hw7_solutions

hw7_solutions - February 27th, 2009 Math 20e - Assignment...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
February 27th, 2009 Math 20e - Assignment #7 Solutions Problem 1 (page 528 #1) . Evaluate R C y dx - xdy where C is the boundary of the square [ - 1 , 1] × [ - 1 , 1] oriented in the counterclockwise direction, using Green’s theorem. Solution. Theorem 2 on page 524 says that the area of a region D with boundary C = ∂D is given by Green’s Theorem as A = 1 2 Z C xdy - y dx. Therefore, - 2 A = R C y dx - xdy . So we just find the area of D = [ - 1 , 1] × [ - 1 , 1], which is 4 and multiply it by - 2 to get the answer: Z C y dx - xdy = - 2 A = - 8 . Problem 2 (page 528 #2) . Find the area of the disk D of radius R using Green’s theorem. Solution. Theorem 2 on page 524 says that the area of a region D with boundary C = ∂D is given by Green’s Theorem as A = 1 2 Z C xdy - y dx. Now we parametrize C by the curve c ( t ) = h R cos t,R sin t i . We also calculate that c 0 ( t ) = h- R sin t,R cos t i . For the purpose of notation, let F ( x,y ) = h- y,x i . Then we have A = 1 2 Z C xdy - y dx = 1 2 Z C h- y,x i · h dx,dy i = 1 2 Z C F ( x,y ) · ds = 1 2 Z 2 π 0 F ( c ( t )) · c 0 ( t ) dt = 1 2 Z 2 π 0 h- R sin t,R cos t i · h- R sin t,R cos t i dt = 1 2 Z 2 π 0 R 2 (cos 2 t + sin 2 t ) dt = 1 2 R 2 Z 2 π 0 dt = πR 2 . 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
(page 528 #3c) . Verify Green’s theorem for the disk D with center (0 , 0) and radius R and the functions P ( x,y ) = xy = Q ( x,y ). Solution. Green’s Theorem states that Z C P dx + Qdy = ZZ D ± ∂Q ∂x - ∂P ∂y ² dxdy. To verify Green’s Theorem, we need to check that these two integrals are indeed equal. First, to take the line integral, we let C be described by the curve c ( t ) = h R cos t,R sin t i , so that c 0 ( t ) = h- R sin t,R cos t i , and thus h dx,dy i = ds = c 0 ( t ) · dt = h- R sin tdt,R cos tdt i . So from our definition of c ( t ) we do the following substitutions below: x = R cos t , y = R sin t , dx = - R sin tdt , and dy = R cos tdt . Z C P dx + Qdy = Z C xy dx + xy dy = Z 2 π 0 ( R cos t )( R sin t )( - R sin tdt ) + Z 2 π 0 ( R cos t )( R sin t )( R cos tdt ) = Z 2 π 0 - R 3 cos t sin 2 tdt + Z 2 π 0 R 3 sin t cos 2 tdt = 0 + 0 = 0 . The fact that the two integrals above are 0 follows from integral formula 76 in the back of your textbook. Next we calculate the double integral using polar coordinates, so that dxdy = r dr dθ . ZZ D ± ∂Q ∂x - ∂P ∂y ² dxdy = ZZ D ( y - x ) dxdy = Z 2 π 0 Z R 0 ( r sin θ - r cos θ ) r dr dθ = Z 2 π 0 ³ r 3 3 (sin θ - cos θ ) ´ R 0 = Z 2 π 0 R 3 3 (sin θ - cos θ ) = R 3 3 [ - cos θ - sin θ ] µ µ µ 2 π 0 = 0 . Problem 4 (page 528 #5) . Find the area bounded by one arc of the cycloid x = a ( θ - sin θ ), y = a (1 - cos θ ), where a > 0, and 0 θ 2 π , and the x axis (use Green’s Theorem). Solution.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/30/2010 for the course MATH 20E 20E taught by Professor Enright during the Fall '09 term at UCSD.

Page1 / 11

hw7_solutions - February 27th, 2009 Math 20e - Assignment...

This preview shows document pages 1 - 3. Sign up to view the full document.

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