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UIUC MATH 241 - Worksheet_031314_sol

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Thursday, March 13 ∗ Solutions ∗ Introduction to multiple integrals1. Evaluate the following integral by reversing the order of integration:Z10Z1pypx3+1d x d y.(Hint: When you change to dxdy, be sure to also change the bounds of integration.)SOLUTION:We are integrating over the region below:y x2y 0x 10.00.20.40.60.81.00.00.20.40.60.81.0Changing the order of integration we getZ10Z1pypx3+1d x d y =Z10Zx20px3+1d y d xR10Rx20px3+1d y d x =R10x2px3+1d x = 2/9[(x3+1)3/2]10=2/9(23/2−1).2. Consider the region bounded by the curves determined by −2x +y2=6 and −x + y = −1.(a) Sketch the region R in the plane.SOLUTION:x 12Iy2- 6Mx  y + 1H5,4LH-1,-2L-20246-2-1012345(b) Set up and evaluate an integral of the formÏRdA that calculates the area of R.SOLUTION:Z4−2Zy+1y2−62d x d y =Z4−2y +1 −y2−62d y =£−1/6y3+1/2y2+4y¤4−2=183. Consider the region R in the first quadrant which lies above the x-axis and between the circlesof radius 1 and 2 centered at (0,0). Without using polar coordinates, evaluateÏRy dA.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 R0) and doubleour answer. R0is shown below.ABx2+ y21x2+ y2 4x 10.00.51.01.52.00.00.51.01.52.0Break up R0into two parts A and B as above. Integrating, we obtainÏRy dA =ÏAy dA +ÏBy dA =Z10Zp4−x2p1−x2y d y d x +Z21Zp4−x20y d y d x=Z10£y2/2¤p4−x2p1−x2d x +Z21£y2/2¤p4−x20d x =Z103/2d x +Z211/2(4 −x2)dx=7/3Now double this value to get 14/3, which is the integral over the entire region R.4. EvaluateZ0−2Zp4−x20(x2+y2)d y d x.Hint: don’t do it directly.SOLUTION:The region over which we are integrating is:x2+ y2 4y 0x 0-2.0-1.5-1.0-0.50.00.00.51.01.52.0Converting to polar we getZ0−2Zp4−x20(x2+y2)d y d x =Zππ/2Z20(r2)r dr dθ = 2π5. The function P(x) = e−x2is fundamental in probability.(a) Sketch the graph of P(x). Explain why it is called a “bell curve.”SOLUTION:-2-1120.20.40.60.81.0(b) Compute I =R∞−∞e−x2d x using the following brilliant strategy of Gauss.i. Instead of computing I , compute I2=µZ∞−∞e−x2d x¶µZ∞−∞e−y2d y¶.ii. Rewrite I2as an integral of the formÎRf (x, y)d A where R is the entire Cartesianplane.SOLUTION:I2=Z∞−∞Z∞−∞e−x2−y2d y d xiii. Convert that integral to polar coordinates.SOLUTION:Z∞−∞Z∞−∞e−x2−y2d y d x =Z2π0Z∞0re−r2dr dθiv. Evaluate to find I2. Deduce the value of I .SOLUTION:Z2π0Z∞0re−r2dr dθ = 2πZ∞0re−r2dr =2π limt→∞Zt0re−r2dr =2π limt→∞h−1/2e−r2it0=π limt→∞(−e−t2+1) = πSo I =pπ.6. ComputeZ∞0Z∞01(1 +x2+y2)2d x d y.SOLUTION:As in the previous problem, let’s convert to polar coordinates.Z∞0Z∞01(1 +x2+y2)2d x d y =Zπ/20Z∞0r(1 +r2)2dr dθ = π/2Z∞0r(1 +r2)2drThis is an improper integral, soπ/2Z∞0r(1 +r2)2dr =π/2 limt→∞Zt0r(1 +r2)2dr =π/4 limt→∞·−11 +r2¸t0=π/4 limt→∞µ−11


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UIUC MATH 241 - Worksheet_031314_sol

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