Multiplying that by the average depth gives the volume in cubic meters. Length times width gives the surface area of the pool. Length x width x average depth x 7.5 = volume (in gallons) Variable Depth Pools: Square and Rectangular If you’d like to find the pool volume in gallons, multiply your results by 7.5, as there are 7.5 gallons for each cubic foot. Multiplying that by the depth gives the volume in cubic meters. Length x width x depth x 7.5 = volume (in gallons) Length x width x average depth = volume (in meters cubed) ![]() Constant Depth Pools: Square or Rectangular For accurate calculations, the pool should be divided into various areas according to the depth. Area of a right triangle: A = (L x W)/2Ĭ alculated the cubic volume by including the surface area and depth of the pool.Area of a square or rectangle: A = L x W.Following are the basic formulas and calculations to determine surface areas: We can use a few simple formulas to calculate pool size. Calculating a pool's area in square meters or feet is the first step in determining information including meters cubed or gallons, maximum capacity of swimmers and other critical information about your pool. And we're done.Calculating the capacity a pool or spa requires the volume and surface area of the pool or spa. So this is equal toġ,436.8 centimeters cubed. In centimeters cubed or cubic centimeters. To just put a pi there, because that might interpret So this is going toīe- so my volume is going to be 4 divided by 3. Some people evenĪpproximate it with 22/7. So I'll do that inĪpproximate pi with 3.14. So we're going toĬentimeters to the third power. The way, let's just apply this radius being 7Ĭentimeters to this formula right over here. And in fact, the sphereĭimensions that is exactly the radius away from the center. The radius of the sphere is half of the diameter. ![]() The volume of a sphere is volume is equal Sphere- and we've proved this, or you will see a proof for this Through the centimeter, that distance right over So we're imagining that weĬan see through the sphere. It a little bit so you can tell that it's ![]() ![]() You could view it asĪ globe of some kind. This isn't just a circle, this is a sphere. The volume of a full sphere is integral -r to r of pi(r^2 - x^2) dx. (By the way, if you take calculus later, you will be able to derive this formula in another way by finding an integral. Therefore, the volume of a full sphere is (4/3) pi r^3. Since we have found that the volume of Figure 2 is (2/3) pi r^3, the same is true for Figure 1, which is a hemisphere of radius r. So Figure 1 and Figure 2 have the same volume. Therefore, these cross sections have equal areas at every height. So the area of the cross section at height h is pi r^2 - pi h^2 = pi(r^2 - h^2). In Figure 2, the cross section is a ring-shaped region with outer radius equal to r (from the cylinder, since each cross section's radius is the cylinder's radius) and inner radius equal to h (from the cone, since in a cone with equal height and radius, each cross section's radius equals its height above the bottom point). So the area of the cross section at height h is pi^2 = pi(r^2 - h^2). In Figure 1, the cross section is a circle with radius sqrt(r^2 - h^2) from the Pythagorean Theorem (hypotenuse is r, one leg is h, and the other leg is the cross section's radius). Once we show that these cross sections have equal areas at every height, then Cavalieri's principle would imply that the volumes of Figure 1 and Figure 2 are equal (since the overall heights of the two figures are equal, specifically to r). Now we need to compare the areas of the horizontal cross sections of Figure 1 and Figure 2 at any given height h above the bottom. Great question!! The 4/3 isn't so obvious and requires some work to derive.įigure 1: the top half of a sphere with radius r.įigure 2: a cylinder with radius r and height r, but with a cone (with point on bottom at the center of the cylinder's bottom base) with radius r and height r removed from it.įrom the volume formulas for a cylinder and a cone, the volume of Figure 2 is
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