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# solut2 - Solutions to homework 2 I.15 The area of a...

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Solutions to homework # 2 I.15. The area of a triangle is equal to half the area of the parallelogram defined by two sides of the triangle and the angle between them. The latter area is the absolute value of the cross product of the two vectors corresponding to the two sides. Taking all three pairs of the sides and comparing the obtained area, we get 1 2 AB sin γ = 1 2 AC sin β = 1 2 BC sin α = Area(Δ ABC ) . Dividing through by 1 2 ABC , we obtain sin α A = sin β B = sin γ C . I.16. Let a , b and c denote the sides of a triangle and let ~ a , ~ b , and ~ c denote the corresponding vectors directed so that ~ a = ~ b + ~ c . Then c 2 = ~ c · ~ c = ( ~ a - ~ b ) · ( ~ a - ~ b ) = ~ a · ~ a - 2 ~ a · ~ b + ~ b · ~ b = a 2 - 2 ab cos γ + b 2 where γ is the angle between ~ a and ~ b . I.25a. By Stokes’ (Greene’s) theorem, I C ~ u · d ~ l = Z Z S ~ ∇ × ~ u dA where the second integral is over the area enclosed by the curve C . For this problem, ~ ∇ × ~ u = ( - xy 2 ) ∂x - ( x 2 y ) ∂y = - y 2 - x 2 . In polar coordinates x = r cos θ , y = r sin θ , dxdy = rdrdφ , so x 2 + y 2 = r 2 and the integral becomes Z 2 π 0 Z 1 0

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solut2 - Solutions to homework 2 I.15 The area of a...

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