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 calchelpplz Apr16-08 10:38 AM

Calculus..volume with cross sections

I need help please on a calculus problem. I am trying to teach myself and am doing fair. I am stuck on this though.

Question:find the volume using CROSS SECTIONS that are squares.
base is bounded by
y = x + 1
y = x squared - 1
cross sections are squares
perpendicular to the x axis

I know to use the area = side squared due to it being a square
I set the 2 equations equal to find the integral goes from -1 to 2

I took the top minus bottom to get equation of (x + 1 ) - ( x squared - 1)
which becomes all squared due to the definition of area of a square

so I get x^4 - x^3 -2x^2 -x^3 +x^2 +2x -2x^2 +2x +4
then I combine like terms and integrate?
I don't get the answer
answer should be 8.1 it says.

thanks in advance.

 Schrodinger's Dog Apr16-08 12:11 PM

If my head hasn't fallen off you should have:

$$[(x+1)-(x^2-1)]^2=-3x^2+4x-2x^3+4x+x^4[/itex] [tex]\int_{-1}^2 -3x^2+4x-2x^3+4x+x^4\;dx=?$$

What do you get for that?

 calchelpplz Apr16-08 01:15 PM

yes, I got that except one of the 4x I had 4 instead

so, if that is correct....
then I do the integral math, find that
then I plug in 2
then I plug in -1
and subtract the two answers?

 HallsofIvy Apr16-08 01:17 PM

Quote:
 Quote by Schrodinger's Dog (Post 1692486) If my head hasn't fallen off you should have: $$[(x+1)-(x^2-1)]^2=-3x^2+4x-2x^3+4x+x^4[/itex] Was that a typo (or has your head fallen off?) [tex][((x+1)- (x^2-1))^2= (-x^2+ x+ 2)^2= x^4-2x^3+ 4x+4$$
You don't need "4x" twice!

Quote:
 $$[\int_{-1}^2 -3x^2+4x-2x^3+4x+x^4\;dx=?$$ What do you get for that?

 Schrodinger's Dog Apr16-08 01:19 PM

Quote:
 Quote by HallsofIvy (Post 1692548) Was that a typo (or has your head fallen off?) $$[((x+1)- (x^2-1))^2= (-x^2+ x+ 2)^2= x^4-2x^3+ 4x+4$$ You don't need "4x" twice!
Yeah I think it was apologies it should have been just +4.

Puts head back on. :redface:

$$\int_{-1}^2 -3x^2+4x-2x^3+4+x^4\;dx=?$$

Quote:
 Quote by calchelpplz (Post 1692546) yes, I got that except one of the 4x I had 4 instead so, if that is correct.... then I do the integral math, find that then I plug in 2 then I plug in -1 and subtract the two answers?
Yes that's correct, sorry for the mistake. :frown:

 calchelpplz Apr16-08 01:33 PM

ok, thanks everyone.

I guess I must have done something wrong but get the answer now.
Of course I should have asked the next question first as now I am stuck on it. Can you post more than one question a day or is that considered a 'no can do' thing?

 Schrodinger's Dog Apr16-08 01:36 PM

Quote:
 Quote by calchelpplz (Post 1692568) ok, thanks everyone. I guess I must have done something wrong but get the answer now. Of course I should have asked the next question first as now I am stuck on it. Can you post more than one question a day or is that considered a 'no can do' thing?
Certainly, as long as you show working, or, at least indicate where you're totally flummoxed, you can post as many as you like. You might not always get a quick reply but be patient.

 calchelpplz Apr16-08 01:40 PM

oh wow that is great.

Do I start a new post then or continue with this one?
I guess you can tell it is my first day here at ye old physics forum

 Schrodinger's Dog Apr16-08 01:46 PM

Quote:
 Quote by calchelpplz (Post 1692584) oh wow that is great. Do I start a new post then or continue with this one? I guess you can tell it is my first day here at ye old physics forum
Welcome to PF.

And yeah just tack it on the end, is probably easier in this case, people will notice the bump generally.

 calchelpplz Apr16-08 02:02 PM

ok, here goes:

Cross sections for volumes method only:
question:
given
y = x cubed
y = 0
x = 1
cross sections are equilateral triangles
perpendicular to the y axis

so first I think (watch out) that when it says perpendicular to the y axis you use y values on your integral but have to solve any equations using x = to calculate your area.

so I solved for x = third root of y
(the top of the graph is this equation then.)

NOW, I get muddled big time.
It seems I have a triangle with the side of it defined as the third root of y.
so I found the height of an equilateral given that as my side
I got x on the square root of 3 all divided by 2.

I would then use this value as my 'x' in the area = 1/2 b*h
so
1/2 (third root of y) (third root of y)(square root of 3) = area?

am I totally off here?

 Dick Apr16-08 02:40 PM

Not totally off, but if b is a base of an equilateral triangle then the height is b*sqrt(3)/2. So (1/2)bh=?. I think you are missing a factor of 2, aren't you?

 calchelpplz Apr16-08 02:53 PM

ok, I see that height is b times square root of 3 all over 2
so area is (square root of 3 )divided by four, times b squared.
is my side of the triangle equal to that top graph which is x = cube root of y?

do I set b = cube root of y and use those values in the area calculation?
Obviously I have to get the equation in there.
I have a hard time with the visual here.

 HallsofIvy Apr16-08 02:56 PM

Quote:
 Quote by calchelpplz (Post 1692604) ok, here goes: Cross sections for volumes method only: question: given y = x cubed y = 0 x = 1 cross sections are equilateral triangles perpendicular to the y axis so first I think (watch out) that when it says perpendicular to the y axis you use y values on your integral but have to solve any equations using x = to calculate your area. so I solved for x = third root of y (the top of the graph is this equation then.) NOW, I get muddled big time. It seems I have a triangle with the side of it defined as the third root of y.
No. One side of the base figure is y= x3 or x= x1/3. The other end of that base is x= 1. The length of the base of a triangle is 1- x1/3, NOT "x1/3".

Quote:
 so I found the height of an equilateral given that as my side I got x on the square root of 3 all divided by 2. I would then use this value as my 'x' in the area = 1/2 b*h so 1/2 (third root of y) (third root of y)(square root of 3) = area? am I totally off here?

 Dick Apr16-08 02:57 PM

I can't tell by reading the problem statement. All I see is that it says that the triangle is perpendicular to the y axis. It doesn't say that y^(1/3) is the side of the triangle, or the height of the triangle or whatever. Unless it says more you'll have to guess or ask for clarification.

 Dick Apr16-08 03:06 PM

Quote:
 Quote by HallsofIvy (Post 1692647) No. One side of the base figure is y= x3 or x= x1/3. The other end of that base is x= 1. The length of the base of a triangle is 1- x1/3, NOT "x1/3".
Uh, good point, if you read it that way. I took the x=1 as a misprint for y=1. calchelpplz, how exactly did you quote that problem?

 calchelpplz Apr16-08 03:09 PM

oops
I was totally seeing it wrong.
Is it that the base is 1 minus (y to the 1/3)? (you had x to the 1/3)
If so I think I see it now.

thanks!

 calchelpplz Apr16-08 03:14 PM

Cross sections for volumes method only:
question:
given
y = x cubed
y = 0
x = 1
cross sections are equilateral triangles
perpendicular to the y axis

no typo............
so is the base equal to 1-(cubed root of y)
and my height is that base (1 - cubed root of y) , times the (cubed root of 3 )over 2?

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