# Integration - areas

1. Jun 5, 2010

### zebra1707

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

Find the area enclosed by the y axis, the line y = 3 and the curve x = y^2

2. Relevant equations

[/b]3. The attempt at a solution[/b]

Area = $$\int$$ 3 to 0 y^2.dy
= (1/3 y^3) 3 to 0
= (1/3 x 27)
= 9 sq units

Im not really confident on the y axis questions, can you please confirm my understanding. If this same area is then rotated 360 degrees about the y -axis, can I seek guidence on how to do this. Cheers P

2. Jun 5, 2010

### Mentallic

What you did is correct. If you aren't comfortable with questions involving the y-axis, you can always change the question so it can be equivalently integrated with the x-axis. For example, x=y2 between y=3 and the y-axis has the same area as y=x2, x=3 and the x-axis

Do you know the formulae (or even better, derive the formulae) for volumes of revolution?

If y=f(x) is rotated about the x-axis between a and b, the volume is $$\pi\int_a^b\left(f(x)\right)^2dx$$

3. Jun 5, 2010

### zebra1707

Thanks Mentallic

Yes, I realise I know the volume formulae. I calculate the volume would be 48.6 units cube - do you mind checking?

Cheers P

4. Jun 6, 2010

### Mentallic

Um, no. Show me your working and I'll point to where you went wrong.

5. Jun 7, 2010

6. Jun 7, 2010

### Mentallic

Yeah exactly, so why are you saying it's 48.6 u3? Whatever happened to the $\pi$? :tongue:

7. Jun 7, 2010

### zebra1707

Sorry, oooopppsss, can I say 48.6pi u3? I got the impression that I had gone horribly wrong. Manythanks for you responses.

Cheers P

8. Jun 7, 2010

### Mentallic

Of course, all you did was convert the fraction to a decimal, so there's no point but yeah, sure I guess :tongue: