ha lvh262 Homework 15 4 karakurt 56295 This print out should have 8 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 1 I 5 2 I 5 2 4 3 I 5 4 10 0 points Which if any of the following are correct A 4 I 5 2 If Z Z I f x y dxdy x2 y 2 1 5 I 5 2 6 I 5 2 correct 2 then I Z 0 1Z 2 f r cos r sin d dr 0 B If I 1 Z Z x2 y 2 1 p 1 x2 y 2 dxdy Explanation In Cartesian coordinates the region of integration is n o p x y 0 y 4 x2 0 x 2 which is the shaded region in y then I 4 3 1 B only 2 neither of them correct r 3 both of them 4 A only Explanation A FALSE when changing to polar coordinates dxdy rd dr B FALSE the integral is the volume of the upper hemi sphere of the sphere of radius 1 centered at the origin hence has value 2 3 not 4 3 002 10 0 points Evaluate the integral Z Z n y o dxdy I 3 2 tan 1 x D when D is the region in the first quadrant inside the circle x2 y 2 4 x 2 On the other hand in polar coordinates the region of integration is n o r 0 r 2 0 2 while tan 1 y x Thus in polar coordinates I Z 0 2 Z 2 0 Z 2 0 h 3 2 rd dr 3 2 i 2 0 5 rdr 2 4 Z 0 2 r dr ha lvh262 Homework 15 4 karakurt 56295 Consequently I 003 5 2 2 10 0 points Evaluate the integral Z Z 2 2 I 2e x y dxdy The presence of the term r now allows this last integral to be evaluated by the subsitution u r 2 For then 1 1 h u i4 1 e 4 I e 0 2 2 004 1 I 10 0 points The paraboloid z x2 y 2 R when R is the region in the first quadrant of the xy plane inside the graph of p 4 y2 x 2 is shown in 1 1 e 4 correct 2 2 I 2 1 e 4 3 I 1 e 2 4 I 1 1 e 2 2 5 I 1 e 4 6 I 2 1 e 2 Explanation In polar cooordinates R is the set n o r 0 r 2 0 2 Find the volume of the solid under this paraboloid and above the circular disk n o 2 2 R x y x y 1 in the xy plane 1 volume 1 8 2 volume 1 4 3 volume 1 2 4 volume 1 4 5 volume 1 correct 2 6 volume 1 8 while I Z Z 2 2e r rdrd R Z Z 2 2re r drd R since x2 y 2 r 2 But then I 2 Z 0 2 Z 2 2 re r d dr 0 Z 0 2 2 re r dr Explanation ha lvh262 Homework 15 4 karakurt 56295 3 The volume of the solid below the graph of 5 volume 16 z x2 y 2 and above R is given by the integral Z Z V x2 y 2 dxdy R In polar coordinates this becomes Z 1 Z 2 V r 3 d dr 0 0 Consequently volume V 005 1 2 10 0 points The solid shown in 6 volume 32 Explanation The paraboloid intersects the xy plane when z 0 i e when x2 y 2 16 0 Thus the solid lies below the graph of 1 z 4 x2 y 2 4 and above the disk n o D x y x2 y 2 16 so its volume is given by the integral Z Z 1 2 2 V 4 x y dxdy 4 D In polar coordinates this becomes Z Z 1 4 2 V r 16 r 2 d dr 4 0 0 Z 4 1 16r r 3 dr 2 0 1 h r 4 i4 8r 2 2 4 0 Consequently volume V 32 is bounded by the paraboloid 1 z 4 x2 y 2 4 and the xy plane Find the volume of this solid 1 volume 8 2 volume 32 correct 3 volume 8 4 volume 16 006 10 0 points The solid shown in ha lvh262 Homework 15 4 karakurt 56295 so that after changing coordinates the integral becomes lies inside the sphere x2 y 2 z 2 16 and outside the cylinder 2 x y 2 4 V Z 3 9 Find the volume of the part of this solid lying above the xy plane 7 7 1 volume 3 7 7 2 volume 3 14 7 correct 3 volume 3 14 7 4 volume 3 5 volume 7 7 6 volume 7 7 Explanation From directly overhead the solid is similar to y 4 Z 2 0 2 Z p 4 r 3 16 r 2 rdrd p 16 r 2 dr i16 2 3 2 16 u 3 9 h using the substitution u r 2 Consequently 14 7 volume V 3 007 10 0 points The plane z 5 and the paraboloid z 8 3x2 3y 2 enclose a solid as shown in z r 3 4 x In Cartesian coordinates this is the annulus n o R x y 9 x2 y 2 16 Thus the volume of the solid above the xyplane is given by the integral Z Z V 16 x2 y 2 1 2 dxdy y x Use polar coordinates to determine the volume of this solid 1 volume R To evaluate V we change to polar coordinates Now n o R r 3 r 4 0 2 2 volume 2 3 volume 1 2 ha lvh262 Homework 15 4 karakurt 56295 and 3 4 volume correct 2 5 volume I 3 5 2 Explanation From the figure we see that paraboloid forms the upper boundary of the solid while the lower one is formed by the plane Thus the volume of the solid is given by the integral Z Z I 8 3x2 3y 2 5 dxdy Z Z 0 1 x2 y 2 dxdy 0 3 Z 2 Z 3 2 h r2 2 0 with domain of integration D in the xyplane To determine D we look at the where the plane z 5 intersects the paraboloid x2 y 2 1 Because D is circular and the integrand depends only on the distance of x y from the origin we change from Cartesian to polar coordinates using the transformations x r cos and y r sin Then Z 2 Z 1 I 3 1 r 2 r drd D 2 5 0 Consequently the solid has z 8 3x 3y volume This intersection occurs when 3 2 keywords volume of solid double integral change coordinates paraboloid plane coordinate transformation curve of intersection i …