Help with Disk/washer and Shell method

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Homework Statement



Region bounded by y=x^2 and y=3x. Set up integrals to determine volume of solid obtained by rotating the region about:

1. the x axis
show both disk/washer method and shell method

2. the line x=4
show both disk/washer method and shell method

The Attempt at a Solution



I think the bounds of the integral are 0,9 for first one
The second one i think is bounded by 0,4.

I just need help figuring out what the Radius is and the height for disk and shell.
 
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"Not sure how to go about solving the problem " isn't a valid attempt.

You must have tried some things? Have you looked in your textbook for a formula of the volume of a solid of revolution??
 
I got the forumlas

Washer v=pi integral from a to b [Rx^2- rx^2]dx
Shell= v=2pi integral from a to b [radius*height]dx
Disk same a washer just without little r
 
anybody?
 
For problem 1 i got for shell equation

y=shell radius
shell height= y^(1/2)-(y/3)
bounded from 12,16

FOr wahser i got
Rx = (y/3)-y^(1/2)
 
cummings15 said:
For problem 1 i got for shell equation

y=shell radius
shell height= y^(1/2)-(y/3)
bounded from 12,16
OK, these are correct for the radius and the "height". What about the thickness of the shell.
Where did you get 12 and 16 for the limits of integration? Did you draw a sketch of the region being revolved?

Where do the curves y = x2 and y = 3x intersect?
cummings15 said:
FOr wahser i got
Rx = (y/3)-y^(1/2)
What does Rx mean? The formula you have here is way off. Draw a sketch of the region and what the washer will look like.
 
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the 12 and 16 are what i got for a,b for the part of equation

i think a will be 0 because of x-axis but i am not sure what b will be
 
cummings15 said:
the 12 and 16 are what i got for a,b for the part of equation
What do you mean, the a and b part of the equation?
cummings15 said:
i think a will be 0 because of x-axis but i am not sure what b will be
Please take another look at my previous post. I asked several questions, of which you answered only one.
 
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