. Differential and integral calculus, an introductory course for colleges and engineering schools. v^~. J xdz — x V? 3 = 0 Then by the table of integrals 2/= Vr2_a;2 Vr2 — X2 1/2^ ?/ W—x2—z/2-f- (r2 — a;2) sin-1 Z/=0 Finally, Vi Jy=o (r2 — z2):c. 7 . | J> _ ^ & _ _ x |> _ ^J. g. A ns. Second Solution. Cut the solid into slices by planes parallel to the:n/-plane. Let the student draw the figure and complete the solution. 219. The Triple Integral in Polar Coordinates. The posi-tion of a point P may be givenby three coordinates, p, 6, 4>, asshown in the figure. We mayregard p as the radi
. Differential and integral calculus, an introductory course for colleges and engineering schools. v^~. J xdz — x V? 3 = 0 Then by the table of integrals 2/= Vr2_a;2 Vr2 — X2 1/2^ ?/ W—x2—z/2-f- (r2 — a;2) sin-1 Z/=0 Finally, Vi Jy=o (r2 — z2):c. 7 . | J> _ ^ & _ _ x |> _ ^J. g. A ns. Second Solution. Cut the solid into slices by planes parallel to the:n/-plane. Let the student draw the figure and complete the solution. 219. The Triple Integral in Polar Coordinates. The posi-tion of a point P may be givenby three coordinates, p, 6, 4>, asshown in the figure. We mayregard p as the radius of thesphere on which P lies and whosecenter is at 0. 6 is the longitudeof P, and 4> is its spherical coordinates, asthey are called, are connectedwith the Cartesian coordinates. 336 INTEGRAL CALCULUS §219 of the point by the equations x = p sin 0 cos 6, y = p sin sin 0, z = p cos 0,as is readily seen from the seek to express the triple integral ///■ f(x, y, z) dV in terms of p, 9, 0. To this end we divide V into small elementsof volume in the following way: First, through OZ we passplanes whose angular distancesapart are A0i; A02, • . ■ • Second, about OZ as axis andwith 0 as vertex we describe cir-cular cones whose vertical semi-angles differ by A0i, A02, ....Third, about 0 as center wedescribe concentric spheres whoseradii differ by Api, Ap2, ....V is thus divided into small elements of volume AV, such asPQRSPQRS in the figure. P is the point whose coordinates are p, 6, 4>. PS = pA0, PQ = p sin 0A0,* PP = Ap, and the product of these three quantities is p2 sin0A0 A0 Ap. Now it can be proved that A7
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