`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,
Inner radius=(1+sec(x))-1=sec(x)
Outer radius=3-1=2
`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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`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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