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:

Step-by-step explanation:

[tex]\text{Given that:}[/tex]

[tex]y = 1+ sec(x) \ \ y =3[/tex]

[tex]\text{we draw the graph and the curves intersect at:}[/tex]

[tex]x = - \dfrac{\pi}{3} \ and \ x = \dfrac{\pi}{3}[/tex]

[tex]\text{Applying washer method;}[/tex]

[tex]f(x) _{outer} - g(x) _{inner} --- (1)[/tex]

[tex]V= \int ^b_a A(x) \ dx --- (2)[/tex]

[tex]\text{outer radius = 3 - 1 = 2}[/tex]

[tex]\text{inner radius =}[/tex] [tex]( 1 + sec(x) ) - 1 = sec (x)[/tex]

[tex]A(x) = \pi ((2)^2 -(sec(x)^2) \\ \\ A(x) = \pi (4 - sec^2 (x)) ---- (3)[/tex]

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

[tex]V = \int ^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ \pi (4- sec^2 (x) ) \ dx[/tex]

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

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

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

[tex]V = 8 \pi(x)^{\dfrac{\pi}{3}}_{0} - 2 \pi \Big( tan \dfrac{\pi}{3} -tan (0)\Big )[/tex]

[tex]V = 8 \pi({\dfrac{\pi}{3}}-{0}) - 2 \pi \Big( tan \sqrt{3}-(0)\Big )[/tex]

[tex]V = 8 \pi({\dfrac{\pi}{3}}) - 2 \pi \Big( \sqrt{3}\Big )[/tex]

[tex]\mathbf{V = 2 \pi \Big(\dfrac{4\pi}{3}- \sqrt{3} \Big)}[/tex]

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