MATH 251 504 Practice Problems for Examination 2 Fall 2006 1 Find the absolute minimum and maximum values of the function f x y xy 6x2 on the closed region bounded by the parabola y 9x2 1 and the x axis 2 Calculate the double integral RR 2 a R 3xex dA where R 1 0 0 1 RR b R 5 cos y 2 dA where R is the region bounded by the lines y x x 0 and y 1 RR 2 2 c R ex y tan 1 xy dA where R is the region described in polar coordinates by r 0 4 0 r RR d R 5x2 y 2xy 5 dA where R is the region between the two parabolas y x2 10 and y 2x2 6 3 Find the volume of the solid which lies between the paraboloid z 6x2 y 2 and the xy plane and above the region x y 0 y 1 y 2 x y 4 Find the center of mass of a lamina occupying the region x y 0 x 5 0 y x2 with density function x y x2 RR 2 5 Determine whether the following is true or false 2 R x sin exy x5 dA 2 where R 0 1 2 0 R0 R1 2 6 Compute 1 x e y xy dy dx 7 Find the volume of the solid bounded by the parabolic cylinder z y 2 and the planes z 4y x 1 and x 2 8 Consider a solid object which occupies the space between the parabolic cylinders z y 2 and z y 2 2 and over the region bounded by the lines x 0 y 0 and y 1 x Suppose the object has density function x y z xyz Find the moments of inertia of the object about the x and y axes 1 Solutions 1 We have fx x y y 12x and fy x and both of these are zero at the point x y 0 0 which is on the boundary of the region For the part of the boundary of the region that lies on the x axis we have the function f x 0 6x2 for 31 x 31 which has maximum value 0 at 0 and minimum value 23 at 13 and 31 On the other part of the boundary we have the function g x f x 9x2 1 9x3 6x2 x for 31 x 13 Then g 0 x 27x2 12x 1 which is zero at x0 12 54 252 Since g 00 x0 0 the value g x0 0 03754 is maximum for g while g 13 g 13 23 is the minimum Thus for f the absolute minimum value is 23 while the absolute maximum value is approximately 0 03754 2 a Z 1 Z 0 1 Z x2 3xe dx dy 1 0 h3 2 0 x2 e 3 dy 1 e 2 x 1 ix 0 b Z 0 Z y Z 2 0 5 cos y dy 1 h Z ix y 5x cos y dy x 0 1 0 0 2 5y cos y 2 dy 1 i0 5 5 sin 1 sin y 2 2 2 1 c Z 4 Z 4 Z r2 re dr d 0 0 h1 2 0 e r2 ir Z d r 0 1 2 1 2 1 2 i 4 e e 16 4 4 0 4 d Z 2 x2 10 2 4 1 2 e 1 d 0 2 2 1 64 4 iy x2 10 1 x2 y 2 xy 6 dx 3 y 2x2 6 2 2x2 6 2 2 Z 2 Z 2 1 5 2 2 2 2 2 2 6 2 6 x x 10 2x 6 dx x x 10 2x 6 dx 2 2 2 3 Z 2 5 5x y 2xy dy dx Z h5 Expand to evaluate the first integral and for the second use substitution 2 3 The volume is given by Z 1Z y Z 2 2 V 6x y dx dy 0 y2 1 h 3 2 2x y x 0 ix y x y 2 Z dy 1 3y 3 2y 6 y 4 dy 0 2 1 i1 37 3 y4 y7 y5 4 7 5 0 140 4 The mass and moments are Z 5 Z x2 Z 2 m x dy dx 5 1 i5 x4 dx x5 54 5 0 0 0 0 Z 5 Z 5 Z x2 Z 5h 2 i 1 4 1 5 i5 54 1 2 2 y x 2 xy dx x dx x Mx x y dy dx 2 10 2 y 0 0 0 2 0 0 0 Z 5 Z x2 Z 5 i 6 5 1 5 My x3 dy dx x5 dx x6 6 6 0 0 0 0 and so the center of mass is 25 1 6 2 5 The statement is true On R the values of the function are bounded between 1 and RR 2 1 and the area of R is 2 Therefore 2 1 area R R x sin exy x5 dA 1 area R 2 6 Using Fubini s theorem Z Z 0Z 1 y 2 e xy dy dx 1 x 1 0 y 1 ye y 2 0 7 The volume is given by Z 2Z 4 2 Z xy dx dy Z 4y y dy dx 1 1 ix 0 1 2 dy xe y x2 y 2 x y 0 1 3 1 y2 1 4 i1 1 3 y dy e y 2 2 8 0 2e 8 y 2 e 0 Z Z 0 1 3 2 h h 1 iy 4 32 2y 2 y 3 dx 3 y 0 3 8 The moment of inertia about the y axis is given by ZZZ Iy 2 2 1 Z Z 1 x Z y 2 2 x z xyz dV E Z 1 Z 0 1 x 0 x3 yz xyz 3 dz dy dx y2 iz y2 2 1 dy dx x3 yz 2 xyz 4 2 4 z y 2 0 0 Z 1 Z 1 x 1 1 3 5 1 9 1 3 2 2 2 4 x y y 2 xy y 2 x y xy dy dx 2 4 2 4 0 0 Z 1h iy 1 x 1 1 1 1 3 2 x y 2 3 x y 2 2 5 x3 y 6 xy 10 dx 12 40 12 40 y 0 0 Z 1 1 3 1 x 1 x 2 2 3 x 1 x 2 2 5 12 40 0 1 1 3 2 3 4 6 10 x 1 x x 1 x x x dx 12 40 3 5 h1 Now expand to evaluate The moment of inertia Ix about the x axis can be similarly R 1 R 1 x R y2 2 2 computed as 0 0 y z 2 xyz dz dy dx y2 4
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