i have a function, y=square root of 9+x^2\nwhat is the volume of the solid that results when the region bounded by the function and the x-axis is rotated about the x-axis?  \n\nalso... \n\nusing method of cylidrical shells to find volume of solid bounded by y=x, x=2, and y=-(x/2) is rotated about the y-asis\n\ni don't need the answers, just help on how to start...\n\nthanks!!!!
123590
Think of the first as the area of a flat round plate sweeping out the inside of the figure you get from the rotation.  What will be its area? πr² where the r is the y value of √(9 + X²).  That changing area will sweep out the volume from whatever the limits are (you don't say), so it's the integral of π(√(9 + X²))², or π(9 + X²).\n\nIn the second case, the thing sweeping out the volume is the surface area of a cylinder as its radius grows.  Thats the circumfrence of the base times the height.  The circumfrence is 2πr where r is the X value and the height is the difference between Y = x and Y = X/2.  So it's the integral from 0 to 2 of:\n\n   2πX(X-X/2).\n\nThinking of an area sweeping thru the volume helps.
419758
I have seen this question sometime this weekend!\nYour question is more complete, though.\nV=pi.r^2h=1000\nS=2pi.r(r+h)\n\nExpress h in terms of r\nh=V/(pi.r^2)\n\nSubstitute in S\nS=2pi.r(r+V/(pi.r^2))\n\nDifferentiate S w.r.t. r,\ndS/dr=2*pi*(V/(pi.r^2+r-2V/(pi.r^2))\n\nEquating this to zero and solve for r:\nr=(V/(2pi))^(1/3) and a negative root\nFor V=1000,\nr=5.41926\nh=1000/(pi.r^2)=10.8385\n\nThe result is not surprising, as the dimensions resemble that of a cube, or a sphere, which have the minimum surface area for a given volume.
313018
What you wrote isn't a cone. It is the part of the hyperboloid of one sheet x^2+y^2-z^2=1 with z>=0. The intersection with the sphere is a circle on the sphere and so the volume is 0.\n\nIf you *meant* the cone z=sqrt(x^2+y^2), you can evaluate the volume in several ways. I would suggest changing to spherical coordinates, where the equation of both the cone and the sphere are very simple. [rho=1 and phi=pi/4] Don't forget the rho^2 sin(phi) term when integrating!
263967
Equate the two volumes.\n\nFirst volume:\n\nV1 = int 2 pi x y dx, x=0-->1\n\nV1 = int 2 pi x^(n+1) dx\n\nV1 = 2 pi [x^(n+2)] / [n+2]\n\nV1 = 2 pi/(n+2)\n\nSecond volume:\n\nV2 = int 2 pi x (1-y) dx, x=0-->1\n\nV2 = int 2 pi x (1-x^n) dx\n\nV2 = int 2 pi (x-x^(n+1)) dx\n\nV2 = 2 pi {(x^2)/2 -[x^(n+2)] / [n+2]}\n\nV2 = 2 pi {1/2 -[1^(n+2)] / [n+2]}\n\nV2 = 2 pi (n/2)/(n+2)\n\nEquating:\n\n2 pi (n/2)/(n+2) = 2 pi/(n+2)\n\nn = 2
322434
I assume delta(x, y, z) = 1 is the mass density ... isn't rho usually used for that?  Anyway, you need to integrate mass times square of perpendicular distance from axis of rotation.\n\nLooks like it will be messy.\n\nMaybe outer x from 0 to 1, middle y from 0 to 1 - x^2, inner z from 0 to (12 - 4x - 3y)/2, of delta(x, y, z) (x^2 + y^2) dz dy dx.
11047
well, the formula for the volume of a cone is 1/3 * base * height.\n\nthe trick to doing this is to find the volume of the large cone, then subtract the volume of the small cone.  what's leftover will be the tumbler-shaped volume.  plug in numbers for the height and base for both the large and small cones, and then subtract. voila!
363502
1. obtain formulas for surface area and volume of cylinder\n2. let volume equal 1000 and solve this volume equation for h\n3. substitute the expression from previous step into surface area    equation to get equation with only r's\n4. differentiate surface area formula with respect to r set this equal to 0 and solve for r i got   r=10/3rd root(pi)\n5. substitute value for r into volume equation solve for h  i got h=10/(pi)^(2/3)\n\nincidentally, i used 2nd derivative test to conclude r is minimum\n\n\n\ni'll get back to u with more
54546
OK, I love a good math problem, here goes...\n\nVolume of cylinder is PI r^2 h.\n\nHeight of cylinder is h.\n\nRadius of cylinder if sqrt (R^2 - h^2).\n\nThus volume of cylinder = PI (R^2 - h^2) h. This is equal to PI R^2 h - PI h^3.\n\nTo find maximum, find derivative and set equal to zero, or\nPI R^2 = 3 PI h^2.\n\nThis in turn simplifies to h = R / sqrt (3).
342130
Easy once you figure out how to set up the equation.  To make things simpler, let's assume the radius = 1.  Since\n\nV = pi(r^2)h'\n\nThen V = pi(1^2)h = pi(h)\n\nNow we're going to reduce the radius by 5%, the new volume will be\n\npi(0.95^2)h =\n\npi(.9025)h\n\nNow we need the multiplier x (for the change in height) such that:\n\nx(pi(0.9025)h) = V = pi(h)\n\nSolving for x gives\n\nx = pi(h)/[pi(0.9025)h\n\nx = 1/(0.9025)\n\nThis is about 1.108 and is the multiplier for the height to restore the volume, right? So you would subtract 1 (the original radius) from that, then multiple by 100 to get the percent change in h needed which yields about 10.80%.  The exact answer would be \n\n100((1/0.9025)-1), again approximately equal to 10.80\n\nLet's test and make sure.  If the original wire is 100mm long with a radius of 1mm, its volume is:\n\n pi(1^2)(100) = 100pi. \n\nReducing radius 5% and increasing the height by the proposed exact amount gives a new volume of:\n\npi(.95^2)(100 * (1/(0.9025)) =\n\n100pi, exactly restoring the volume.\n\nGary H
454505
hmm im not quite in college yet, but i hope to be help.\ndo you know the values for f, h, v, and a?\nif you do, id substitute it in.\nthen square both sides. \nand then solve.
363112
Ok, this is hard to explain...\n\n1. Graph the function: |9x-2|\n2. Draw a vertical rectangle in each x interval of one (a rectangle between 0 and 1, 1 and 2, etc.; it makes a total of six rectangles.\n. Add the volume of the rectangles together\n\nor\n\n3b. Use the sum function (sigma) to add the volume of an infinite number of rectangles together between 0 and 6. If your confused, just do 3a.\n\n4. You're done.
552199
i think the volume is constant, so\nV= pi r^2 h, so\nh = V/(pi r^2)\nthen\n dh/dt = V/pi * (-2)/r^3 * dr/dt =\n- 128*2/(1,8)^3 * 0.05 which i think is a) \nBtw it's pi not pie....lol
385242
Simple application of L = I w  with w = 6 rad/m\n\nCase 1:Center-of-mass case \nI = 1/2 MR Sq: giving \nL= (1/2 M R Sq)(w) =1/2 x3 x0.2x 0,2 x 6 \n=0.36 kg m Sq .Ans 1\n\nCase 2: Case of midway-center and rim:\n(Using II-Axis Theorem.)\nI = 1/2 MR Sq + M (R/2)Sq = 3/4 MR Sq giving \nL = 3/4MR Sq x w =3/4 x 3 x 0.2 x 0.2x 6 = 0.54 kg m Sq Ans 2 \nGot it?
340932
I will answer the first question and maybe you can figure out the second two for yourself. giving you the dimensions of r and h you can find the ratio and the rest of the information. When you approach the other problems try to find the area of the shape you are trying to solve for and substitute for any second variable by soving for h in the volume then plugging into the total area before you take the derivative.\n\nArea of the square tops: 2 * 2r^2\nArea of the side: 2 * pi(r^2)\nTotal Area after multiplying out: 8r^2 + 2pi(rh)\n\nThen find a substitute for for h because you can't take the derivative of two variables. you can get the substitute from the volume. \n\nVolume = 1000=pi (r^2h) ... h= 1000/2pi (r^2)\n\nArea = 8r^2 + 2pi (r) (1000/2pir^2)\nArea = 8r^2+2000/r\nArea' = 16r-(2000/r^2)\n16r^3=2000 ...r=3.969 h=20.206\n\nGOOD LUCK!!!
176883
well for f of g, you know the answer is \n       5/ (4(square root of x) -2)    right? \nTo find the domain, you just need to find all the x values the function exists at-- (the word domain means x-values.)\nIt should exist everywhere except where\n    4(square root of x)-2 \nequals zero, because then the denominator would equal zero, and that is undefined.\n\nif you need more help just ask!
636870
Domain = all x-values included in the function\nRange = all y-values included in the function\n\nBoth will be written as an inequality.  Here's an example:\n\ny = square root(x-3) + 2\n\nGraph this equation on your calucluator or sketch a graph by hand by plugging in x coordinates starting with 3.  The vertex of this graph is at (3, 2).  Therefore, the domain is:\n\nx is greater than or equal to 3\n\nThe range is:\n\ny is greater than or equal to 2.\n\nIt really helps to visualize the problem when doing these problems.\n\nGood luck!
419644
Call the side of the square you cut out x, then the box you make will have a square base of side (24 - 2x) and the height will be x. \n\nThe area of this box is then (24 - 2x)^2*x = 2x*(12 - x)^2. Call this a function of x, and we need to find the maximum.\n\n2x^3 -48x^2 + 288x = f(x)\n\n6x^2 - 96x + 288 = f'(x) = 0\n\nx^2 - 16x + 48 = 0\n\n(x - 12)(x - 4) = 0\n\nSo this has the 2 answers x = 12 and x = 4.\n\nx = 12 gives a minimum (volume = 0), so x = 4 is a maximum,  1024\n---------------\nI just looked at the above answers. JimBurnel made a careless error, the volume should be x(24 - 2x)^2, not x(24 - 2x). The rest of his logic is right, but the equation is wrong.\n\nAnd Paritosh Vasava assumes by "largest" you want surface area, I assume you mean volume. Of course the largest surface area comes when the squares you cut out are of size 0, your box is then 24*24*0, area = 576, volume = 0.