# Calculusvolume with cross sections

1. Apr 16, 2008

### calchelpplz

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?
answer should be 8.1 it says.

2. Apr 16, 2008

### Schrodinger's Dog

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?

Last edited: Apr 16, 2008
3. Apr 16, 2008

### calchelpplz

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

4. Apr 16, 2008

### HallsofIvy

Staff Emeritus
$$[((x+1)- (x^2-1))^2= (-x^2+ x+ 2)^2= x^4-2x^3+ 4x+4$$
You don't need "4x" twice!

5. Apr 16, 2008

### Schrodinger's Dog

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

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

Yes that's correct, sorry for the mistake.

Last edited: Apr 16, 2008
6. Apr 16, 2008

### calchelpplz

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?

7. Apr 16, 2008

### Schrodinger's Dog

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.

8. Apr 16, 2008

### calchelpplz

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

9. Apr 16, 2008

### Schrodinger's Dog

Welcome to PF.

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

10. Apr 16, 2008

### calchelpplz

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. Apr 16, 2008

### Dick

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. Apr 16, 2008

### calchelpplz

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. Apr 16, 2008

### HallsofIvy

Staff Emeritus
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".

14. Apr 16, 2008

### Dick

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. Apr 16, 2008

### Dick

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?

16. Apr 16, 2008

### calchelpplz

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. Apr 16, 2008

### calchelpplz

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?