Calculusvolume with cross sections

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Homework Help Overview

The discussion revolves around finding the volume of a solid with cross sections defined as squares and equilateral triangles. The first problem involves determining the volume using squares as cross sections, with the base bounded by the equations y = x + 1 and y = x^2 - 1. The second problem focuses on a solid with equilateral triangle cross sections, defined by the curves y = x^3, y = 0, and x = 1, with cross sections perpendicular to the y-axis.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss setting up integrals based on the area of squares and equilateral triangles. There are attempts to clarify the equations and bounds for integration, as well as the correct expressions for the areas of the cross sections. Questions arise about the correct interpretation of the problem statements and the relationships between the variables involved.

Discussion Status

Participants are actively engaging with the problems, sharing their calculations and questioning each other's reasoning. Some have expressed confusion about specific steps, while others are providing clarifications and corrections. There is a collaborative effort to ensure understanding of the concepts involved, though no consensus has been reached on the final solutions.

Contextual Notes

Participants note the importance of correctly interpreting the geometry of the cross sections and the bounds of integration. There are discussions about potential typos in the problem statements and the implications of those on the calculations. Some participants express uncertainty about the rules for posting multiple questions in the forum.

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

[tex][(x+1)-(x^2-1)]^2=-3x^2+4x-2x^3+4x+x^4[/itex]<br /> <br /> [tex]\int_{-1}^2 -3x^2+4x-2x^3+4x+x^4\;dx=?[/tex]<br /> <br /> What do you get for that?[/tex]
 
Last edited:
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?
 
Schrödinger's Dog said:
If my head hasn't fallen off you should have:

[tex][(x+1)-(x^2-1)]^2=-3x^2+4x-2x^3+4x+x^4[/itex][/tex]
[tex] Was that a typo (or has your head fallen off?)<br /> [tex][((x+1)- (x^2-1))^2= (-x^2+ x+ 2)^2= x^4-2x^3+ 4x+4[/tex]<br /> You don't need "4x" twice!<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> [tex][\int_{-1}^2 -3x^2+4x-2x^3+4x+x^4\;dx=?[/tex]<br /> <br /> What do you get for that? </div> </div> </blockquote>[/tex]
 
HallsofIvy said:
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[/tex]
You don't need "4x" twice!

Yeah I think it was apologies it should have been just +4.

Puts head back on. :redface:

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

calchelpplease said:
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:
 
Last edited:
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?
 
calchelpplease said:
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.
 
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
 
calchelpplease said:
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.
 
  • #10
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?
 
  • #11
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?
 
  • #12
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.
 
  • #13
calchelpplease said:
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".

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?
 
  • #14
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.
 
  • #15
HallsofIvy said:
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. calchelpplease, how exactly did you quote that problem?
 
  • #16
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!
 
  • #17
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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