Test help please :(

Hello, and welcome to FMH! :)

Have you been given permission to seek outside help for this test? If so, do you have any work to show for any of these questions? We actually prefer that you post one question per thread (so the threads don't become convoluted), and show what you've done so far so we know where you're stuck.
 
Well, for the first one, let's begin by graphing the region to be revolved about the \(x\)-axis:

fmh_0134.png

Can you give the volume of an arbitrary disk?
 
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For some \(x\) on \([0,1]\), what will be the radius of the disk at that location?
 
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Honestly sir I'm way behind because of baseball season. And wasn't able to go to much of my Cal 2 class.
 
ng3 said:
Honestly sir I'm way behind because of baseball season. And wasn't able to go to much of my Cal 2 class.
Click to expand...

Picture the region in the graph I posted revolving about the \(x\)-axis, and we are going to slice up the resulting volume into a bunch of really thin disks all of equal thickness. The slicing will be parallel to the \(y\)-axis. Can you see that the radius \(r\) of an arbitrary disk is equal to the distance from the upper bound of the region to the lower bound, that is:

[MATH]r=6x^2-0=6x^2[/MATH]
And so, the volume of such a disk is given by:

[MATH]dV=\pi r^2\,dx=\pi(6x^2)^2\,dx=36\pi x^4\,dx[/MATH]
And so, to find the total volume of the solid of revolution, we add up all the disks via integration:

[MATH]V=36\pi\int_0^1 x^4\,dx[/MATH]
Does this make sense?
 
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MarkFL said:
Picture the region in the graph I posted revolving about the \(x\)-axis, and we are going to slice up the resulting volume into a bunch of really thin disks all of equal thickness. The slicing will be parallel to the \(y\)-axis. Can you see that the radius \(r\) of an arbitrary disk is equal to the distance from the upper bound of the region to the lower bound, that is:

[MATH]r=6x^2-0=6x^2[/MATH]
And so, the volume of such a disk is given by:

[MATH]dV=\pi r^2\,dx=\pi(6x^2)^2\,dx=36\pi x^4\,dx[/MATH]
And so, to find the total volume of the solid of revolution, we add up all the disks via integration:

[MATH]V=36\pi\int_0^1 x^4\,dx[/MATH]
Does this make sense?
Click to expand...
YES!
 
ng3 said:
idk
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You cannot evaluate the definite integral in the formula for the volume I posted?

[MATH]V=36\pi\int_0^1 x^4\,dx[/MATH]
As far as definite integral go, this is arguably one of the simplest types. I don't say that to shame you or make you feel bad, I'm trying to get a feeling for what you can do when it comes to these problems.
 
MarkFL said:
You cannot evaluate the definite integral in the formula for the volume I posted?

[MATH]V=36\pi\int_0^1 x^4\,dx[/MATH]
As far as definite integral go, this is arguably one of the simplest types. I don't say that to shame you or make you feel bad, I'm trying to get a feeling for what you can do when it comes to these problems.
Click to expand...
Mark, Once again you have proven that you have infinite patience....

...or is it only (36/5)*[math]\pi [/math] ..... just an irrational amount either way....
 
ng3 said:
idk
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If you don't know how to answer this question I have grave doubts about how ready you are for this test. I'd speak to your instructor about this. Not to mention spending your time on baseball and not on your class...

-Dan
 
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