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quester [9]
3 years ago
7

Use the shell method to write and evaluate the definite integral that represents the volume of the solid generated by revolving

the plane region about the y-axis. y = x5/2 y = 32 x = 0
Mathematics
1 answer:
harina [27]3 years ago
5 0

Answer:

The volume of the solid is 714.887 units³

Step-by-step explanation:

* Lets talk about the shell method

- The shell method is to finding the volume by decomposing

 a solid of revolution into cylindrical shells

- Consider a region in the plane that is divided into thin vertical

 rectangle

- If each vertical rectangle is revolved about the y-axis, we

 obtain a cylindrical shell, with the top and bottom removed.

- The resulting volume of the cylindrical shell is the surface area

  of the cylinder times the thickness of the cylinder

- The formula for the volume will be:  V = \int\limits^a_b {2\pi xf(x)} \, dx,

  where 2πx · f(x) is the surface area of the cylinder shell and

  dx is its thickness

* Lets solve the problem

∵ y = x^{\frac{5}{2}}

∵ The plane region is revolving about the y-axis

∵ y = 32 and x = 0

- Lets find the volume by the shell method

- The definite integral are x = 0 and the value of x when y = 32

- Lets find the value of x when y = 0

∵ y = x^{\frac{5}{2}}

∵ y = 32

∴ 32=x^{\frac{5}{2}}

- We will use this rule to find x, if x^{\frac{a}{b}}=c, then=== x=c^{\frac{b}{a}} , where c

 is a constant

∴ x=(32)^{\frac{2}{5}}=4

∴ The definite integral are x = 0 , x = 4

- Now we will use the rule

∵ V = \int\limits^a_b {2\pi}xf(x) \, dx

∵ y = f(x) = x^(5/2) , a = 4 , b = 0

∴ V=2\pi \int\limits^4_0 {x}.x^{\frac{5}{2}}\, dx

- simplify x(x^5/2) by adding their power

∴ V = 2\pi \int\limits^4_0 {x^{\frac{7}{2}}} \, dx

- The rule of integration of x^{n} is ==== \frac{x^{n+1}}{(n+1)}

∴ V = 2\pi \int\limits^4_0 {x^{\frac{9}{2}}} \, dx=2\pi[\frac{x^{\frac{9}{2}}}{\frac{9}{2}}] from x = 0 to x = 4

∴ V=2\pi[\frac{2}{9}x^{\frac{9}{2}}] from x = 0 to x = 4

- Substitute x = 4 and x = 0

∴ V=2\pi[\frac{2}{9}(4)^{\frac{9}{2}}-\frac{2}{9}(0)^{\frac{9}{2}}}]=2\pi[\frac{1024}{9}-0]

∴ V=\frac{2048}{9}\pi=714.887

* The volume of the solid is 714.887 units³

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Step-by-step explanation:

If we expressed a number as:

\\ N = a * 10^{b} (1)

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Then, <em>b</em> represents the <em>order of magnitude </em>of such a number (<em>Order of magnitude (2020), </em>in Wikipedia).

The order of magnitude can be defined as "...the smallest power of ten needed to represent a quantity" (Weisstein, Eric W. "Order of Magnitude". From MathWorld--A Wolfram Web Resource).

Having gathered all this information, we can proceed as follows:

<h3>First case</h3>

The<em> La Plata river dolphin</em> weighs between 30 and 50kg and the sperm whale weighs between 35,000 and 40,000kg.

Then, considering (1) and (3) to express the dolphin and whale's weight (since in this way the order of magnitude is the same as the exponent part in the <em>scientific notation</em>):

\\ 30kg \leq Dolphin_{weight} \leq 50kg

\\ 3*10^{1}kg \leq Dolphin_{weight} \leq 5*10^{1}kg

\\ 35000kg \leq Whale_{weight} \leq 40000kg

\\ 3.5*10^{4}kg \leq Whale_{weight} \leq 4.0*10^{4}kg

Since the range for the weights are in the same order of magnitude for both dolphin and whale (considering the definition above):

\\ Dolphin_{weight} = 10^{1}\;(order\;of\;magnitude=1)

\\ Whale_{weight} = 10^{4}\;(order\;of\;magnitude=4)

Then

\\ \frac{Whale_{weight} = 10^{4}}{Dolphin_{weight} = 10^{1}}

\\ \frac{Whale_{weight}}{Dolphin_{weight}} = \frac{10^{4}}{10^{1}}

\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{4-1}

\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{3}

Thus

<em>A sperm whale is </em><em>3</em><em> </em><em>orders of magnitude</em><em> heavier than a La Plata river dolphin.</em>

<h3>Second case</h3>

Following the same reasoning, we can conclude that <em>the radius of the larger ball is </em><em>one (1) order of magnitude</em><em> bigger than the radius of the smaller ball:</em>

\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = \frac{10^{1}}{10^{0}}

\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = 10^{1-0} = 10^{1}

<h3>Third case</h3>

For this case, we need to calculate <em>the volume of a sphere</em> for both radii (1cm and 10cm).

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Then, the volume of the <em>ball of radius 1cm</em> is:

\\ V_{radius=1} = \frac{4}{3}*\pi*(1cm)^{3}

\\ V_{radius=1} \approx 4.19*10^{0}cm^{3}

And, the volume of the <em>ball of radius 10cm</em> is:

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\\ V_{radius=10} \approx 4.19*10^{3}cm^{3}

Thus

\\ \frac{10^{3}}{10^{0}} = 10^{3}

As a result, <em>the volume of the larger ball is </em><em>3 orders of magnitude</em><em> bigger than the volume of the smaller ball</em>.

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