Calculus 3 Solution_to_the_Final_Exam

# Calculus 3 Solution_to_the_Final_Exam - Solution to the...

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Solution to the Final Exam May 10, 2006 1. To evaluate the integral 1 0 1 - x 2 0 cos( x 2 + y 2 ) dydx, we first note that the domain of integration is a y –simple region bounded by the curves x = 0, x = 1, y = 0, and y = 1 - x 2 . Since y = 1 - x 2 describes the upper half of the unit circle centered at the origin, the domain of integration is the sector of this circle in the first quadrant (as 0 x 1). It is thus natural to evaluate this integral in polar coordinates. We have 1 0 1 - x 2 0 cos( x 2 + y 2 ) dydx = π/ 2 0 1 0 cos( r 2 ) rdrdθ = π/ 2 0 sin( r 2 ) 2 1 r =0 = π/ 2 0 sin 1 2 = sin 1 2 θ π/ 2 θ =0 = π sin 1 4 2. Let D be part of the unit ball x 2 + y 2 + z 2 1 inside the cone z = x 2 + y 2 , with density δ ( x, y, z ) = z . (a) The mass of D is given by m = D δ ( x, y, z ) dV = D z dV. We see from the shape of D that it is natural to evaluate this triple integral in spherical coordinates. Clearly, 0 ρ 1 and 0 θ 2 π . The range of values of φ is determined by the shape of the cone. Note that z = x 2 + y 2 z = r ρ cos φ = ρ sin φ tan φ = 1 . Thus, 0 φ π/ 4, and so m = π/ 4 0 2 π 0 1 0 ( ρ cos θ ) ρ 2 sin φ dρ dθ dφ = π/ 4 0 2 π 0 1 0 ρ 3 cos φ sin φ dρ dθ dφ.

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