Review Sheet Exam 3 Solutions

# Review Sheet Exam 3 Solutions - Math 13 Solutions to Exam 3...

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Unformatted text preview: Math 13 Solutions to Exam 3 Review Spring 2007 1. Set up (but do not evaluate) a triple inte- gral in the order indicated for the volume of the solid E if (a) E is bounded below by the paraboloid z = x 2 + y 2 and above by the plane 2 x + z = 3 , in the order dz dy dx . (b) E is bounded above by the sphere x 2 + y 2 + z 2 = 4 and below by the plane y + z = 2 , in the order dz dxdy . (a) A sketch of the region is: x y z and its projection onto the xy-plane is found by the intersection of the two surfaces: z = x 2 + y 2 = 3 − 2 x x 2 + 2 x + y 2 = 3 ( x + 1) 2 + y 2 = 4 , a circle centered at ( − 1 , 0) with radius 2, which has a sketch: 1 − 1 − 2 − 3 1 2 − 1 − 2 x y or, solving for y , y = ± radicalbig 4 − ( x + 1) 2 . Therefore the volume of the solid re- gion is: V = integraldisplay 1 − 3 integraldisplay √ 4 − ( x +1) 2 − √ 4 − ( x +1) 2 integraldisplay 3 − 2 x x 2 + y 2 dz dy dx . (b) A sketch of the region is: x y z and the projection is found by the in- tersection of the two surfaces: z 2 = 4 − x 2 − y 2 = (2 − y ) 2 = 4 − 4 y + y 2 0 = x 2 + 2 y 2 − 4 y 2 = x 2 + 2 y 2 − 4 y + 2 = x 2 + 2( y − 1) 2 whose graph and region is: 1 − 1 1 2 V = integraldisplay 2 integraldisplay √ 4 y − 2 y 2 − √ 4 y − 2 y 2 integraldisplay √ 4 − x 2 − y 2 2 − y dz dxdy . 1 2. Rewrite the integral integraldisplay 1 − 1 integraldisplay 1 x 2 integraldisplay 1 − y f ( x,y,z ) dz dy dx as an interated integral in the order dxdy dz . To develop a sketch of the region, consider the projection onto the xy-plane, or x 2 ≤ y ≤ 1 and − 1 ≤ x ≤ 1: x y 1 − 1 1 and a projection onto the yz-plane is in the first quadrant bounded by z = 1 − y : 1 1 y z Together, these can be used to produce the three-dimensional sketch of the solid region: x y z z = 1 − y y = x 2 The integral in the desired order is: integraldisplay 1 integraldisplay 1 − z integraldisplay √ y − √ y dxdy dz . 3. The region E is bounded by the surfaces z = 0 , x = 0 , x = 2 y and z = 1 − y 2 and is graphed below: x y z 2 1 z = 1 − y 2 x = 2 y Write the integral integraldisplayintegraldisplayintegraldisplay E f ( x,y,z ) dV in each of the following orders: dz dy dx , dy dxdz , dxdz dy . The projection of the region E onto the coordinate planes are: 1 2 1 2 x y y = x/ 2 y = 1 x z z = 1 − x 2 / 4 1 2 1 y z z = 1 − y 2 1 2 1 The three desired integrals are: 2 integraldisplayintegraldisplayintegraldisplay E f dV = integraldisplay 2 integraldisplay 1 x/ 2 integraldisplay 1 − y 2 f dz dy dx , integraldisplayintegraldisplayintegraldisplay E f dV = integraldisplay 1 integraldisplay 2 √ 1 − z integraldisplay √ 1 − z x/ 2 f dy dxdz , integraldisplayintegraldisplayintegraldisplay E f dV = integraldisplay 1 integraldisplay 1 − y 2 integraldisplay 2 y f dxdz dy ....
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Review Sheet Exam 3 Solutions - Math 13 Solutions to Exam 3...

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