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The quotients formed by the area of these polygons divided by the square of the circle radius can be made arbitrarily close to π as the number of polygon sides becomes large, proving that the area inside the circle of radius r is πr 2, π being defined as the ratio of the circumference to the diameter (C/d).
The lemniscate is symmetric to the midpoint of the line segment F 1 F 2. The area enclosed by the lemniscate is a 2 = 2c 2. The lemniscate is the circle inversion of a hyperbola and vice versa. The two tangents at the midpoint O are perpendicular, and each of them forms an angle of π / 4 with the line connecting F 1 and F 2.
A different technique, which goes back to Laplace (1812), [3] is the following. Let = =. Since the limits on s as y → ±∞ depend on the sign of x, it simplifies the calculation to use the fact that e −x 2 is an even function, and, therefore, the integral over all real numbers is just twice the integral from zero to infinity.
A polar rose is a mathematical curve that looks like a petaled flower, and that can be expressed as a simple polar equation, = (+) for any constant γ 0 (including 0). If k is an integer, these equations will produce a k -petaled rose if k is odd , or a 2 k -petaled rose if k is even.
For example, assuming the Earth is a sphere of radius 6371 km, the surface area of the arctic (north of the Arctic Circle, at latitude 66.56° as of August 2016 [7]) is 2π ⋅ 6371 2 | sin 90° − sin 66.56° | = 21.04 million km 2 (8.12 million sq mi), or 0.5 ⋅ | sin 90° − sin 66.56° | = 4.125% of the total surface area of the Earth.
Graphs of roses are composed of petals.A petal is the shape formed by the graph of a half-cycle of the sinusoid that specifies the rose. (A cycle is a portion of a sinusoid that is one period T = 2π / k long and consists of a positive half-cycle, the continuous set of points where r ≥ 0 and is T / 2 = π / k long, and a negative half-cycle is the other half where r ...
If the polar line of C with respect to a point Q is a line L, then Q is said to be a pole of L. A given line has (n−1) 2 poles (counting multiplicities etc.) where n is the degree of C. To see this, pick two points P and Q on L. The locus of points whose polar lines pass through P is the first polar of P and this is a curve of degree n−1.
This last integrand () ˙ can be recognized as the derivative of the earlier integrand (()) ˙ (with respect to r), and shows that a polar planimeter computes the area integral in terms of the derivative, which is reflected in Green's theorem, which equates a line integral of a function on a (1-dimensional) contour to the (2-dimensional ...
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