Dy sin 2 2 2 40 24 4 hence greens theorem holds true

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: sin x) dx+cos x dy = 1−t −sin 1−t 2 2 γ3 γ3 0 1 π π t2 2 π 2 π = cos 1 − t − 1 dt = − t − − cos 1 − t + sin 1 − t 2 2 2π 2 π 20 2 π − +1− . 4 π +t − π , −1 2 0 = 1−t MATB42H Now γ F1 dx + F2 dy = −1 + 0 + F1 dx + F2 dy + F1 dx + F2 dy + F1 dx + F2 dy = γ1 γ2 γ3 π 2 2 π =− − . − +1− 4 π 4π ∂F1 ∂ F2 − ∂x ∂y On the other hand, R 1 page 3 Solutions # 5 π /2 dA = π /2 1 − sin x − 1 R dA = 1 πy π − cos 2 2 0 π y/2 0 0 π y/2 21 π πy πy πy 2 πy 2π π 2 =− − + =− − . dy = − + − sin 2 2 π 2 40 2π4 4π Hence Green’s Theorem holds true for this example. − sin x − 1 d...
View Full Document

Ask a homework question - tutors are online