Please see attachment.

My workings are -

I am stuck on Volume part. (b) and (c)

Please help.

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- April 19th 2009, 07:01 PMShyamVolume
Please see attachment.

My workings are -

I am stuck on Volume part. (b) and (c)

Please help. - April 19th 2009, 07:13 PMderfleurer
Easiest way to do this is to use shell method (I find).

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). - April 19th 2009, 07:24 PMCalculus26
For Part c see attachment -- I answered a similar problem yesterday

- April 19th 2009, 07:30 PMCalculus26
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

outer radius 2^x+1 and inner radius 4-x^2 - April 19th 2009, 07:32 PMCalculus26
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

- April 19th 2009, 07:38 PMderfleurer
Ah, sorry. Shouldn't have rushed that answer, I suppose.

But, with shell method, the idea is to integrate (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

- April 19th 2009, 07:46 PMShyam
- April 19th 2009, 07:47 PMCalculus26
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 - April 19th 2009, 07:51 PMCalculus26
No you have to integrate over the same interval a to b which as defleurer pointed are the intersection point of the 2 curves

- April 19th 2009, 07:58 PMderfleurer