Mathematics, 30.06.2019 01:20 mgreenamb

Use the method of cylindrical shells to find the volume v generated by rotating the region bounded by the curves about the given axis. y = 4ex, y = 4e−x, x = 1; about the y-axis

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Volume is the measure of the amount of 3D space occupied by a given figure. The volume of intended solid of revolution is $V = 24e^2\pi \: \rm unit^3$ approx.

How to use cylindrical shell method to find the volume?

Volume of a cylinder with radius r units, and height h is $\pi r^2h$ cubic units.

When ta region is drawn, and rotated(say about y axis), there is formed a solid due to revolution around the axis of revolution. The volume of this solid can be found by various techniques.

Let we discuss cylindrical shell technique.

Suppose that revolution is being done on y axis. Cut dy amount of height in that solid, and in that short height region, you can find its volume by: Subtracting bigger cylinder's volume from smaller cylinder's volume. Now integrate such volumes over y axis.

For the given case, the diagram of the region which is being rotated is plotted below.

When we rotate it around y axis, there are 2 cylinders on each small partition dy, smaller cylinder is due to the curve $y = 4e^{-x}$ , and the bigger cylinder is due to the curve y = 4e^x

Their volume is given as:

$\pi (4e^x)^2 \times dy - \pi (4e^{-x}x)^2 \times dy = 16e^2\pi(e^x - e^{-x})dy$

This dy height cut will be integrated from y = 4/e to y = 1 (as the region is bounded by x = 1)

Thus,

the final volume is:

$V = \int^{4/e}_1 16e^2\pi(e^x - e^{-x})dy = [16e^2\pi(e^x + e^{-x})]^{4/e}_1\\\\V = 16e^2\pi [e^{4/e} + e^{-4/e} -(e+e^{-1} )]\\\\V \approx 16e^2\pi(1.5) = 24e^2\pi \: \rm unit^3$

Thus, The volume of intended solid of revolution is $V = 24e^2\pi \: \rm unit^3$ approx.

Learn more about cylindrical shell integration here:

link

Answer from: kidzay

The volume is $\frac{16\pi}{e}$

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: $V=\int\limits^b_a {2\pi xf(x)} \, dx$

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

thickness

* Lets solve the problem

- To find the volume V generated by rotating the region bounded

by the curves y = 4e^x and y = 4e^-x about the y-axis by use

cylindrical shells

- Consider that the height of the cylinder is y = (4e^x - 4e^-x)

- Consider that the radius of the cylinder is x

- The limits are x = 0 and x = 1

- Lets take 2π and 4 as a common factor out the integration

∴ $V=\int\limits^1_0 {2\pi x(4e^{x}-4e^{-x})} \, dx$

∴ $V=2\pi(4)\int\limits^1_0 ({xe^{x}-xe^{-x})} \, dx$

- To integrate $xe^{x}$ and $xe^{-x}$ we will use

integration by parts methods $\int\ {uv'=uv-\int{v}\,u' }\,$

∵ u = x

∴ u' = du/dx = 1 ⇒ differentiation x with respect to x is 1

∵ v' = dv/dx = e^x

- The integration e^x is e^x ÷ differentiation of x (1)

∴ $v=\int\ {e^{x}}\, dx= e^{x}$

∴ $\int\ {xe^{x}} \, dx=xe^{x}-\int\ e^{x}\, dx=xe^{x}-e^{x}$

- Similar we will integrate xe^-x

∵ u = x

∴ u' = du/dx = 1

∵ v' = dv/dx = e^-x

- The integration e^-x is e^x ÷ differentiation of -x (-1)

∴ $v=\int\ {e^{-x}} \, dx=-e^{-x}$

∴ $\int\ {x}e^{-x}\, dx=-xe^{-x}+\int\ {e^{-x}} \, dx=-xe^{-x}-e^{-x}$

∴ V = $8\pi \int\limits^1_0 ({xe^{x}-xe^{-x})} \, dx=8\pi[xe^{x}-e^{x}+xe^{-x}+e^{-x}]$ from 0 to 1

- Lets substitute x = 1 minus x = 0

∴ $V=8\pi[(1)(e^{1})-(e^{1})+(1)(e^{-1})+(e^{-1})-(0)(e^{0})+(e^{0})-(0)(e^{0})-(e^{0})]$

∴ $V=8\pi[e^{1}-e^{1}+e^{-1}+e^{-1}-0+1-0-1]=8\pi[2e^{-1}]=16\pi e^{-1}$

∵ $e^{-1}=\frac{1}{e}$

∴ $V=\frac{16\pi}{e}$

Use the method of cylindrical shells to find the volume v generated by rotating the region bounded b

Answer from: Quest

the answer to this is a. a^2 + b^2 = c^2

Answer from: Quest

Look, when you want to do something like ${x}^{2}$ put it like x^2. x2 could mean x² or x₂. so to clear confusion, know the notation. to find the product multiply the x's and you get -8x and the y's and you get 6y. since it's all being multiplied it is just -8x6y, or answer a. although the better answer is -48xy(combining -8 and 6) because you should multiply constants whenever you can.