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### test-2-sols

Course: MATH 2350, Fall 2008
School: Texas Tech
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Word Count: 581

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2350H1 MATH SOLUTIONS TO SECOND IN-CLASS EXAM Problem 1: Find the centroid of the quadrilateral with corners at (0, 0), (0, 1), (3, 1) and (3, 3). Answer: Call the quadrilateral Q. Q is the domain between the lines 1 2 y = 3 x and y = 3 x + 1, for 0 x 3. Thus, it is of Type I, and we nd 3 2 x+1 3 dA = Q 0 3 1 3x 2 x+1 3 dy dx = 15 , 2 27 , 2 x dA = Q 0 3 1 3x 2 x+1 3 x dy dx = dy A = Q 0 1 3x y dy dx =...

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2350H1 MATH SOLUTIONS TO SECOND IN-CLASS EXAM Problem 1: Find the centroid of the quadrilateral with corners at (0, 0), (0, 1), (3, 1) and (3, 3). Answer: Call the quadrilateral Q. Q is the domain between the lines 1 2 y = 3 x and y = 3 x + 1, for 0 x 3. Thus, it is of Type I, and we nd 3 2 x+1 3 dA = Q 0 3 1 3x 2 x+1 3 dy dx = 15 , 2 27 , 2 x dA = Q 0 3 1 3x 2 x+1 3 x dy dx = dy A = Q 0 1 3x y dy dx = 6. Thus, x= 9 27/2 = , 15/2 5 y= 6 4 = . 15/2 5 1 2 MATH 2350H1 SOLUTIONS TO SECOND IN-CLASS EXAM Problem 2: Let S be the domain consisting of all those points in space that are inside the sphere with center (0, 0, 0) and radius 1, and above the cone given by z = 1 x2 + y 2 . (a) Set up triple integrals in cylindrical and spherical coordinates for the volume of S. [Hint: The spherical equation for the cone is = 1/(cos() + sin()).] (b) Select one of the triple integrals from (a), and use it to nd the volume of S. Answer: (a) In cylindrical coordinates, the cone is z = 1 r and the sphere is r2 + z 2 = 1. S consists of those points above the unit disc that are above the cone and below the upper hemisphere. Thus, in cylindrical coordinates, we have 2 2 1 1r dV = S 0 0 1r r dz dr d. In spherical coordinates, the sphere is given by = 1 and the cone by = 1/(cos()+sin()). Since S sits in the upper half-space (z 0), we get /2 2 0 1 dV = S 0 1/(cos()+sin()) 2 sin() d d d. (b) Cylindrical coordinates look easiest: 2 2 1 1r dV = S 0 2 0 1 1r r dz dr d r( 0 2 0 = = 0 2 1 r2 1 + r) dr d 1 [ 1 (1 r2 )3/2 2 r2 + 1 r3 ]1 d 0 3 3 1 6 = 0 d = . 3 (Here, r 1 r2 dr was found by substituting u = 1 r2 .) MATH 2350H1 SOLUTIONS TO SECOND IN-CLASS EXAM 3 Problem 3: Let S be the solid enclosed by the cylinder + x2 z 2 = 1 between the planes y = 1 and y = 1. (a) Set up the triple integral using that S is x-simple. (b) Set up the triple integral using that S is y-simple. (c) Set up the triple integral using that S is z-simple. S f dV in rectangular coordinates, f dV in rectangular coordinates, f dV in rectangular coordinates, S S Answer: (a) The two halves of the cylinder are given by x = 1 z 2 on the domain 1 y, z 1. Thus, we get 2 1 1 1z f dV = S 1 1 f dx dy dz. 1z 2 (b) S consists of everything between the two planes y = 1 over the unit disc in the xz-plane. Therefore, 2 1 1x 1 f dV = S 1 f dy dz dx. 1x2 1 (c) The two halves of the cylinder are given by z = the domain 1 x, y 1. Thus, we get 2 1 1 1x 1 x2 on f dV = S 1 1 f dz dy dx. 1x2 4 MATH 2350H1 SOLUTIONS TO SECOND IN-CLASS EXAM Problem 4: Let C1 be the unit circle, and let C2 be the circle with 1 center ( 2 , 1 ) and radius 1/ 2. Set up and evaluate a double integral 2 to nd the area of the crescent D consisting of those points that are inside C...

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