25. Review Double integrals Integrate function f ( x, y ) over a - PDF document

25. Review Double integrals Integrate function f ( x, y ) over a region R : �� f d A. R Computes the volume of the graph of f lying over R . Example 25.1. Evaluate � x 2 � 1 xe y 1 − y d y d x. 0 0 We cannot caculate this directly. First we figure out the region of integration. 0 ≤ x ≤ 1. Given x , we have 0 ≤ y ≤ x 2 . So we have the region R between x = 0 and x = 1 under the graph of y = x 2 . Then we switch the order of integration. � 1 � x 2 � 1 � 1 xe y xe y xe y �� 1 − y d y d x = 1 − y d y d x = 1 − y d x d y. 0 0 R 0 √ y The inner integral is � 1 � 1 xe y x 2 e y = e y (1 − y ) � 2(1 − y ) = 1 2 e y . 1 − y d x = 2(1 − y ) √ y √ y So the outer integral is � 1 � 1 1 � 1 = e − 1 2 e y d x = 2 e y . 2 0 0 We can use the double integral to calculate the mass, centre of mass and moment of inertia: Example 25.2. A metal plate is in the shape of a circle of radius 20 cm. Its density in g/cm 2 at a distance of r cm from the centre of the circle is 10 r + 3 . Find the total mass as an integral. � 2 π � 20 �� M = δ d A = (10 r + 3) r d r d θ. R 0 0 Line integrals Integrate a vector field � F over an oriented curve C . � � F · d � r. C Represents the work done. 1

One can compute directly, by parametrising C . Let C = C 1 + C 2 + C 3 be the curve which starts at (0 , 0) goes along the x -axis to (1 , 0), goes around the unit circle until (0 , 1) and comes back to the origin. y (0 , 1) C 2 C 3 x (0 , 0) C 1 (1 , 0) Figure 1. The curve C Let � F = − x 3 ˆ ı + x 2 y ˆ  . � � � � � � � � F · d � r = F · d � r + F · d � r + F · d � r. C C 1 C 2 C 3 Note that � � F · d � r = 0 , C 3 as � F = � 0 along the y -axis. Parametrise C 1 by x ( t ) = t , y ( t ) = 0. � F = �− t 3 , 0 � and d � r = � 1 , 0 � d t. So � � 1 � 1 � 1 � − 1 = − 1 − t 3 d t = � �− t 3 , 0 � · � 1 , 0 � d t = 4 t 4 F · d � r = 4 . C 1 0 0 0 Parametrise C 2 by x ( t ) = cos t , y ( t ) = sin t . F = �− cos 3 t, cos 2 t sin t � � and d � r = �− sin t, cos t � d t. So � π/ 2 � �− cos 3 t, cos 2 t sin t � · �− sin t, cos t � d t � F · d � r = C 1 0 � π/ 2 2 cos 3 t sin t d t = 0 � π/ 2 � − cos 4 t/ 2 = = 1 / 2 . 0 2

In total we get 1 / 4. We can also use Green’s theorem: � �� � curl � F · d � r = F d A C R � π/ 2 � 1 r 3 cos θ d r d θ. = 0 0 The inner integral is � 1 � 1 � 1 = 1 r 3 cos θ d r = 4 r 4 cos θ 4 cos θ. 0 0 So the outer integral is � π/ 2 � π/ 2 1 � 1 = 1 4 cos θ d θ = 4 sin θ 4 . 0 0 What about the same question, but now let us compute the flux. � � � � � � � � F · ˆ n d s = F · ˆ n d s + F · ˆ n d s + F · ˆ n d s. C C 1 C 2 C 3 Once again the flux across C 3 is zero. Along C 1 the normal vector is  . So the flux is zero, since � − ˆ F is parallel to ˆ ı along the x -axis. Along C 2 , we have ˆ n d s = � d y, − d x � . So � π/ 2 � �− cos 3 t, cos 2 t sin t � · � cos t, sin t � d t � F · ˆ n d s = C 1 0 � π/ 2 − cos 4 t + cos 2 t sin 2 t d t = 0 = − π 8 . Or we could apply the normal form of Green’s theorem: � �� � div � F · ˆ n d s = F d A C R �� − 2 x 2 d A = R � π/ 2 � 1 − 2 r 3 cos 2 θ d r d θ. = 0 0 The inner integral is � 1 � 1 � − 1 = − 1 − 2 r 3 cos θ d r = 2 r 4 cos 2 θ 2 cos 2 θ. 0 0 3

So the outer integral is � π/ 2 � π/ 2 − 1 � − t 4 − 1 = − 1 2 cos 2 θ d θ = 8 sin(2 θ ) 8 π. 0 0 Let F = (3 x 2 − 2 y sin x cos x )ˆ ı + ( a cos 2 x + 1)ˆ � . For which values of a is � F a gradient vector field? M y = − 2 sin x and N x = − 2 a cos x sin x. These are equal if and only if a = 1. For this value of a , what is the integral over the curve C , y ( t ) = t 3 − 1 , x ( t ) = t 2 and 0 ≤ t ≤ 1? Find a potential function f ( x, y ). We want f x = 3 x 2 − 2 y sin x cos x f y = cos 2 x + 1 . and Integrate the first equation with respect to x , f ( x, y ) = x 3 − y cos 2 x + g ( y ) . Use the second equation to determine g ( y ), − cos 2 x + dg dg dy = cos 2 x + 1 so that dy = 1 . Hence g ( y ) = y + c . So f ( x, y ) = x 3 − y cos 2 x + y, will do. � � � F · d � r = ∇ f · d � r = f (1 , 1) − f (0 , 0) = 1 . C C 4