ha lvh262 HomeWork 15 3 karakurt 56295 This print out should have 25 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points Evaluate the integral Z 1 Z x2 I x 5y dy dx 0 5 I 9 20 Consequently I 002 3 1 1 4 2 4 10 0 points Which if any of the following are correct A For all continuous functions f Z 1Z 2 Z 2Z 1 f x y dx dy f x y dx dy 0 0 0 1 both of them 2 A only 3 B only 4 neither of them correct 003 Explanation After integration with respect to y we see that x2 Z 1 5 2 dx I xy y 2 0 0 Z 1 5 4 3 x x dx 2 0 1 1 4 1 5 x x 4 2 0 0 0 B False incorrect reversal of the order of integration when integrating over the upper triangle in the square 0 1 0 1 13 3 I 20 7 20 0 Explanation A False incorrect reversal of the order of integration when integrating over a rectangle 3 correct 4 4 I B For all continuous functions g Z 1Z y Z 1Z x g x y dx dy g x y dy dx 0 11 1 I 20 2 I 1 0 0 10 0 points Evaluate the double integral Z 1Z y I 6x y dx dy 1 1 I 2 I 4 15 2 correct 15 3 I 4 I y2 7 15 1 3 5 I 1 15 Explanation Treating I as an iterated integral integrating first with respect to x with y fixed we see that Z 1Z y I 6x y dx dy 1 y2 Z 1 1 3x2 xy y y2 dy ha lvh262 HomeWork 15 3 karakurt 56295 Thus 2 But then Z I 1 2 4 3 y y dy 1 Z 1 1 y 2 y 3 dy 3 1 3 y y5 y4 y 3 3 5 3 4 1 Consequently I 004 Consequently o 0 y 1 8 15 19 30 10 0 points when D n o x y 0 x 1 x y x 1 I 2 3 2 I 4 7 3 I 0 13 correct 30 y 1 I 0 y 6x y dx dy Now 4 I 16 correct 21 5 I 8 9 Explanation The double integral can be rewritten as the repeated integral I Z 1 Z 0 y 13 30 D Explanation The integral can be written as the repeated integral Z Z Z 2 2 i1 y 2 y 5 2 y 3 2 5 3 0 Evaluate the double integral Z Z I 8x3 y 2 dxdy 5 4 I 6 h3 005 11 15 5 I 10 0 points when A is the region n x y y x y 3 I 0 3y y 3 2 2y 2 dx I A 2 I 1 2 15 Find the value of the double integral Z Z I 6x y dxdy 1 I I Z y 6x y dx h 2 3x xy 3y y 3 2 2y 2 i y y x x 8x3 y 2 dy dx integrating first with respect to y Now Z x ix h8 16 6 3 3 3 2 x y x 8x y dy x 3 3 x ha lvh262 HomeWork 15 3 karakurt 56295 Consequently Consequently 16 I 3 1 Z 006 x6 dx 0 16 21 I 10 0 points Evaluate the double integral Z Z 4y dxdy I 2 2 D x 1 when D is the region n x y 0 x 1 in the xy plane o 0 y x 007 D when D n o x y 0 x 1 x y x 1 5 2 I 4 ln 2 2 I 0 3 I 2 2 3 I correct 5 4 I 4 5 5 I 3 5 5 I 1 1 correct 2 Explanation As an iterated integral integrating first with respect to y we see that Z I 0 Now Z 0 x 1 nZ 0 x o 4y dy dx x2 1 2 4y dy 1 2 i x h 2y 2 In this case Z 1 I 2 0 x2 1 2 0 Explanation The double integral can be rewritten as the repeated integral I h 1 i1 x dx 2 x2 1 2 x 1 0 x x n 4 x y 3xy 2 o dy dx integrating first with respect to y Now x x x 2 2 x 1 2 Z 1 Z 0 Z x2 Evaluate the double integral Z Z 4 2 I x y 3xy dA 1 I 4 I 2 ln 2 1 2 10 0 points 1 I ln 2 6 I 3 n o x4 y 3xy 2 dy h1 2 x4 y 2 xy 3 ix x 2x4 x 2 x2 remember Consequently I 2 Z 0 1 x4 dx 2 5 ha lvh262 HomeWork 15 3 karakurt 56295 008 when D is the bounded region enclosed by y x and y x2 10 0 points 1 I 11 correct 12 2 I 3 4 when D is the bounded region enclosed by the graphs of 3 I 7 12 y x2 4 I 5 12 5 I 1 4 Evaluate the double integral Z Z I 2x cos y dxdy D y 0 4 x 1 1 I sin 1 1 Explanation The area of integration D is the shaded region in the figure 2 I 1 cos 1 correct 3 I 2 1 sin 1 4 I 2 1 cos 1 5 I 2 sin 1 1 6 I 2 cos 1 1 7 I cos 1 1 8 I 1 sin 1 Explanation After integration with respect to y we see that Z 1h ix2 I 2x sin y dx 0 0 Z 1 2x sin x2 dx 0 h cos x2 i1 0 using substitution in the second integral Consequently I 1 cos 1 009 10 0 points Evaluate the double integral Z Z I 3x 4 dA D To determine the limits of integration therefore we have first to find the points of intersection of the line y x and the parabola y x2 These occur when x2 x i e when x 0 and x 1 Thus the double integral can be written as a repeated integral Z 1 Z x I 3x 4 dy dx x2 0 integrating first with respect to y After integration this inner integral becomes h ix 3x 4 y 2 3x 4 x x2 x 4x x2 3x3 Thus I Z 0 1 4x x2 3x3 dx 3 1 2x x3 x4 3 4 2 1 0 ha lvh262 HomeWork 15 3 karakurt 56295 Consequently I 010 not drawn to scale the double integral can be written as the repeated integral Z 5 Z 4 x 1 I dy dx x y 1 1 1 11 12 10 0 points Find the volume V …