# Volume

• Apr 19th 2009, 06:01 PM
Shyam
Volume
My workings are -
I am stuck on Volume part. (b) and (c)
• Apr 19th 2009, 06:13 PM
derfleurer
Easiest way to do this is to use shell method (I find).

$2\pi\int\limits_{a}^{b}(y + 1)Rdy$

Where b and a are the y values where the two functions intersect, and R is the equation for the height of the shell. Basically f(y) - g(y).
• Apr 19th 2009, 06:24 PM
Calculus26
For Part c see attachment -- I answered a similar problem yesterday
• Apr 19th 2009, 06:30 PM
Calculus26
If you use the shell method you'd need 2 integrals

and you'd have to invert y = 2^x to get x = ln(y)

and y =3 -x^2 to get x = (3-y)^1/2

Don't know where defleurer got his results

The easiesies way is to use washers

• Apr 19th 2009, 06:32 PM
Calculus26
Aah! Defluerer didn't see the R in the integral -- I see what you are saying but it would still take 2 integrals and inverting the fns
• Apr 19th 2009, 06:38 PM
derfleurer
Ah, sorry. Shouldn't have rushed that answer, I suppose.

But, with shell method, the idea is to integrate $2\pi rh$ (surface area) is it not? And your height would be f(y) - g(y) (by which I did mean inverting the functions). Your radius is then the distance you are from the axis of rotation. Were that axis of rotation simply y=0, than the radius would just be y. In this case, our axis of rotation is y=-1 and thus our radius should be y + 1. Thus, i'm still thinking this should work

$2\pi\int\limits_{a}^{b}(y + 1)(f(y) - g(y))dy$
• Apr 19th 2009, 06:46 PM
Shyam
Quote:

Originally Posted by Calculus26
If you use the shell method you'd need 2 integrals

and you'd have to invert y = 2^x to get x = ln(y)

and y =3 -x^2 to get x = (3-y)^1/2

Don't know where defleurer got his results

The easiesies way is to use washers

Is the part (b) like this?

${\text{Volume }} = \pi \int\limits_{ - \sqrt 3 }^1 {\left( {3 - x^2 + 1} \right)^2 dx} - \pi \int\limits_0^1 {\left( {2^x + 1} \right)^2 } dx$
• Apr 19th 2009, 06:47 PM
Calculus26
Certainly the shell methid works

but it will take 2 integrals one from a to 2 and one from 2 t0 3 as f and g change at the line y = 2
• Apr 19th 2009, 06:51 PM
Calculus26
No you have to integrate over the same interval a to b which as defleurer pointed are the intersection point of the 2 curves
• Apr 19th 2009, 06:58 PM
derfleurer
Quote:

Originally Posted by Calculus26
Certainly the shell methid works

but it will take 2 integrals one from a to 2 and one from 2 t0 3 as f and g change at the line y = 2

Ah, yes, I see now. Should've have caught that one sooner.

Nevermind what I said. Play it safe and stick with washer. =p