notes001 (87)

notes001 (87) - Z C & x 2 & y ds =...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
November 5, 2010 NAME______________________________ correct notation) in order to get full credit. 1) Change the Cartesian integral Z 2 0 Z p 4 y 2 0 x 2 + y 2 ± dx dy into an equivalent polar integral and then evaluate the integral. Solution: By drawing a picture of the region of integration, we see that Z 2 0 Z p 4 y 2 0 x 2 + y 2 ± dx dy = Z 2 0 Z 2 0 r 3 = 2 ± . 2) Evaluate the line integral Z C x 2 y ± ds where C is the curve x 2 + y 2 = 4 starting at the point (2 ; 0) and ending at the point p 2 ; p 2 ± . Solution: The curve can be parameterized as r ( t ) = 2 cos ( t ) i + 2 sin ( t ) j 0 ± t ± ± 4 . Thus r 0 ( t ) = 2 sin ( t ) i + 2 cos ( t ) j and j r 0 ( t ) j = 2 and we have
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Z C & x 2 & y ds = Z &= 4 & 4 cos 2 ( t ) & 2 sin ( t ) (2) dt = + 2 p 2 & 2 . 3) Let V ( x; y ) be the velocity eld V ( x; y ) = & y i + x j . 1 Find the &ow along and the &ux across the curve r ( t ) = cos ( t ) i + sin ( t ) j & t & & . Solution: Here we have M ( x; y ) = y , N ( x; y ) = x , dx=dt = sin ( t ) and dy=dt = cos ( t ) . The &ow is thus Z & (( sin ( t )) ( sin ( t )) + (cos ( t )) (cos ( t ))) dt = & and the &ux is Z & (( sin ( t )) (cos ( t )) (cos ( t )) ( sin ( t ))) dt = 0 . 2...
View Full Document

Page1 / 2

notes001 (87) - Z C & x 2 & y ds =...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online