Page 1 Math 251 copyright Joe Kahlig 15C 1 a False This would be true for ex y b True c True d 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 e False To be true the inside integral needs to start at z 0 not z 1 2 a x2 y 2 4y this is a circle b 2 sec cos 2 z 2 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 3r3 sin2 drd 10 0 Z 2 Z 2y 4 Change the limits of the integral to get 0 integrating this gives e4 2 ey dxdy 0 1 5 The correct region has been shaded in the graph x 3 y x Z 3 5 Z 10 x p 42 22 1 dydx 4 21 x x y 10 p y 9 y 2 so we get zx 0 and zy p 9 y2 s Z 2 Z 4 x2 r y2 9 1 dydx dydx 2 9 y 9 y2 0 0 r 9 r drd 9 r2 sin2 6 The surface is given by z Z 2Z 0 4 x2 0 Z 2 Z 2 or 0 0 Page 2 Math 251 copyright Joe Kahlig 15C 7 Step 1 find the intersection of the sphere and the paraboloid The paraboloid gives z 4 x2 y 2 substituting this into the sphere equation gives z 4 z 2 6 Solving gives z 1 and z 2 Since this is the top of the sphere z 1 is what we want The intersection is x2 y 2 1 2 6 or x2 y 2 5 Also note that y 0 means that 0 Z Z 5 Z 6 r2 r dzdrd Z r2 4 0 0 3 Z 12 4y 12 2y z 2 Z f x y z dxdzdy 8 0 0 y 2 Z Z 5Z 50 r2 zr2 sin dzdrd 9 a 2 2 Z 0 Z r 4 Z b 2 Z 2 10 Z 3 2 Z 4 sin2 cos sin d d d r2 cos dzdrd 1 8 y 2 11 2 50 0 6 Z r sin 40 Z 0 0 y2 k x 14 2 y 2 dxdy
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