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tresset_1 [31]
3 years ago
13

PRECALC!! NUMBER 8 & 9!! DUE IN 20 MINUTES!! All help is appreciated!!

Mathematics
1 answer:
yarga [219]3 years ago
4 0

Answer:

? Post the questions...

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Please help!!!!!! Please!
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Answer:

you have to divide is bye 4 to get the answer

Step-by-step explanation:

1.56

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Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the re
alukav5142 [94]

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)}

8 0
2 years ago
Combining and Distributing
Viktor [21]
What's the question v
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