Emeritus
Hello yaakob7. Welcome to PF !

Mostly it's just a matter of rewriting the integrand.

The substitution, u = 2a/x will also help. Do that first.

3. Jan 25, 2012

### yaakob7

Did you mean u=2a-x??

I tried but the expression remains almost the same:

$V=2\pi\int^{2a}_{0}\sqrt{u}\sqrt{(2a-u)^3}dx=2\pi\int^{2a}_{0}(2a-u)\sqrt{u}\sqrt{(2a-u)}dx$

Im stuck there. I can`t find a simplify function that really works...

4. Jan 25, 2012

### nazfagge

Good topic I read and i understanding

5. Jan 25, 2012

### SammyS

Staff Emeritus
Hello nazfagge. Welcome to PF !

I'm glad that you understand this.

6. Jan 25, 2012

### SammyS

Staff Emeritus
Well, u = 2a/x was a typo.

I meant: Let u = x/(2a). Then x = (2a)u , and dx =(2a)du.

You can then factor 2a out of a whole lot of stuff.