Volume Between Two Curves Help!! (1 Viewer)

[SUSYKINS]

ok so i've been using excel's 50 tips for HSC book and i've come across this question:

the area of the curves y = 3x^2 and y=4-x^2 for x>0 is rotated around the y-axis. find the volume of the solid formed.

the answer is 2(pi) units^2

im not sure how to do it because the formula they've used it really strange.

 

[Aerath]

 

[12o9]

Sorry about the handwriting =/.

nvm Aerath beat me T.T.

 

[tommykins]

回复: Re: Volume Between Two Curves Help!!

You get extra points for that smile at the end.

 

[Aerath]

Re: 回复: Re: Volume Between Two Curves Help!!

And clearer working out.

 

[VenomP]

Can I ask why you have to separate it into two different areas to solve it?

Can't it all be done at once by integrating: f(x) - g(x)?

 

[Dr.Chau]

Can I ask why you have to separate it into two different areas to solve it?

Can't it all be done at once by integrating: f(x) - g(x)?

nah babe, you can't. part of the curve is positve and the other bit is negative. i'm not very good at maths, and i haven't seen the questions, but that's what i think.

 

[VenomP]

nah babe, you can't. part of the curve is positve and the other bit is negative. i'm not very good at maths, and i haven't seen the questions, but that's what i think.

I thought that didnt matter for areas between two curves.

Or am I thinking of something else?

 

[Dr.Chau]

I thought that didnt matter for areas between two curves.

Or am I thinking of something else?

i dunno, as i said, me is well shit at maths.

 

[12o9]

Can I ask why you have to separate it into two different areas to solve it?

Can't it all be done at once by integrating: f(x) - g(x)?

By doing that, you're getting the area under f(x) minus g(x) hence giving you a totally different area/answer

 

[lolokay]

Can I ask why you have to separate it into two different areas to solve it?

Can't it all be done at once by integrating: f(x) - g(x)?

it's a revolution about the y-axis, so you would need pi[f(y)² - g(y)²] to be able to inegrate one function, but the shaded area is between functions of x, not y, so you have to inegrate the y functions individually where they are the closest to the y axis.

edit: and because the volume you're finding is not between the points of intersection of the 2 y functions