fa02prelim3sol - Math 192 Prelim 3 SOLUTIONS Nov. 21, 2002...

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Unformatted text preview: Math 192 Prelim 3 SOLUTIONS Nov. 21, 2002 1. (a) The region is a triangle bounded by the y-axis, the line y = 3, and the line y = x . The integral is Z 3 Z y x 2 + y 3 dx dy = Z 3 y 3 3 + y 4 dy = 3 4 12 + 3 5 5 . (b) The region is bounded by the y-axis, the line y = , and the curve x = y . The integral is Z Z y 2 y cos( x ) dx dy = Z y sin( y 2 ) dy = . (c) The region is bounded by the x-axis, the line x = 2, and the curve y = x 2 . Beginning by using Fu- binis theorem, the integral is Z 4 Z 2 y ye- x 5 dx dy = Z 2 Z x 2 ye- x 5 dy dx = Z 2 (1 / 2) x 4 e- x 5 dx = 1 10- e- 32 10 . 2. Note that f ( x,y ) = 4( y- x 3 ) i + 2(2 x- y ) j . If the professor is neither ascending nor descending, then he must be walking along a level curve. Since level curves are perpendicular to the gradient, we just need to solve the equation f ( t,t 2 ) v ( t ) = 0, where v is the professors velocity vector. Since v ( t ) = i +2 t j , we have 4( t 2- t 3...
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