Calculate the volume with a rational funtion

[yaakob7]

Homework Statement

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

 

[SammyS]

Homework Statement

Calculate the volume (V) generated by the given function:
\displaystyle 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:

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

now:

\displaystyle 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:

\displaystyle 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

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.

 

[yaakob7]

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.

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...

 

[nazfagge]

Good topic I read and i understanding

 

[SammyS]

Good topic I read and i understanding

Hello nazfagge. Welcome to PF !

I'm glad that you understand this.

 

[SammyS]

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

I'm stuck there. I can`t find a simplify function that really works...

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.