Application of double integral.

�� �� 23. Inclusion-Exclusion Proposition 23.1. Let D = D1 ∪ D2 be a bounded region and let f : D −→ R be a function. If f is integrable over D1 and over D2, then f is integrable over D and and D1 ∩ D2, and we have �� �� �� �� f(x,y)dx dy = f(x,y)dx dy+ f(x,y)dx dy− f(x,y)dx dy. DD1 D2 D1∩D2 Example 23.2. Let D = { (x,y) ∈ R2 | 1 ≤ x 2 + y 2 ≤ 9 }. Then D is not an elementary region. Let D1 = { (x,y) ∈ D | y ≥ 0 } and D2 = { (x,y) ∈ D | y ≤ 0 }. Then D1 and D2 are both of type 1. If f is continuous, then f is integrable over D and D1 ∩ D2. In fact D1 ∩ D2 = L ∪ R = { (x,y) ∈ R2 |− 3 ≤ x ≤−1, 0 ≤ y ≤ 0 } ∪{ (x,y) ∈ R2 | 1 ≤ x ≤ 3, 0 ≤ y ≤ 0 }. Now L and R are elementary regions. We have �� � 3 �� 0 � f(x,y)dx dy = f(x,y)dy dx =0. R 10 Therefore, by symmetry, f(x,y)dx dy = f(x,y)dx dy =0 LR and so �� �� �� f(x,y)dx dy = f(x,y)dx dy + f(x,y)dx dy. DD1 D2 1 � To integrate f over D1, break D1 into three parts. �� � 3 �� δ(x) � f(x,y)dx dy = f(x,y)dy dx D1 −3 γ(x) � −1 �� √9−x� 2 = f(x,y)dy dx −30 � 1 �� √9−x� 2 + f(x,y)dy dx 2−1 √1−x� 3 �� √9−x� 2 + f(x,y)dy dx. 10 One can do something similar for D2. Example 23.3. Suppose we are given that �� � 1 �� 2y � f(x,y)dx dy = f(x,y)dx dy. D 0 y What is the region D? It is the region bounded by the two lines y = x and x =2y and between the two lines y =0 and y =1. Change order of integration: �� � 1 �� x �� 2 �� 1 � f(x,y)dx dy = f(x,y)dx dy+ f(x,y)dx dy. D 0 x/21 x/2 Example 23.4. Calculate the volume of a solid ball of radius a. Let B = { (x,y,z) ∈ R3 | x 2 + y 2 + z 2 ≤ a 2 }. We want the volume of B. Break into two pieces. Let B+ = { (x,y,z) ∈ R3 | x 2 + y 2 + z 2 ≤ a 2 ,z ≥ 0 }. Let D = { (x,y) ∈ R2 | x 2 + y 2 ≤ a 2 }. Then B+ is bounded by the xy-plane and the graph of the function f : D −→ R, given by f(x,y)= a2 − x2 − y2 . 2 �� � � �� � It follows that vol(B+)= a2 − x2 − y2 dy dx � aD�� √a−x�� 22 = a222 dy dx √a− x− y� −aa �� −√a2−−xx2 � 2 � 22 y= −a −√a−x1 − 2 − x2 √a2 − x2 dy dx. 22 aNow let’s make the substitution y dyt = so that dt = .√a2 − x2 √a2 − x2 � �� 1 � 2vol(B+)= a √1 − t2(a − x 2)dt dx �−a −1 �� 1 � 2 = a (a − x 2) √1 − t2 dt dx −a −1 Now let’s make the substitution t = sin u so that dt = cos udu. π a 2 vol(B+)= (a 2 − x 2) π cos 2 u du dx 2�−a − a π =(a 2 − x 2)2 dx −�a �π 2 x3 a = a 2 x − 3 −a 3a= π(a 3 − 3) 2πa3 = . 3 Therefore, we get the expected answer 4πa3 vol(B) = 2 vol(B+)= . 3 Example 23.5. Now consider the example of a cone whose base radius is a and whose height is b. Put the central axis along the x-axis and 3 � � � � 0 the base in the yz-plane. In the xy-plane we get an equilateral triangle of height b and base 2a. If we view this as a region of type 1, we have xx γ(x)= −a 1 − and δ(x)= a 1 − . bb We want to integrate the function f : D −→ R, given by � ��2x f(x,y)= a2 1 −− y2 . b So half of the volume of the cone is � b �� a(1−) �� �2 � π � b ��2 )bbx x xx 2 a2 1 − b − y2 dy dx = a1 − b dx2−a(1− 0 � b πa2 2xx2 = 2 0 1 − b + b2 dx = πa2 2 � x − x2 b + x3 3b2 �b 0 = 1 6(πa2b). Therefore the volume is 1 3(πa2b). 4 MIT OpenCourseWarehttp://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.