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

Answer:

Step-by-step explanation:

$\text{Given that:}$

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

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

$x = - \dfrac{\pi}{3} \ and \ x = \dfrac{\pi}{3}$

$\text{Applying washer method;}$

$f(x) _{outer} - g(x) _{inner} --- (1)$

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

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

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

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

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

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

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

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

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

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

$V = 8 \pi({\dfrac{\pi}{3}}-{0}) - 2 \pi \Big( tan \sqrt{3}-(0)\Big )$

$V = 8 \pi({\dfrac{\pi}{3}}) - 2 \pi \Big( \sqrt{3}\Big )$

$\mathbf{V = 2 \pi \Big(\dfrac{4\pi}{3}- \sqrt{3} \Big)}$