Engineering Calculus Notes 221

# Engineering Calculus Notes 221 - 209 2.5. INTEGRATION ALONG...

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: 209 2.5. INTEGRATION ALONG CURVES which, while not impossible, is a lot harder. However, we can also express √ C as the graph of x = y and parametrize in terms of y ; this yields 1 √ 2y ds = 2 + 1 dy 1 + 1 dy 4y = so 1 y y ds = 0 C 1 1 + 1 dy 4y y + y 2 dy 4 = 0 which, completing the square, 1 y+ = 0 1 = 8 1 0 1 8 2 − 1 dy 64 (8y + 1)2 − 1 dy and the substitution 8y + 1 = sec θ changes this into 1 64 arcsec 9 0 1 (tan θ sec θ − ln(sec θ + tan θ ))arcsec 9 0 128 √ √ 95 1 = − ln(9 + 4 5). 32 128 (sec3 θ − sec θ ) dθ = Exercises for § 2.5 Practice problems: 1. Set up an integral expressing the arc length of each curve below. Do not attempt to integrate. ...
View Full Document

## This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

Ask a homework question - tutors are online