Tell me what I did wrong >.<

SvenBee · 2008-04-06 03:40:41

I got a my grades posted today on a math test and there was one problem that I thought I did correctly, but it turns out I did it wrong.

Set up the integral using the shell method for:

Some please just explain how to set up this integral.

LuisRodg · 2008-04-06 09:01:11

The formula to find the volume of a solid using the shell method is this:

So to set up your integral it would be like this:

Last edited by LuisRodg (2008-04-06 09:06:42)

SvenBee · 2008-04-06 10:53:06

Thank you very much.

mleiner · 2008-04-06 10:59:04

∫ x = 1 to 4 pi(1/x^2)^2 dx

mleiner · 2008-04-06 11:00:47

it's because the volume of a disc is pi r^2 and the radius is the upper y - the lower y and 1/x^2 - 0 = 1/x^2

SvenBee · 2008-04-06 11:10:25

Luis, I thought that when you had a horizontal axis of revolution that your equation is in terms of x???

LuisRodg · 2008-04-06 13:30:24

Your right. My bad.

So:

So the integral that we setup is in this form:

To be honest, im not sure myself I did it right. Feel free to find any errors on my part because I want to know myself if I did it right. This solids of revolutions were always tricky

Last edited by LuisRodg (2008-04-06 13:39:49)

SvenBee · 2008-04-06 13:56:39

you went from:

to:

Can you rationalize the denominator while it's a term of the binomial??

I went from:

to:

So there's some difference in the final answer here.

I'm confused XD

LuisRodg · 2008-04-06 14:03:06

Yes I can go from:

to:

because algebraically they are equal.

What your doing is the same thing.

Last edited by LuisRodg (2008-04-06 14:08:11)

SvenBee · 2008-04-06 14:05:30

Brain fart.... how?

Dragonshade · 2008-04-06 14:05:39

Luis is right

LuisRodg · 2008-04-06 14:08:54

SvenBee wrote:

Brain fart.... how?

Its just algebra. Nothing to do with calculus.

Do you see it now?

Last edited by LuisRodg (2008-04-06 14:09:14)

SvenBee · 2008-04-06 14:12:27

Yeah, the algebra was bogging me down, I've been doing too much match today. >.<

Anyways

But in the answer key I am seeing

LuisRodg · 2008-04-06 14:15:09

Depends on how many digits your willing to calculate pi with.

SvenBee · 2008-04-06 14:16:38

AHHH!!!! There it is. Lol, all those innocent brain cells sacrificed just because I didn't realize the digits for pi. argh. Thank you soo much.

LuisRodg · 2008-04-06 14:17:50

No problem. Pleasure to help you.

Thanks Dragonshade for confirming I was right.

SvenBee · 2008-04-06 14:20:06

If I was going to change the axis from x to something like x = 2 the integral would be the same except for y, which would now be (2-y). Correct?

LuisRodg · 2008-04-06 14:22:15

You mean changing x=1, x=4 to x=2, x=4?

SvenBee · 2008-04-06 14:24:02

No, i mean instead of revolving around the x-axis I revolve the entire thing around y=2 (not x=2 like in the original post, oops, smell the sizzling brain cells yet?).

LuisRodg · 2008-04-06 14:27:27

You cant rotate it around y=2. It cuts the graph. You would have to rotate it around the x-axis (y=0) or around y = negative

SvenBee · 2008-04-06 14:29:24

You could rotate it around x=1 without cutting the graph. But if you did that the first part of the integral would look like (1-y)?

LuisRodg · 2008-04-06 14:32:25

Rotating the graph around x=1 or y=1 or y=2 is completely different.

Can you please specify? Im confused.

SvenBee · 2008-04-06 14:39:11

SvenBee wrote:

If I was going to change the axis from x to something like x = 2 the integral would be the same except for y, which would now be (2-y). Correct?

What I said there was incorrect (I don't think I can blame this on a long day any longer legitimately). If i rotate the graph of

with boundaries at y=0, x=1, and x=4 around the axis at x=1, then my integral will change from to correct?

LuisRodg · 2008-04-06 14:42:36

What you have to realize is that in the initial problem you said to rotate around the x-axis which is the same as rotating around y=0. In which case we setup the integral as dy. If you now rotate around x=1 then the integral changes to dx and it changes completely.

SvenBee · 2008-04-06 14:45:29

I think I need to call it a night I posted x=1 something like 3 times in a row, and the whole time I meant y=1, I'm really sorry. You don't need to bother with it anymore I'll get a fresh look tomorrow.

I appreciate your help very much and thanks for putting up with my "coming down from a 48+ hour caffeine induced cram session".

>.>

Here's a kirby for your effort: (>^^)> and thanks again.

Math Is Fun Forum

#1 2008-04-06 03:40:41

Tell me what I did wrong >.<

#2 2008-04-06 09:01:11

Re: Tell me what I did wrong >.<

#3 2008-04-06 10:53:06

Re: Tell me what I did wrong >.<

#4 2008-04-06 10:59:04

Re: Tell me what I did wrong >.<

#5 2008-04-06 11:00:47

Re: Tell me what I did wrong >.<

#6 2008-04-06 11:10:25

Re: Tell me what I did wrong >.<

#7 2008-04-06 13:30:24

Re: Tell me what I did wrong >.<

#8 2008-04-06 13:56:39

Re: Tell me what I did wrong >.<

#9 2008-04-06 14:03:06

Re: Tell me what I did wrong >.<

#10 2008-04-06 14:05:30

Re: Tell me what I did wrong >.<

#11 2008-04-06 14:05:39

Re: Tell me what I did wrong >.<

#12 2008-04-06 14:08:54

Re: Tell me what I did wrong >.<

#13 2008-04-06 14:12:27

Re: Tell me what I did wrong >.<

#14 2008-04-06 14:15:09

Re: Tell me what I did wrong >.<

#15 2008-04-06 14:16:38

Re: Tell me what I did wrong >.<

#16 2008-04-06 14:17:50

Re: Tell me what I did wrong >.<

#17 2008-04-06 14:20:06

Re: Tell me what I did wrong >.<

#18 2008-04-06 14:22:15

Re: Tell me what I did wrong >.<

#19 2008-04-06 14:24:02

Re: Tell me what I did wrong >.<

#20 2008-04-06 14:27:27

Re: Tell me what I did wrong >.<

#21 2008-04-06 14:29:24

Re: Tell me what I did wrong >.<

#22 2008-04-06 14:32:25

Re: Tell me what I did wrong >.<

#23 2008-04-06 14:39:11

Re: Tell me what I did wrong >.<

#24 2008-04-06 14:42:36

Re: Tell me what I did wrong >.<

#25 2008-04-06 14:45:29

Re: Tell me what I did wrong >.<

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