Correct Values for Shell Method: y=6sqrt(x),y=6,x=0,abt x=1

[Shutterbug424]

The region bounded by y=6[sqrt(x)], y=6, and x=0 about the line x=1

I chose p(x) to be (1-x) and h(x) to be [6-6sqrt(x)]. Is this correct? The answer I got was 14pi/5.

 

[tkhunny]

Please figure out that notation is not standardized. "p(x)" and "h(x)" may not mean much.

Well, is it right? Do it with disks and see if you get the same answer.

 

[Shutterbug424]

Please figure out that notation is not standardized. "p(x)" and "h(x)" may not mean much.

Well, is it right? Do it with disks and see if you get the same answer.

I'm using the equation V=2pi * the integral from a to b of [p(x)h(x)dx], where p(x) is the distance from the center of the rectangle to the axis of revolution. And h(x) is the height of said rectangle.

I am less concerned with having the correct answer than I am with whether or not I am using the Shell Method correctly (establishing the correct p(x) and h(x)). Any help would be appreciated.

 

[mmm4444bot]

... I am less concerned with having the correct answer than ...

Hmmm. Is this your way of telling us that you are not sure how to determine the volume using discs, also? (This is a rhetorical question.)

If not, then why did you post your answer? (This is not a rhetorical question, but it is conditional.)

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Yes. You properly set up the dimensions of the approximating rectangles for cylindrical shells.

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... V = 2pi * the integral from [0] to [1] of p(x) * h(x) dx

I would like to know (to satisfy my own curiosity) whether or not during your evaluation of the volume you ever simplified the equation to what follows.

\(\displaystyle V \;=\; 12 \cdot \pi \cdot \int_0^1 (1 - x) \cdot (1 - \sqrt{x})\, dx\)

Cheers,

~ Mark

 

[tkhunny]

And really, for good practice and for building confidence, if you have the time, do it both ways!!

\(\displaystyle \pi \cdot \int_{0}^{6}\;1^{2}\;-\;\left[1-\left(\frac{y}{6}\right)^{2}\right]^{2}\;dy\)

 

[Shutterbug424]

... I am less concerned with having the correct answer than ...

Hmmm. Is this your way of telling us that you are not sure how to determine the volume using discs, also? (This is a rhetorical question.)

If not, then why did you post your answer? (This is not a rhetorical question, but it is conditional.)

-

Yes. You properly set up the dimensions of the approximating rectangles for cylindrical shells.

-

... V = 2pi * the integral from [0] to [1] of p(x) * h(x) dx

I would like to know (to satisfy my own curiosity) whether or not during your evaluation of the volume you ever simplified the equation to what follows.

\(\displaystyle V \;=\; 12 \cdot \pi \cdot \int_0^1 (1 - x) \cdot (1 - \sqrt{x})\, dx\)

Cheers,

~ Mark

Thanks all! I was becoming frustrated with this particular problem when I kept getting a different answer when I applied the washer method. It turned out to be a simple sign distribution error.

And to satisfy your curiosity, I did pull the constant out front. I can't remember specifics at the moment, but that sounds about right.[/code][/code]

 

[mmm4444bot]

... frustrated ... kept getting a different answer [than when applying] the washer method. It turned out to be a simple sign distribution error.

Okay. I would like to suggest that, in the future with similar situations, you post something like the following. The sooner we understand where you're coming from, the sooner we can provide an applicable response.



"I need to find the volume of the region bounded by y=6*sqrt(x), y=6, and x=0 rotated about the line x=1. I got 14?/5 using washers, but my answer using cylyndircal shells is not the same. Here is my work; I cannot find an error. Is 14?/5 the correct volume?"

< insert work here >



Cheers,

~ Mark