How do you solve an elliptic integral of the first kind?
This is the general integral 0 to pi over 2 d theta over the square root of 1 minus k square sine square of theta is equal to f of k pi over 2.
What is elliptic integral of first kind?
Abstract. The complete elliptic integral of the first kind K(k) is defined for 0<k<1 0 < k < 1 by K(k):=∫π20dθ√1−k2sin2θ. K ( k ) := ∫ 0 π 2 d θ 1 − k 2 sin 2 The real number k is called the modulus of the elliptic integral.
How do you find the complete elliptic integral?
- (1 − k2 sin2 θ)
- (3.2)
- The parameter k is called the modulus of the elliptic integral and φ is the amplitude angle.
- The complete elliptic integral is obtained by setting the amplitude φ = π/2 or sin φ = 1, the maximum range on the upper bound of integration for the elliptic integral.
- F.
- π
- )
What is elliptic integral of the second kind?
Incomplete elliptic integral of the second kind
Substituting t = sin θ and x = sin φ, one obtains the Legendre normal form: Equivalently, in terms of the amplitude and modular angle: where a is the semi-major axis, and e is the eccentricity.
What are elliptic integrals used for?
Elliptic integrals can be viewed as generalizations of the inverse trigonometric functions and provide solutions to a wider class of problems. For instance, while the arc length of a circle is given as a simple function of the parameter, computing the arc length of an ellipse requires an elliptic integral.
What is incomplete elliptic integral?
The incomplete elliptic integral of the second kind is defined as follows: E ( φ | m ) = ∫ 0 φ 1 − m sin 2 θ d θ Note that some definitions use the elliptical modulus k or the modular angle α instead of the parameter m. They are related as m = k2 = sin2α.
What do you mean by elliptic integral?
noun. : the integral as to x of a function rational in x and the square root of a polynomial of third or fourth degree in x.