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Unformatted text preview: Solution to the Final Exam May 10, 2006 1. To evaluate the integral Z 1 Z √ 1 x 2 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 Z 1 Z √ 1 x 2 cos( x 2 + y 2 ) dydx = Z π/ 2 Z 1 cos( r 2 ) rdrdθ = Z π/ 2 sin( r 2 ) 2 1 r =0 dθ = Z π/ 2 sin 1 2 dθ = 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 = ZZ D Z δ ( x,y,z ) dV = ZZ D Z 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 = q x 2 + y 2 ⇒ z = r ⇒ ρ cos φ = ρ sin φ ⇒ tan φ = 1 . Thus, 0 ≤ φ ≤ π/ 4, and so m = Z π/ 4 Z 2 π Z 1 ( ρ cos θ ) ρ 2 sin φdρdθ dφ = Z π/ 4 Z 2 π Z 1 ρ 3 cos φ sin φdρdθ dφ....
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This note was uploaded on 04/02/2008 for the course MATH 234 taught by Professor Dickey during the Spring '08 term at University of Wisconsin.
 Spring '08
 DICKEY
 Calculus, Multivariable Calculus

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