Final Exam Solution

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

Ask a homework question - tutors are online