This can be done by using integrals.

No if we draw this function inx,y cartesian plane. It will be curve (or line) from y(x=2) to y(x=3). Thne if we rotate this line around the x-axis we will get a volume enclosed by that rotated curve.

Bow how to find the volume? A volume can be considered as an integration of tiny areas. For this suppose z is any value in(2,3) range. ife we have dx extension from this (x+dx) you get a small line from y(x) to y(x+dx). Now if we rotate this you also get a small cylindrical volume, the volume of this can be approximated as,

`dv = pi*(y(x))^2 dx`

beacuse y(x) is the radius of that small cylinder.

now if we integrate all these tiny cylinders from x=2 to x=3, we will be able to find the volume of our rotated curve.

so,

`V = int_2^3 7sqrt(25 -x^2) dx`

now if we calculate the value of this integral we can get the answer.

To calculate the value of this integral, we have to make substitution as,

x = 5sin(t)

then `dx = cos(t) dt.`

and the limits, x = 2 becomes,

2 = 5 sin(t)

`t = sin^(-1)(2/5) = alpha`

similarly, at x=3, 3 = 5sin(t)

`t = sin^(-1)(3/5) =beta`

therefore our integral changes to,

`V = int_alpha^beta 7sqrt(25 -25sin^2(t)) cos(t)dt`

`V = 7int_alpha^beta 5cos(t)*cos(t)dt`

`V = 35int_alpha^beta cos^2(t)dt`

now from trignometry we know,

`cos^2(t) = (1-cos(2t))/2`

so,

`V = 35/2 int_alpha^beta (1-cos(2t))dt`

`35/2 int(1-cos(2t))dt = 35/2 (t-(sin(2t))/2) = 35/2(t -sin(t)cos(t))`

`V = 35/2(beta -sin(beta)cos(beta) -alpha+sin(alpha)cos(alpha))`

since `sin(beta)` is `3/5` then `cos(beta)` becomes `4/5 `

since `sin(alpha)` is `2/5` then `cos(beta)` becomes `sqrt(21)/5`

`V = 35/2 (beta-alpha+2/5*sqrt(21)/5 -3/5*4/5)`

beta = 0.6435 radians and alpha = 0.4115 radians

So,

V = 35/2 (0.6435 - 0.4115+2/5*sqrt(21)/5 -3/5*4/5)

V = 2.0756.

The volume is 2.0756