how do u calculate the volume of a curve when rotated between point A and B?
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.
911301
If x = the length of a side of the square base and y = the height, then volume = lengthxwidthxheight \n\n32000 = x^2y\n\ny = 32000/x^2\n\nThere are 5 surfaces, 4 of which are rectangles with area xy and one square with area x^2\n\nSo surface area = 4xy + x^2\n\nBut y = 32000/x^2 so substitute to get\n\nSA = 4x(32000/x^2) + x^2\n\nThat is 128000/x + x^2, or, 128000x^-1 + x^2\n\nFind the derivative and set it = to zero.\nSA' = -128000x^-2 + 2x = 0\n\n2x = 128000/x^2\n\n2x^3 = 128000\n\nx^3 = 64000\n\nFind the cube root of 64000; that's x\n\nPlug in to get y
945543
A short element of the curve is dL=sqrt(dr^2+(r*dt)^2), r*dt elementary length of the arc with radius r and polar angle t (your theta), dr elementary increase of r for dt.\nThus dL=sqrt(r’^2+r^2)*dt.\nIf  r=a*(1-cost), then dL=a*sqrt((sint)^2+(1-cost)^2)=a*sqrt(2)*sqrt(1-cost)*dt. \nFortunately 1-cost=2*(sin(t/2))^2 or dL=2a*sin(t/2)*dt, L=-4a*cos(t/2) for t=0 until 2pi, thus L=-4*a*(-1-1)=8*a – Oh! check it poor kid!\nIf r=(sin(t/2))^2=0.5*(1-cost), then a=0.5 for t=0 until pi. It must be done likewise. Thus L=-4*0.5*(0-1)=2
1119363
Easy... fill the cone with some liquid.... then pour this liquid into a measuring cylinder and measure the amount of liquid... you'll get the volume of the cone.\n\nMathematically it is 1/3 the volume of Cylinder\nthat is 1/3(pie*r*r*h)
856302
Surface Area = (2 •  p   • r²) + (2 • p  • r • height)\n\nWhere (2 • p • r² ) is the surface area of the "ends" and\n(2 • p • r • height) is the lateral area (the area of the "side").\n\nVolume   =   p • r² • height   =   ¼ • p • d² • height\n\np = Pi
387389
general formula is:\nvolume= pi times the integral of a to b (f(x))^2 dx
877943
volume of cyllinder = area of circle times height\n\n= Pi x radius x radius  x height\n\nremember its just the same as finding the volume of any prism.\nfind the area of the shape of the end then multiply by its length!!
657525
If you have the equation of the curve then integrate it on the lower and upper limits of the variable. In some cases you can find devide the identical parts and find out area of a single identical part and multiply by the number of parts made (sine wave).\n\nIf shape is a simple square circle etc you can find area by simple formula.
911487
The equations of motion for the angular displacement of each disk are given by:\n\nDisk A: R1(t) = (1/2)*alpha1*t² + w1*t +Ri1\nDisk B: R2(t) = (1/2)*alpha2*t² + w2*t +Ri2\n\nFor part (a), we're looking for the time, T, for which R1(T) = R2(T)\n\n(1/2)*alpha1*T² + w1*T +Ri1 = (1/2)*alpha2*T² + w2*T +Ri2\n\nFrom the problem statement, we obtain the following information:\n\nRi1 = Ri2\nalpha1 = 0\nw1 = 9.5 rad/s\nalpha2 = 2.2 rad/s²\nw2 = 0\n\nSubstituting these values and simplifying, we obtain:\n\nw1*T= (1/2)*alpha2*T²\n\nw1= (1/2)*alpha2*T\n\nT = 2*w1 / (alpha2)\n\nT = 2(9.5 rad/s) / (2.2 rad/s²)\n\nT = 8.64 s\n\nTechnically, part (b) is impossible to say without knowing how the reference line is defined, but I'll give it a whirl. If the reference line is measured from the center of each disk to its outer edge, then yes.\n\nHowever, if the reference line is placed across the diameter of the disk, then no. In this case, the reference lines will be aligned when the angular displacement of Disk A is 180° or pi radians ahead of Disk B. To solve for this time, simply add pi to Disk B's equation of motion:\n\n(1/2)*alpha1*T² + w1*T +Ri1 = (1/2)*alpha2*T² + w2*T +Ri2 + pi\n\nSubstituting out known values again leads us to:\n\nw1*T= (1/2)*alpha2*T² + pi\n\nor:\n\n(1/2)*alpha2*T² - w1*T + pi = 0\n\n1.1*T² - 9.5*T + pi = 0\n\nUsing the quadratic formula, we find that \n\nT = .344 seconds or 8.292 seconds\n\nThus, if the reference line is directed across the entire disk, then they are momentarily aligned twice before having the same angular displacement.\n\n(This added after a while)\n{BTW, monohan47130 - no I didn't. Please read what I wrote. If the reference lines are defined along the radius only, then you are correct - there is only one point at which the disks have the same angular displacement. But, reference lines are typically made along the entire circle (diameter). This means that they will be aligned 3 times total in 8.64 seconds :) }
150092
t=to\n_____\n√1-v2/c2
389859
The volume of a sphere can be found by the formula: \nvolume = 4/3 π r3 \n\nhttp://grapevine.abe.msstate.edu/~fto/tools/vol/sphere.html\n\nTry this website to help you out:\nhttp://www.gomath.com/algebra/sphere.php
612367
Volume of cylinder = Area of the base X height = πr2h.
839189
The required depth = 0.3536m, or about 35.4cm\n\nThis can be solved with or without reference to the angle between the trough sides. The volume V = 6 A where A is the area of a triangular end. So we just have to find when A is a maximum.\n\nA = b*h/2 (where h is the required depth)\n\nAlso, b = 2*sqrt(0.5^2 - h^2)\n\nso A = h*sqrt(0.5^2 -h^2)\n\nNow find h, such that: dA/dh =0\n\nor, more simply:\n\nThe angle betwen the 2 pieces which maximizes the volume is 90 degrees(see footnote). Once you know this you can determine h since h = 0.5*cos(45) = 0.3526\n\nFootnote: Express the the end area in terms of the angle between the 2 pieces of wood. Use the area formula: A = (1/2)a*b*sin(x) where x is the angle between the two pieces. In your case, a=b= 0.5m and so  A = 0.5*0.5*sin(x)/2 = 0.125sin(x).\nSo the volume V=  6*0.125*sinx(x) = 0.75sin(x) cubic m. This will be a maximum when x = 90 degrees since we know the sine function value is a maximum when x=90 degrees, but formally:\n\ndV/dx = 0.75 d/dx sin(x) = 0.75cos(x)=0 when cos(x) =0 when x = 90degrees\nMax volume = 0.75sin(90) = 0.75 m^3
741778
This is not tricky at all. The volume is 1-1/2 cylinders of "height" 50,000 meters and radius 4 meters.\n\nThe formula for the volume of a cylinder is\n\nV = pi * r^2 * h
1175612
you can use this formula m1/v1 = m2v2 whereas\nm1v1 is the old conc. volume and m2v2 is the new concentration an volume
540563
weigh the graduated cylinder full with water.\ninvert the cylinder (still full) in a tub of water.\nblow bubbles into the cyliner to find the volume of the air you blew in.\nremove and reweigh the cyliner.\n\nit may be a little tricky, but it's not too bad.
449048
P1 x V1 / T1 = P2 x V2 / T2\n\nYou have all you need.  Calculate it.