Final Exam Solution

2x2 1 x2 y 2 x2 y 2 2 2x3 sec2 x sec2 2 2 4 4 x

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Unformatted text preview: lim a η→0 a+η ϕ1 (x) f (x, y ) dy dx. ϕ1 (x) In general the first and third integrals are not necessarily equal. For parts (b)–(e) let f (x, y ) = x 2 −y 2 . (x2 +y 2 )2 1 (b) (5 points) Show that for fixed x > 0 the improper integral 0 f (x, y ) dy = 2 1 Hint: You may use f (x, y ) = − x2 +y2 + (x22xy2 )2 and y = x tan θ. + 1 . 1+x2 Solution: Fix x > 0. Let y = x tan θ. Then dy = x sec2 θ dθ and x2 + y 2 = x2 (1 + tan2 θ) = x2 sec2 θ. 2x2 1 + x2 + y 2 (x2 + y 2 )2 2x3 sec2 θ x sec2 θ −2 2 + 4 4 x sec θ x sec θ 2 x sec θ 2x3 sec2 θ −2 2 + 4 4 x sec θ x sec θ 12 − + cos2 θ xx sin 2θ 11 1+ −+ xx 2 f (x, y ) dy = − = = = = 1 1−η f (x, y ) dy = lim η →0 0 f (x, y ) = lim η →0 η dy dθ dθ = sin θ cos θ x sin 2θ sin θ cos θ = . 2x x 1−η η y where θ = arctan . x y Use a right triangle with one of the acute angles being θ = arctan x . Let the side “opposite” to that angle be y and the one “adjacent” to it be x. Then cos θ = √ x 2 and sin θ = √ y 2 . Then, 2 2 x +y 1 0 y f (x, y ) dy = lim 2 η →0 x + y 2 x +y 1−η = lim y =η η →0 x2 1−η η 2− 2 x + η2 + (1 − η ) = 1 . 1 + x2 Math 3010 Final Exam - Page 4 of 14 (c) (1 point) What would be an expression for tions necessary). Solution: 1 0 f (x, y ) dx for fixed y > 0 (no solu- 1 0 Dec. 8, 2012 f (x, y ) dx = − 1 . 1 + y2 (d) (2 points) Let D = [0, 1] × [0, 1]. Evaluate each of the following improper integrals if it exists. If it does not...
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This note was uploaded on 01/19/2014 for the course MATH 3010 taught by Professor Magpantay during the Winter '13 term at York University.

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