Question

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.

y=1+ secx, y =3; about y=1

Answer 1

Explanation:

`\text{Given that:}`

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

`\text{we draw the graph and the curves intersect at:}`

`x = - (\pi)/(3) \ and \ x = (\pi)/(3)`

`\text{Applying washer method;}`

`f(x) _(outer) - g(x) _(inner) --- (1)`

`V= \int ^b_a A(x) \ dx --- (2)`

`\text{outer radius = 3 - 1 = 2}`

`\text{inner radius =}`
`( 1 + sec(x) ) - 1 = sec (x)`

`A(x) = \pi ((2)^2 -(sec(x)^2) \\ \\ A(x) = \pi (4 - sec^2 (x)) ---- (3)`

`\text{The volume V =}\int ^{(\pi)/(3)}_{-(\pi)/(3)} \ \ A(x) \ dx`

`V = \int ^{(\pi)/(3)}_{-(\pi)/(3)} \ \ \pi (4- sec^2 (x) ) \ dx`

`V = 2 \pi \int ^{(\pi)/(3)}_(0)( 4 - sec^2 (x)) \ dx`

`V = 2 \pi \int ^(\pi/3)_(0) 4 . \ dx - 2 \pi \int ^(\pi/3)_(0) sec^2 (x) \ dx`

`V = 2 \pi(4) \int ^(\pi/3)_(0) 1 . \ dx - 2 \pi \Big( tan (x)\Big )^{(\pi)/(3)}_(0)`

`V = 8 \pi(x)^{(\pi)/(3)}_(0) - 2 \pi \Big( tan (\pi)/(3) -tan (0)\Big )`

`V = 8 \pi({(\pi)/(3)}-{0}) - 2 \pi \Big( tan √(3)-(0)\Big )`

`V = 8 \pi({(\pi)/(3)}) - 2 \pi \Big( √(3)\Big )`

`\mathbf{V = 2 \pi \Big((4\pi)/(3)- √(3) \Big)}`