How do I find the volume of integration when it is rotated around a line that is not an axis?
963250
OK, I had to pull out my calc book for this.\n\nI think the "disk method" would be the "volume by cross-sections method" in my book.  Let me know if not.\n\nSo, do you have a mental picture of what's going to happen?\n\nx^4 meets y = 1 at x = 1 and x = -1, so the thing that's going to be rotated looks a bit like a squashed, colored-in letter U.\n\nThe method of cross sections (or disks) involves trying to find the area of an infinitely thin cross section, rotated around some other line.\n\nSo, if you can imagine rotating around y=2, going out from that line, you first run into y=1 and then into y=x^4.  That is, y=1 is the "inside" function and y=x^4 is the "outside" function.\n\nSo imagine for a moment that y=1 isn't there and you're just rotating y=x^4 about y=2.  Then the radius of the disk would be "2-x^4", and the area would be pi times (2-x^4)^2, right?\n\nNow subtract out the "inner ring" formed by rotating y=1 around y=2.  The radius would be 1, so the area would be just pi.\n\nSo... the function for the area of the cross section, A(x) would be:\n\nA(x) = pi[(2-x^4)^2 - 1] = pi[x^8 - 4x^4 + 4 - 1]\nA(x) = pi[x^8 - 4x^4 + 3]\n\nAnd then you would just integrate this from x = -1 to x = 1.\n\nIf you need more help, I could take it further.\n\nDisclaimer: I haven't had calc since 1991.
146505
There are several different cases of this problem, depending on what axis or line you rotate the curve around.  I would suggest adding some more details to the problem if you want a more specific answer (i.e. axis of rotation, limits, function).\n\nHere are some of the cases:\nLet y=f(x) be the function of the curve.\n\n1. Rotation about the x-axis:\nV=pi*Int(  (f(x))^2  )\nWhere Int() is the integral of whatever's inside from the limits of A to B, in relation to x. Also, "^" means "raised to the power of." Plus, "*" is the multiplication symbol, and pi is the constant we all know about (3.14.....)\n\n2. Rotation about the y-axis:\nV=2*pi*Int(x*f(x))\nWhere the symbols mean the same thing as above.\n\nNote: These are only specific cases of the problem. If your problem is different, then these formulas may be altered slightly.  Also, these equations give the volume of the space enclosed by the revolution of one curve. If you have more than one curve, then the formulas will be different too.  Oh yeah, and if you're not in a calculus class, then I have no idea how to do this with other non-calculus methods.
856026
Use the gas law\nPV=nRT
247410
There are different formulas for different shapes. This page has a list of them: http://www.math.com/tables/geometry/volumes.htm
1087751
The amount of air is the difference in volumes.\nA = V(r+1) - V(r) = (4/3)pi (r+1)^3 - (4/3)pi r^3\nA = (4/3)pi((r+1)^3 - r^3)
658163
The previous two answerers found the volume of the region under that surface and above R. That is not what is being asked for.\n\nThe quantity you have to integrate is \nsqrt(1+(dz/dx)^2 +(dz/dy)^2), where the derivatives are partial derivatives.\n\nIn this case, dz/dx=-1 and dz/dy=-2y, so we have to integate\nsqrt(2+4y^2) over the region R.\nThis gives\nint(0,2)(0,x) sqrt(2+4y^2)dydx.\nThis is unpleasant, so we try to interchange thew order of integration:\nint(0,2)(y,2) sqrt(2+4y^2)dxdy\n=int(0,2) (2-y)sqrt(2+4y^2) dy.\nThis is still rather messy, but can be broken into two integrals, the first of which is done by trig substitution and the second by dsimple substitution u=2+4y^2.\n\nI'll let you do the rest.
715400
Angle of line from horizontal position.
926939
if a vector has an angle of A above horizontal and V is magnitude of vector then\nx component = V cos A\ny component = V sin A
796480
use v^2 =  GM/r\n\nv = root GM/2r\n\norbit = 2 pi r\n\nfor part 2 use suvat equations, s=vu+.5at^2 etc and it should fall out
466950
Volume = Area x height\nArea = r x r x pi\npi is approximated by the fraction 22/7\nTherefore Volume of cylinder = 4 x4 x 22/7 x7 = 352 units^3
840169
u= x\ndv=cosx dx\n\ndu =dx\nv= sinx\n\n\nintegral of xcosx dx = xsinx - integral of sinx dx\n= xsinx + cosx +C\n`
888834
Integrate from -π/2 to π/2 and double it
171039
integrate it by parts : \nint[U.dv] = [U.V] - int[V.du]\nwhere U = x, dv = sin^-1
412752
This looks like it's been rotated 90 degrees counterclockwise.\n\nSo, I guess it would be as follows:\n\nx = -(9/25)(y-3)^2 + 7
307198
It is linear on intervals, on (- infinite, 0) and on (0, +infinite) but not on the whole R. \n\nTry plotting it, you get a reversed V, not a straight line.
147855
Try http://integrals.wolfram.com/
158690
substitute x=e^y\nthen use a table of integration formulas?