Thursday October 25 Solutions Introduction to multiple integrals 1 Evaluate the following integral by reversing the order of integration Z 1Z 0 1 p p y x 3 1 d x d y Hint When you change to dx dy be sure to also change the bounds of integration SOLUTION We are integrating over the region below 1 0 0 8 0 6 y x2 0 4 x 1 0 2 y 0 0 0 0 0 0 2 0 4 0 6 0 8 Changing the order of integration we get Z 1Z 0 R 1 R x2 p 0 0 x3 1 d y d x R1 0 1 p y p x3 1 d x d y x2 p Z 1Z 0 0 x3 1 d y d x p x 2 x 3 1 d x 2 9 x 3 1 3 2 10 2 9 23 2 1 1 0 2 Consider the region bounded by the curves determined by 2x y 2 6 and x y 1 a Sketch the region R in the plane SOLUTION 5 H5 4L 4 3 x 1 2 2 Iy 2 6M 1 x y 1 0 1 H 1 2L 2 2 0 2 4 6 dA that calculates the area of R b Set up and evaluate an integral of the form R SOLUTION Z 4 2 Z y 1 y 2 6 2 Z dx dy 4 2 y 1 4 y2 6 d y 1 6y 3 1 2y 2 4y 2 18 2 3 Consider the region R in the first quadrant which lies above the x axis and between the circles of radius 1 and 2 centered at 0 0 Without using polar coordinates evaluate y dA R SOLUTION Notice that both the function y and the region R are symmetric about the y axis so we can integrate y over the half of R which lies in the first quadrant Call this R 0 and double our answer R 0 is shown below 2 0 A 1 5 x2 y 2 4 x 1 1 0 x2 y 2 1 B 0 5 0 0 0 0 0 5 1 0 1 5 2 0 Break up R 0 into two parts A and B as above Integrating we obtain R 1 Z y dA 0 y 2 2 Z 1Z A y dA p4 x 2 p 1 x 2 B 2 Z dx y dA 1 y 2 2 0 p 4 x 2 p 1 x 2 p4 x 2 0 y dy dx Z dx p 4 x 2 Z 2Z 0 y dy dx 1 1 0 Z 3 2 d x 1 2 1 2 4 x 2 d x 7 3 Now double this value to get 14 3 which is the integral over the entire region R 4 Evaluate 0 Z p 4 x 2 Z 2 0 x 2 y 2 d y d x Hint don t do it directly SOLUTION The region over which we are integrating is 2 0 x2 y 2 4 1 5 x 0 1 0 0 5 y 0 0 0 2 0 1 5 1 0 0 5 0 0 Converting to polar we get Z 0 Z 2 0 p 4 x 2 2 2 x y d y d x Z Z 2 0 2 r 2 r d r d 2 2 5 The function P x e x is fundamental in probability a Sketch the graph of P x Explain why it is called a bell curve SOLUTION 1 0 0 8 0 6 0 4 0 2 2 b Compute I R e 1 1 2 x 2 d x using the following brilliant strategy of Gauss Z Z y 2 x 2 2 e dy e dx i Instead of computing I compute I ii Rewrite I 2 as an integral of the form R f x y d A where R is the entire Cartesian plane SOLUTION Z Z 2 2 I2 e x y d y d x iii Convert that integral to polar coordinates SOLUTION Z Z e x 2 y 2 Z dy dx 2 Z 0 0 2 r e r d r d 2 iv Evaluate to find I Deduce the value of I SOLUTION Z 0 2 Z 0 2 r e r d r d 2 Z 0 2 r e r d r 2 lim Z t 0 2 t h i 2 2 t r e r d r 2 lim 1 2e r lim e t 1 p So I t t 0 Z Z 6 Compute 0 0 1 1 x 2 y 2 2 d x d y SOLUTION As in the previous problem let s convert to polar coordinates Z 0 Z 0 1 dx dy 2 1 x y 2 2 Z 2 Z 0 0 r d r d 2 1 r 2 2 Z 0 r dr 1 r 2 2 This is an improper integral so 2 Z 0 r d r 2 lim t 1 r 2 2 0 4 lim t t Z t 1 r d r 4 lim t 1 r 2 2 1 r2 0 1 1 4 1 t2
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