Homework Help: 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: