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Snowcat [4.5K]
3 years ago
10

HELP ME ASAP PLSSSSSSSS

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
7 0

Answer:

The answer is B.50

Step-by-step explanation

Replace x with 50

  1. (50+20)=70
  2. +50 from the other lone x
  3. Find z by subtracting 180 by the external angle (150) and get 30
  4. Add to get 180 which is what a triangle adds up to
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Use the method of cylindrical shells to find the volume v generated by rotating the region bounded by the curves about the given
kondor19780726 [428]

Answer:

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 

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

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* 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}

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Step-by-step explanation:

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Multiply the numerator and denominator by 7

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