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HARVARD MATH 21A - VOLUME SPHERE/TORUS

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4/5/2004 VOLUME SPHERE/TORUS Math21a, O. KnillWe first calculate the volume of a sphere of radius R in different ways. Then we show how to calculate thevolume of the torus in three different ways. The page serves more as an illustration for the variety of toolswhich are available. We did not cover all the material (yet) to understand all of the methods.SEVEN WAYS TO COM-PUTE THE VOLUME OFTHE SPHERE1) IN RECTANGULAR COORDINATES. The volume isV =Z Z ZRdxdydz =ZL−LZ√L2−x2−√L2−x2Z√L2−x2−y2−√L2−x2−y2dzdydxAfter computing the most inner integral, we have V =RL−LR√L2−x2−√L2−x22(L2−x2−y2)1/2dydx. Define a2= L2−x2and use the integral in the box below:V = πZL−L(L2− x2) dx = 2πL3− 2πL3/3 = 4πL3/3 .Substitutionxa= sin(u), dx = a cos(u)du gives:Ra−a√a2− x2dx=Rπ/2−π/2aq1 −sin2(u)a cos(u) du = a2Rπ/2−π/2cos2(u)du =a2π/2 .2) IN CYLINDRICAL COORDINATES. At height z, we parameterize a disc of radius L2− z2, so that theintegral isZL−LZ√L2−z20Z2π0rdθdrdz =ZL−L(L2− z2)2πdz = 4πL3− 2πL3/3 = 4πL3/3 .3) IN SPHERICAL COORDINATES.ZR0Z2π0Zπ0ρ2sin(φ) dφdθdρ = 2πZR0ρ2dρZπ0sin(φ) dφ = 4πL3/3 .4) WITH CAVALIERI. Cavalieri cuts the hemisphere at height z to obtain a disc of radius√L2− z2with areaπ(L2− z2). He looked now at the complement of a cone of height L and radius L which when cut at height zgives a ring of outer radius L and inner radius z. The ring has area π(L2− z2). Cavalieri concludes (Cavalieriprinciple) that the volume of that body is the same as the volume of the hemisphere. Since the difference ofthe volume of the cylinder and the cone which is πL3−πL3/3 the hemisphere has the volume 2πL3/3 and thesphere has volume 4πL3/3.5) CAS. Integrate[1, {x, −L, L}, {y, −Sqrt[L2−x2], Sqrt[L2−x2]}, {z, −Sqrt[L2−x2−y2], Sqrt[L2−x2−y2]}]6) IN LAS VEGAS. The Monte Carlo Method is to shoot randomly onto the cube [−L, L] × [−L, L] × [−L, L]and see how many times we hit the sphere. Here an experiment with Mathematica:R := (2Random[] − 1); k = 0; Do[x = R; y = R; z = R; If[x2+ y2+ z2< 1, k + +], {10000}]; k/10000Assume, we hit 5277 of 10000 the measured fraction of the volume of the sphere with the volume of the cube 8is 0.5277. The volume of 1/8’th of the sphere is π/6 = 0.5247) USING GAUSS THEOREM (see later) The vector field F (x, y, z) = (x, y, z) has divergence 3 Gauss theoremtells that 3V is the flux of the vector field through the surface which is L times the surface area 4πL2. Therefore,V = 4πL3/3FIVE WAYS TO COM-PUTE THE VOLUME OFTHE TORUS1) WITH TORAL COORDINATES.T (r, θ, φ) = (x, y, z) = ((b + r cos(φ)) cos(θ), (b + r cos(φ)) sin(θ), r sin(φ)) parameterizes the torus.The Jacobean is det(T0) =∂(x,y,z)∂(r,θ,φ)= r(b + r cos(φ). The torus is the image of the cube [0, a] × [0, 2π] × [0, 2π]under the map T . The change of variables formula givesZa0Z2π0Z2π0r(b + r cos(φ)) dφdθdr = (2π)(2π)Za0br dr = 2π2a2b2) USING CYLINDRICAL COORDINATES.If we fix the z coordinate, we obtain an annulus with inner radius b −√a2− z2and outer radius b +√a2− z2.This annulus has the area π(b +√a2− z2)2−π(b −√a2− z2)2. Therefore, the volume is 4πbRa−a√a2− z2dz =4πb(πa/2) = 2π2a2b.3) USING PAPPUS CENTROID THEOREM. ”The volume of a solid of revolution generated by the revolutionof a region S in the x − z plane around the z axis is equal to the product of the area of S and the arc length2πb of the circle on which the center of S moves”.In the case of the torus, the length of the curve is 2πb. The area of the lamina is A = πa2. Therefore, thevolume is 2π2a2b.PROOF OF THE CENTROID THEOREM. We use a coordinate change transformation. In Polar coordinates,the lamina S with center of mass (b, c) is parametrized by r and z. Introduce new coordinates T (u, v) = (u+b, v+c) = (r, z) so that (0, 0) is the center of mass in the new coordinates. The Jacobean of this coordinate change is 1.The volume of the solid of revolution is V = (2π)R RSr drdz = (2π)R RR(u + b) dudv = 2πbR Rb dudv = 2πbA,where we used thatR RRu dudv = 0 because (u, v) = (0, 0) = (R RRu dudv/A,R RRv dudv/A) is the center ofmass of R.4) MONTE CARLO AGAIN. Lets assume b = 2 and a = 1. If (x, y, z) is a random point in [−3, 3] × [−3, 3] ×[−1, 1] then (r − 2)2+ z2≤ 1 is the condition to be in the torus, where r2= x2+ y2.Assume, we hit 5484 of 100000 the measured fraction of the volume 72 of the box we estimate 72∗5484/100000 =3.94848 for the actual volume 4π2= 39.4784.5) CAS. Integrate[r, {r, 1, 3}, {theta, 0, 2Pi}, {z, −Sqrt[1 − (r − 2)2], Sqrt[1 − (r −


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HARVARD MATH 21A - VOLUME SPHERE/TORUS

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