Find Area Enclosed by Y-Axis, y=3 & x=y^2

[zebra1707]

Homework Statement

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

Homework 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

 

[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

 

[zebra1707]

Thanks Mentallic

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

Cheers P

 

[Mentallic]

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

 

[zebra1707]

\pi\int_a^b\left(f(x)\right)^2dx

y^2 = x and line y = 3

becomes y^4

= pi [y^5/5] 3 to 0

= pi [243/5]


Example
http://www.intmath.com/Applications-integration/4_Volume-solid-revolution.php

Cheers P

 

[Mentallic]

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

 

[zebra1707]

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

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

Cheers P

 

[Mentallic]

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