# Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 4 + sec x, -pi/3≤ x≤pi/3, y = 6; about y = 4 Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 4 + sec x, -pi/3≤ x≤pi/3, y = 6; about y = 4   V=______________ This can be done by using integrals.

First we shall change the function, by using a substitution, y = z +4

therefore our function becomes,

z =sec(x) and the y limit changes to f =2 and the totational axis changes to f =0, which is the x-axis and it makes...

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This can be done by using integrals.

First we shall change the function, by using a substitution, y = z +4

therefore our function becomes,

z =sec(x) and the y limit changes to f =2 and the totational axis changes to f =0, which is the x-axis and it makes the calulcation easier.

No if we draw this function inx,y cartesian plane. It will be curve (or line) from z(x=-pi/3) to z(x=pi/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 x is any value in (-pi/3,pi/3) range. if we have dx extension from this (x+dx) you get a small line from z(x) to z(x+dx). Now if we rotate this you also get a small cylindrical volume, the volume of this can be approximated as,

dv = pi*(z(x))^2 dx

` `

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

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

`V = int_((-pi)/3)^(pi/3) sec(x)dx`

`intsec(x)dx = ln(sec(x)+tan(x))`

so V is,

`V = ln(sec(pi/3)+tan(pi/3)) -(ln(sec((-pi)/3)+tan((-pi)/3)))`

`V = ln((sec(pi/3)+tan(pi/3))/(sec((-pi)/3)+tan((-pi)/3)))`

`V = ln((3.732)/(0.2679))`

V = 2.634

So the volume generated is 2.634

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