A horizontal tank is 68" long and 24" in dia. How can I compute the number of gal. in an 8" depth. (Arc)?
477733
You need to integrate part of a circle.  Imagine this cylindrical tank centered at (0,-4) in order to accomodate the 8" depth.  First, you want to find your X values of integration.  Imagine that the liquid in the tank (which is upside down now) makes a horizontal line.  The distance from the center to the end of one of these lines is the radius, or 12".  The distance vertically is 4", or the radius minus the depth.  Therefore, you can find the horizontal distance by using right triangle properties. The answer is the square root of 128, each direction.  So you will be integrating from -(128)^(1/2) to 128^(1/2).\n\nNext, you must set up the integral.  The equation for this circle would be X^2 + (Y+4)^2 = 144.  Solving for Y gives us\n\nY = (144-X^2)^(1/2) - 4\n\nIntegrating this would give us\n\n[1/2(144-x^2) + 72arcsin(x/12)-4x] from the negative square root of 128 to the positive square root of 128.\n\nevaluating this gives us approximately 102.75 in^2\n\nMultiply this area by the length of the tank to get the volume...\n\n102.75 in^2 * 68 in = 6886.89 in^3\n\n1 in^3 = .004329 gal, so...\n\n3437.36 in^3 * .004329 gal/in^3 =30.24 gal.\n\nAs far as I can see, that seems logical.\n\nNote: I had the wrong integral formula to begin, so this has been edited.
904637
n + 8in. < 24in.\n    - 8in.    - 8in.\n          n< 16in.
948718
Width = 6 - 2x\nLength = 8 - 2x\nDepth = x\n\nVolume = x(6 - 2x)(8 - 2x)\n= 4x^3 - 28x^2 + 48x ← kindricko's work to here\n\nJust to add to kindricko's answer\n\nVolume is a positive quantity\nSo x cannot be negative or V will be negative\nSo the smallest value of x will be 0 (your graph will confirm this)\n\nAlso none of the dimensions of the box can be negative\nSo 6 - 2x cannot be less than 0 or even equal 0 (as then the box becomes a rectangle)\nie 6 - 2x > 0\nie x < 3\n\nSo the possible values for x lie in the domain\n0 < x < 3 and your graph should confirm this.\nNow somewhere in between the volume will reach a maximum.\nWhat is the value for x there, and so what is the maximum possible volume? (You can read this from your graph too if you draw it accurately enough)
1137727
I like the joke!  To play along, I define:\nr = metric r  and \nd = metric d\n\nV = (pi/3)*(d^2)*(3r - d)\n\nThis will work for the tank being any fraction of a sphere, not just 1/2
983864
First find the surface area of the tank:\n\nS.A. = 2x(lb+bh+hl) = 2(12x9 + 9x4 + 4x12) = 384 meter square\n\nSince it says 5 per meter and it is 2 meters wide, the Rs. 5 applies to 2x1 = 2 meter square of the sheet\n\nSo if ur paying Rs 5 for every 2m.sq., then u pay Rs. 2.5 for 1m.sq.\n\nTherefore, total cost = 384 x 2.5 = Rs 960
1044477
the smaller tank is 900L and the bigger tank is 1200L\n\nTank 1 = T\nTank 2 = T+300\n\n(2/3) T = (1/2) (T+300)\n(2/3) T = (1/2) T + 150\n(4/6) T = (3/6) T + 150\n(1/6) T = 150\nT = 900
453835
multiply the dimensions, length times width timed depth to get  volume, and then multiply the volume  times the volume of one gallon.  I gallon is 0.1336805 cubic feet. \n\nassuming your question is in feet and inches:\n\n18ft*34ft 3in *8ft = 4943.952 ft^3\n\n4943.952 ft^3 * 1gal/0.1336805ft^3 = 36,983.34 gallons
365724
V = Pi*r^2 * h\n\n90 ft diameter and 6 ft deep\n\nor 45ft radius\n6ft high\n\nV = Pi * (45)^2 * 6\nV = 2025pi * 6\nV = 12150pi\n\nV = about 38170.35ft^3\n\nIn gallons, that would be about 285500 gallons
640019
12 sides = 4 feet \n1 side = 4/12 = 1/3 feet\nVolume = (1/3)^3 = 1/27 feet
551690
T= L + 2B\n\nT is total surface area\nL is Perimeter x Height\nB is base area
361907
If the tank is a cylinder, the Volume is πr²h.  If you get the volume in cubic feet, multiply by 7.5 to get gallons.
925743
This is pretty straightforward.\nFirst you have to write the formula fo rthe cost of the tank as a function of, say, its radius, r:\n\nThe volume of the cylinder is 100 cubic feet.\nIt is also pi*r^2 times the height, h\n\nso let's solve for h\n100=pi*r^2*h\nh = 100/(pi*r^2)\n\nnext, the cost:\nthe rectangle cost is 2.40*(h*2*pi*r),\nor 2.4*100/(pi*r^2)*2*pi*r\nor, simplifying, 480/r\nThe cost of the base and top are:\n1.50*2 pieces * pi*r^2\n\nSo the total cost is\n480/r+3*pi*r^2\n\nNow you need to differentiate this and find the roots, and make sure they are minima, not maxima!
409279
The perimeter is the distance around the edge so in this case you would add 6, 8 and 10 together to get 24 inches.
865361
The formula is F=D²/16d\nWhere D is the diameter (10 feet) and d is the depth (5 feet)\nF = 100/80 = 1.25
257174
AN-3 = 3/16"  =  1/8" hose I.D.\nAN-4 = 1/4"   =  7/32   "   "\nAN-6 = 3/8"   =  11/32  "   "\nAN-8 = 1/2"   =  7/16   "   "\nAN-10 = 5/8"  =  9/16   "   "\nAN-12 = 3/4"  =  11/16  "   "\nAN-16 =  1"   =  7/8    "   "\nAN-20 = 11/4" =  11/8   "   "\nAn-24 = 11/2" =  11/2   "   "\n\nI don't have the metric conversions. Hope this Helps!
655504
can't answer the question, as you haven't indicated the amount of curvature to the oval, or any kind of an idea for how the bottom of the pool slopes, or even how deep it is.\n\nfor every 1 foot of depth, a 12 X 24 rectangular pool has 2154.39 gallons of water.  this should give you an idea of the total required by your "oval" pool.\n\n24 X 12 X 1 = 288 cubic feet\n\nthen convert cubic feet to gallons.  i did this the hard way, but found an online calculator that confirms i did it right, and is MUCH easier than the hard way ;)\n\nUPDATE:\nusing your graciously provided depth measure, the total required for a RECTANGULAR pool 12 X 24 is 11,131.01 gallons.  not enough info to answer for an oval, but at least you have an idea.
1162775
if u evr seen a water tank they r usually round\nhttp://www2.warnerbros.com/main/homepage/homepage.html\n\nvolume = sqr ft x ht\nsqr ft  (of a circle) = pi x half the diameter(radius) squared\n\nvol = pi x (1/2x38) ^2x 100\nvol=3.1415926535897932384626433832795 x (1/2 x 38)^2 x 100\nvol=113,411.49479459153590850142613639 cubic feet