Let R be the region in the first quadrant bounded by the graph of y=sqrt{x-2} and the line y=2.(a). Find the volume of the solid generated when R is revolved about the x-axis.
(b). Find the volume of the solid generated when R is revolved about the line y=-2.

Answers

Answer 1
Answer:

Volume of the solid generated when R is revolved about the x-axis is 10π and  the volume of the solid generated when R is revolved about the line y = -2 is 40π/3.

What is Graph?

Graph is a mathematical representation of a network and it describes the relationship between lines and points.

The volume of the solid generated when R is revolved about the x-axis,

V=\int\limits^a_b\pi {y^(2) } \, dx

where a and b are the x-coordinates of the points of intersection of the curve y = √(x-2) and the line y = 2.

Solving y = √(x-2) and y = 2 for x, we get:

x = 6 and x = 2

Limits of integration are a = 2 and b = 6. Substituting y = √(x-2) into the formula for the volume, we get:

V = \int\limits^6_2\pi\sqrt{(x-2)^(2) } \, dx

V= π [(6²/2 - 2(6)) - (2²/2 - 2(2))]

=10π

Volume of the solid generated when R is revolved about the x-axis is 10π.

b. The volume of the solid generated when R is revolved about the line y = -2

V=\int\limits^a_b\pi {(y+2)^(2) } \, dx

Substituting y = √(x-2) into the formula for the volume, we get:

V=\int\limits^2_6\pi (√(x-2)+2)^(2) \, dx

We can simplify this by using the identity:

V =40π/3

Therefore, the volume of the solid generated when R is revolved about the line y = -2 is 40π/3.

Hence, Volume of the solid generated when R is revolved about the x-axis is 10π and  the volume of the solid generated when R is revolved about the line y = -2 is 40π/3.

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Answer 2
Answer: A.) Find the volume of the solid generated when R is revolved about the x-axis.

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Answers

Answer:

The solution to the inequality is;

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Step-by-step explanation:

We want to find the solution to the inequality;

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We can have this as;

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Step-by-step explanation:

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The two cylinders are similar. If the ratio of their surface areas is 9/1.44, find the volume of each cylinder. Round your answer to the nearest hundredth. A.small cylinder: 152.00 m3
large cylinder: 950.02 m3
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Answers

r1² π : r2² π = 9 : 1.44
r1 : r 2 = 3 : 1.2
r2 = 1.2 · r1  / 3
r 2 = 0.4 · r1
If the two cylinders are similar than also: h 2 = 0.4 · h 2
V 1 = r1²   · π  · h1
V 2 = ( 0.4 r1)² π · 0.4 · h1 = 0.064 r1² · π · h1
V 2 ( small cylinder )= 0.064 · V 1 ( large cylinder ) 
851.22 m³= 13.300.25 m³ · 0.064
Answer: C )
The two cylinders are similar. If the ratio of their surface areas is 9/1.44, find the volume of each cylinder. Round your answer to the nearest hundredth.

Answer: out of all the available options the one that shows the correct volumes for each cylinder is answer choice C) small cylinder: 851.22 m³ & large cylinder: 13,300.25 m³.

I hope it helps, Regards.

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Answers

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Where x is the number of times this growth/decay occurs, a = initial amount, and r = fraction by which this growth/decay occurs.

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More about the exponent link is given below.

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It begins with 4

After 1 hour= 4 x 2= 8
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