aa r X i v : . [ m a t h . G M ] F e b TILTED CONE AND CYLINDER, CONE AND TILTED SPHERE
MEHMET KIRDAR
Abstract.
We consider two classical volume problems related to elliptic integrals. One of these is remadefrom C. E. Rhodes’s Note in AMM in 1932. The other is inspired. Introduction
In this note, I discuss two classical volume problems. The first problem which, I saw in [1] , dates backto 1932, has a neat solution formula. I will reproduce the formula for the case k < Tilted cone and cylinder
Consider the cylinder x + y = 1 and the cone z = cot α p ( x − k ) + y , 0 ≤ k ≤ . We want to findthe volume of the bounded region inside the cylinder, under the cone and above z = 0. Here α is thefixed angle of the cone, the angle between the cone and its axis, 0 ≤ α ≤ π . Let the origin be O = (0 , , , the vertex of the cone be T = ( k, ,
0) and let P = (cos θ, sin θ, ≤ θ ≤ π, be a point one the unit circle of the xy -plane. Let the angle between T P and positive sideof the x -axis be φ , 0 ≤ φ ≤ π .If T P = R then by law of cosines, R = p − k sin φ − k cos φ . The perpendicular from P to xy -plane cuts the cone with height z = R cot α . Therefore, the parameterization of the region in tiltedcylindrical coordinates is 0 ≤ r ≤ R, ≤ φ ≤ π and 0 ≤ z ≤ r cot α . And in tilted coordinates volumedifferential is dV = rdrdφdz . With two successive integrations, the volume integral can be reduced to V = 2 cot α π Z R dφ .Now, R = (cid:0) − k sin φ (cid:1) − k cos φ − k cos φ + 3 k cos φ sin φ + 3 k (cid:0) cos φ (cid:1) q − k sin φ and since π Z ( − k cos φ − k cos φ + 3 k cos φ sin φ ) dφ = 0 we have V = 4 cot α π Z (3 k + 1 − k sin φ ) q − k sin φdφ. Date : 17 Feb. 2021
Mathematics Subject Classification. [2020] 51M25, 33E05.
Key words and phrases.
Cone, cylinder, sphere, elliptic, integral.
By the definition of the elliptic integral of the second kind E ( k ) = π Z p − k sin φdφ , V becomes V = 4(3 k + 1) cot α E ( k ) − k cot α π Z sin φ q − k sin φdφ. Next, E ( k ) = π Z sin φ p − k sin φdφ must be computed in terms of elliptic integrals. Let ∆ = p − k sin φ , S = sin φ and C = cos φ . The tricky identity is S ∆ = 2 k − k ∆ + 1 − k k
1∆ + (cid:20) −
13 (1 − S )∆ + k S C (cid:21) . Integrating from 0 to π π/ Z h − (1 − S )∆ + k S C i dφ = (cid:2) − SC ∆ (cid:3) π/ = 0, we find E ( k ) = 2 k − k E ( k ) + 1 − k k K ( k ) , and V = 49 cot α (cid:2) ( k + 7) E ( k ) + 4( k − K ( k ) (cid:3) where K ( k ) = π Z (cid:0) − k sin φ (cid:1) − dφ. This formula is obtained in [1]. There, formula for k >
Cone and tilted sphere
Let us find the volume of the bounded region between the tilted sphere ( x + k ) + y + z = 1 , ≤ k ≤ z = cot α p x + y . The sphere in spherical coordinates is ρ + 2 kρ cos θ sin φ + k − φ = α . Thus,the volume of the region0 ≤ ρ ≤ − k cos θ sin φ + q − k + k cos θ sin φ, ≤ θ ≤ π, ≤ φ ≤ α is found as V = π Z α Z ( ( k cos θ sin φ − k cos θ sin φ − k cos θ sin φ )+ (cid:16) k cos θ sin φ + − k sin φ (cid:17) p − k + k cos θ sin φ ) dφdθ. Since π Z ( k cos θ sin φ − k cos θ sin φ − k cos θ sin φ ) dθ = 0 and due to symmetry, we find ILTED CONE AND CYLINDER, CONE AND TILTED SPHERE 3 V = α Z π Z (cid:18) k cos θ sin φ + 43 (1 − k ) sin φ (cid:19) q − k + k cos θ sin φdθdφ. Let us define K = k sin φ p − k cos φ and thus V = α Z (cid:18) k sin φ + 43 (1 − k ) sin φ (cid:19) p − k cos φE ( K ) dφ − α Z k sin φ p − k cos φ π Z sin θ p − K sin θdθ dφ. From the first problem, π Z sin θ p − K sin θdθ = E ( K ) = 2 K − K E ( K ) + 1 − K K K ( K ) , hence, V = 49 α Z (8 k sin φ + 7(1 − k ) sin φ ) p − k cos φE ( K ) dφ − α Z sin φ p − k cos φK ( K ) dφ. We can now insert infinite series of E ( K ) and K ( K ) and do term by term integration to obtaina formula which involves trigonometric integrals. Let us recall that E ( K ) = π X n =0 c n − n K n and K ( K ) = π X n =0 c n K n where c n = (cid:18) (2 n )!2 n ( n !) (cid:19) . Putting these and K in the last equation we obtain V = 2 π ∞ X n =0 c n k n − n α Z k sin n +3 +(3 − k + 8 n ) sin n +1 φ (1 − k cos φ ) n − dφ. The zeroth term of the series gives the following approximation of the volume for small k :2 π (cid:18) (1 + 2 k ) √ − k − (1 + 4 k − k cos α ) √ − k cos α cos α + (2 − k ) arcsin k − arcsin( k cos α ) k (cid:19) and this gives the exact value 2 π − cos α ) in the limit k → . I thank Paul Bracken who introduced me with AGM and elliptic integrals.
References [1] C. E. Rhodes A Geometric Interpretation of Landen’s Transformation The American Mathematical Monthly Vol. 39,No. 10 (Dec., 1932), pp. 594-596
Department of Mathematics, Faculty of Arts and Science, Namık Kemal University, Tekirda˘g,Turkey
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