Page 1 Math 251 copyright Joe Kahlig 15C 1 a True b False This would be true for ex y c False To be true the inside integral needs to start at z 0 not z 1 d True e True When using the values of the only requirement is that the starting value and ending value can not have a difference of more than 2 2 a x2 y 2 6y this is a circle b 4 sec cos 4 z 4 this is a plane 3 Note Since f x y is not a constant function the volume of each of the 5 solids formed with the base being the leaf and the height being f x y will not be equal Note 2 you can not integrate from 0 to 2 since on parts of this interval the radius will be negative I have set up the integral for the leaf that is centered on the positive x axis Z 10 Z cos 5 5r3 cos2 drd 10 0 Z 3 Z 2y 4 Change the limits of the integral to get 0 integrating this gives e9 2 ey dxdy 0 1 5 The correct region has been shaded in the graph x 1 y x Z 1 4 Z 8 x p 32 52 1 dydx 9 35 x x y 8 p y 25 y 2 so we get zx 0 and zy p 25 y 2 s Z 3 Z 9 x2 r y2 25 1 dydx dydx 2 25 y 25 y 2 0 0 r 25 r drd 25 r2 sin2 6 The surface is given by z Z 3Z 0 9 x2 0 Z 2 Z 3 or 0 0 Page 2 Math 251 copyright Joe Kahlig 15C 7 First find the intersection of the sphere and the paraboloid The paraboloid gives z 5 x2 y 2 substituting this into the sphere equation gives z 5 z 2 25 Solving gives z 4 and z 5 Since this is the top of the sphere z 4 is what we want The intersection is x2 y 2 4 2 25 or x2 y 2 9 Also note that y 0 means that 0 Z Z 3 Z 25 r2 r dzdrd 0 Z r2 5 0 4 Z 16 4y 16 2y z 2 Z f x y z dxdzdy 8 0 0 y Z 3Z Z 18 r2 zr2 cos dzdrd 9 a 0 0 Z r 4 Z Z 18 b 0 Z 2 Z 0 6Z Z 11 3 3 3 1 Z 15 y 2 4y 2 0 r sin 40 10 0 4 sin2 cos cos d d d r2 sin dzdrd k x 20 2 y 2 dxdy
View Full Document
Unlocking...