Please see attachment.
My workings are -
I am stuck on Volume part. (b) and (c)
Please help.
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
Ah, sorry. Shouldn't have rushed that answer, I suppose.
But, with shell method, the idea is to integrate $\displaystyle 2\pi rh$ (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
$\displaystyle 2\pi\int\limits_{a}^{b}(y + 1)(f(y) - g(y))dy$