Recent content by 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...

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!

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...

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...

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

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?

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?

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...