`y = 1 + sec(x), y = 3` Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer. (about y = 1)

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`y=1+sec(x),y=3`

Refer the image. From the graph, the curves intersects at x=-pi/3 and x=pi/3.

Using washer method,

A cross section is a washer of cross sectional area A(x) with,

`A(x)=pi(2^2-(sec(x))^2)`

`A(x)=pi(4-sec^2(x))`

Volume of the solid obtained by rotating the region bounded by the given curves about y=1 (V) is,

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

`V=int_(-pi/3)^(pi/3)pi(4-sec^2(x))dx`

`V=2piint_0^(pi/3)(4-sec^2(x))dx`

`V=2pi[4x-tan(x)]_0^(pi/3)`

`V=2pi((4*pi/3-tan(pi/3))-(4*0-tan(0)))`

`V=2pi(4*pi/3-sqrt(3)-0)`

`V=2pi((4pi)/3-sqrt(3))`

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Related Questions

scisser | Student

`y=1+sec(x), y=3` intersect when
`1 + sec x = 3 `
`=gt cos x = 1/2 `
`=gt x = -pi/6, pi/6` .

Using the washer method,
`V = pi * int_(-pi/3)^(pi/3) ((3 - 1)^2 - (1+ sec x - 1)^2) dx `
=` pi * int_(-pi/3)^(pi/3)( 4 - sec^2(x)) dx `
`= pi * (4x - tan x)|_(-pi/3)^(pi/3)`
`= 2 pi * (4pi/3 - sqrt(3)). `

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