Use the method of cylindrical shells to find the volume of thesolid, generated by the enclosed region when rotating throughone revolution.y=6x-2x, y=x, Rotation about x-0Answer the following:1111Sketch the graphs and show all information to determine thevolume using the method specified above.aSet-up the formulation for the volume of the approximatingrectangle (i.e. elemental volume).bSpecify the domain of the enclosed region..cSet-up the integral and if possible, simplify it.d

Question
Asked Sep 12, 2019

Hi, could you let me know if the integral for part (c) of the attached image is V=integral from 0 to 3 of 2pi x((6x-2x^2)-x^2))dx ? Thank you.

Use the method of cylindrical shells to find the volume of the
solid, generated by the enclosed region when rotating through
one revolution.
y=6x-2x, y=x, Rotation about x-0
Answer the following:
11
11
Sketch the graphs and show all information to determine the
volume using the method specified above.
a
Set-up the formulation for the volume of the approximating
rectangle (i.e. elemental volume).
b
Specify the domain of the enclosed region.
.c
Set-up the integral and if possible, simplify it.
d
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Use the method of cylindrical shells to find the volume of the solid, generated by the enclosed region when rotating through one revolution. y=6x-2x, y=x, Rotation about x-0 Answer the following: 11 11 Sketch the graphs and show all information to determine the volume using the method specified above. a Set-up the formulation for the volume of the approximating rectangle (i.e. elemental volume). b Specify the domain of the enclosed region. .c Set-up the integral and if possible, simplify it. d

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Step 1

Consider the enclosed region by  equation 1 and equation 2

6x-2x equation (1)
equation (2)
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6x-2x equation (1) equation (2)

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Step 2

In the graph blue represent to equation 2 and red represent to equation 1

|(2, 4)
4
2
2
-2
(0, 0)
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|(2, 4) 4 2 2 -2 (0, 0)

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Step 3

Now solve the both equation for the inte...

6x-2x22
=X
6х — х? + 2х?
3x? — 6х %— 0
3x (х- 2) -0
Then,
solutins are x=0, x-2
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6x-2x22 =X 6х — х? + 2х? 3x? — 6х %— 0 3x (х- 2) -0 Then, solutins are x=0, x-2

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