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. 
If my head hasn't fallen off you should have:
[tex][(x+1)(x^21)]^2=3x^2+4x2x^3+4x+x^4[/latex] [tex]\int_{1}^2 3x^2+4x2x^3+4x+x^4\;dx=?[/tex] What do you get for that? 
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? 
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[tex][((x+1) (x^21))^2= (x^2+ x+ 2)^2= x^42x^3+ 4x+4[/tex] You don't need "4x" twice! Quote:

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Puts head back on. :redface: [tex]\int_{1}^2 3x^2+4x2x^3+4+x^4\;dx=?[/tex] Quote:

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

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. 
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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.

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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! 
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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