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# 1703 - MATH 203 SOLUTIONS TO FINAL EXAMINATION SPRING 2005...

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MATH 203 SOLUTIONS TO FINAL EXAMINATION SPRING 2005 1 . (a) An ellipsoid with vertices ( ± 2 , 0 , 0), (0 , ± 2 , 0) and (0 , 0 , ± 8). The traces are a circle of radius 2 and an ellipse with semimajor axis 8 on the z -axis and semiminor axis 2. (b) Let F = x 2 4 + y 2 4 + z 2 64 ; F = x 2 , y 2 , z 32 ; F (1 , - 1 , 4 2) = 1 2 , - 1 2 , 2 8 . The plane has equation 1 2 ( x - 1) - 1 2 ( y + 1) + 2 8 ( z - 4 2) = 0 or x - y - 2 4 z = 4 2 . (a) Proof: 1 , 2 , - 3 = c 1 , 2 , - 1 for any nonzero real number c . Note: Verifying that the dot product of the two vectors above is not zero, shows the line is not perpendicular to the normal to the plane, i.e, that the line and the plane are not parallel. (b) - 1 - 3 t = 3 + t + 2(2 + 2 t ); t = - 1; substitute into equations of line: (2 , 0 , 2) (c) The normal 1 , 2 , - 1 to the plane is the direction of the line; a point on the line is (0 , 0 , 4); the symmetric form is x 1 = y 2 = z - 4 - 1 . 3 . 2 2 0 4 x 2 x 2 y dy dx = 2 0 x 2 y 2 | 4 y = x 2 dx = 2 0 16 x 2 - x 6 dx = ( 16 x 3 3 - x 7 7 ) | 2 0 = 2 7 ( 1 3 - 1 7 ) = 512 21 ; or 2 4 0 y 0 x 2 y dx dy = 2 4 0 x 3 y 3 | y 1 / 2 x =0 dy = 2 3 4 0 y 5 / 2 dy = 2 3 · 2 7 y 7 / 2 | 4 0 = 512 21 . 4 . (a) f = - e - xyz yz, xz, xy + 1 , 0 , 0 ; f (2 , 0 , 1) = - 0 , 2 , 0 + 1 , 0 , 0 = 1 , - 2 , 0 . D u f = 1 , - 2 , 0 · 3 , 1 , 1 / 11 = 1 11 . CAUTION: (3 i - 2 j ) / 11 is wrong! (b) (i) f (2 , 0 , 1) = 1 , - 2 , 0 ; (ii) f (2 , 0 , 1) = 5. 5 . The boundary surfaces define two bounded regions of equal volume: one region in the first octant and one region for which all x - and y -coordinates are negative. For the region in the first octant, z = 4 - x 2

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1703 - MATH 203 SOLUTIONS TO FINAL EXAMINATION SPRING 2005...

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