Double Integrals over General Regions

Math 215 Examples Double Integrals over General Regions Key Concepts Regions of Type I and II A region in the xy-plane is called type I if it is defined as the region between two functions of x : A region in the xy-plane is called type II if it is defined as the region between two functions of y : Integrating over Regions of Type I Let R be the type I region lying between y = g(x) and y = h(x) for a ≤ x ≤ b . Then b ∬ f (x, y) dA = ∫ h(x) ∫ a f (x, y) dy dx. g(x) R To visualize this, imagine the slice at a fixed value of x. This slice begins at g(x) and ends at h(x). Shown below-left is the region in the xy-plane, with the slice drawn as a blue arrow, and below-right is the slice drawn in a 3D-coordinate system. (Note that the origin in the 3D-coordinate system is at the bottom right of the box.) The area under this slice is h(x) Area(x) = ∫ f (x, y) dy. g(x) As with rectangular regions, then, to compute the original double integral, we sweep through the entire volume as these slices range from x = a to x = b : This yields the final result of b ∬ f (x, y) dA = ∫ b Area(x) dx = ∫ a h(x) ∫ a f (x, y) dy dx. g(x) R Integrating over Regions of Type II Let R be the type II region lying between x = g(y) and x = h(y) for a ≤ y ≤ b . Then b ∬ f(x, y) dA = ∫ h(y) ∫ a f(x, y) dx dy. g(y) R This situation is analogous to the one above for type I regions, but now our slices are at fixed y -values. The slice at y begins at g(y) and ends at h(y): The area under this slice is h(y) Area(y) = ∫ f(x, y) dx. g(y) As these slices of area range through from y = a to y = b , we sweep through the entire volume: Hence b ∬ f(x, y) dA = ∫ b Area(y) dy = ∫ a R Illustrated Example Example Let R be the region in the xy-plane bounded by the curves y = x and y = √− x . Evaluate 3 a h(y) ∫ g(y) f(x, y) dx dy. 2 ∬ x y dA. R Worked Solution Let us first look at this region R in the xy-plane: − − The intersection points occur when x = √x , which we can solve to find x = 0 and x = 1 . Notice that for 0 ≤ x ≤ 1 , the curve y = √x is on top and y = x is on the bottom. Hence R is the type I region defined by 0 ≤ x ≤ 1 and x ≤ y ≤ √− x. 3 3 We can now write the desired double integral as an iterated integral: 1 √x 2 ∬ x y dA = ∫ 2 ∫ x y dy dx 3 0 x R We evaluate the inner integral treating x as constant: √x 1 2 ∫ x y dy = 2 x y √x 2∣ ∣ 2 3 x 1 = 3 x 3 (x 8 −x ) 2 Using this, we have 1 2 ∬ x y dA = ∫ √x 2 ∫ x y dy dx 3 0 x R 1 1 = ∫ 3 (x 8 − x ) dx 2 0 1 4 9 ∣ = (x /8 − x /18) ∣ 0 = (1/8 − 1/18) − (0 − 0) = 5/72. Visualizing the Example Our inner integral ∫ √x 3 x 2 x y dy gives the area of the slice at any x value. For example, at x = 0.6 , we have the slice shown below. Plugging x = 0.6 into this integral, we see that the area of this slice is √0.6 √0.6 2 ∫ 0.6 y dy = ∫ 0.6 3 0.36y dy 0.216 = 0.18y √0.6 2∣ ∣0.216 2 = 0.18(0.6 − 0.216 ) = 0.0996. Note that this is exactly what we get when we plug x = 0.6 into 1 2 3 (x 8 −x ) , as we computed in general that √x 2 ∫ 3 x As we let x range from 0 to 1, these slices sweep out the entire volume of 5/72: 1 x y dy = 3 (x 2 8 − x ). 3 Further Questions 1. What is the area under the slice at x = 0.4 ? What is the area under the slice at x = 1 ? 2. Suppose we switched the bounds on the inner integral and instead integrated from √− x to x . How would this mistake affect the final answer? 3 3. The region R in the example is also a type II region. What functions x = g(y) and x = h(y) bound it? 4. Evaluate the integral in the example by treating R as the type II region you described in Question 3. Using the Mathematica Demo All graphics on this page were generated by the Mathematica notebook 15_3DoubleIntegralsOverGeneralRegions. This notebook generates images and animations like those on this page for any type I region with y = g(x) and y = h(x) . The notebook also provides many additional options, including the option to attempt to treat the region as type II. As an exercise, use the notebook to provide a clear graphical answer to Question 3. You can do this simply by setting regionType = 2 and re-executing the code in the Graphics section. Then feel free to change the bounds and the function f (x, y) itself, keeping in mind the limitations described in the notebook.