y ln x x2 i a 100 points 5 dxdy 1 y2 over the

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Unformatted text preview: 3 }. when A= (x, y ) : 0 ≤ x ≤ 1, 0≤y≤1 . 1. I = 7 5 9 ln 3 + 2 2 2. I = 7 5 9 ln 3 − 2 2 3. I = 7 9 ln 3 − 5 2 11 4. I = π 6 4. I = 7 2 5 ln 3 − 9 2 5 5. I = π 3 5. I = 7 3 ln 3 − 1. I = 3 π 2 4 2. I = π correct 3 3. I = 2π (x, y ) : 0 ≤ x ≤ 1, 0≤y≤1 is a rectangle with sides parallel to the coordinate axes, the double integral can be represented as the iterated integral 1 1 I= 0 0 5 + x2 dx dy . 1 + y2 Now 1 0 5 2 Explanation: Since the area of the rectangle A is 6, the average value of f over A is given by Explanation: Since A= correct 1 1 5 + x2 dx = 5 x + x3 2 2 1+y 1+y 3 I= 0 7(x + y ) ln x dxdy. A The integral with respect to y can be carried out immediately, whereas the integral with respect to x would require integration by parts, so this suggests that I should be written as a repeated integral 1 . 1 6 I= 7 6 3 3 (x + y ) ln x dy dx, 1 0 rogers (grr459) – HW13 – rusin – (55220) integrating first with respect to y . After integration the inner integral becomes Consequently, 3 1 9 xy + y 2 ln x = 3x ln x + ln x. 2 2 0 Now is the time to integrate by parts, for then 3 3x ln x + 1 9 ln x dx 2 volume = 7 cu. units . keywords: double integral, volume of solid, integral over rectangle, volume under plane 3 9 9 32 x ln x − x2 + x ln x − x 2 4 2 2 = 4 3 1 008 . 10.0 points Evaluate the iterated integral Thus 9 5 ln 3 − 2 2 I=7 . 1 2 I= y 0 007 16 − x2 dx dy . 10.0 points Find the volume of the solid lying under the plane z = 7 − x − 2y and above the rectangle A= 0 (x, y ) : 1 ≤ x ≤ 2 , 0≤y≤2 . 1. volume = 5 cu. units 2. volume = 7 cu. units correct 3. volume = 4 cu. units 4. volume = 8 cu. units 5. volume = 6 cu. units Explanation: The volume of the solid is given by the double integral π1 + 32 √ π 3 2. I = 8 + 3 2 1. I = 2 π1 − 32 √ π 3 4. I = 2 correct + 3 2 √ π 3 5. I = 2 − 3 2 √ π 3 6. I = 8 − 3 2 3. I = 8 Explanation: To integrate 2 V= A y (7 − x − 2y ) dxdy . 0 As a repeated integral this becomes 2 1 0 (7 − x − 2y ) dy dx 2 = 1 7y − xy − y 2 1 (10 − 2x) dx = dx = 4 cos u du , 2 0 dx 2 = with respect to x while keeping y fixed we use the substitution x = 4 sin u. For then 2 V= 16 − x2 dx 10x − x2 and x=0 =⇒ u=0 x=2 =⇒ u= 2 1 . π . 6 rogers (grr459) – HW13 – rusin – (55220) 5 Consequently, Thus π /6 2 16 − x2 dx = 4y y 0 4 cos2 u du 31 1 . −= 42 4 I= 0 π /6 = 8y (1 + cos 2u) du 010 0 since Find the value of the integral cos 2u = 2 cos2 u − 1 . Consequently, I= 1 π /6 1 sin 2u dy 2 0 0 √ 3 121 π , =8 + y 0 6 4 2 I=8 y u+ and so √ 3 π + I=2 3 2 009 A (x, y ) : 0 ≤ y ≤ x − 4, 4 ≤ x ≤ 6 . 22 3 2. I = 23 3 Evaluate the integral x2 I= 0 0 1. I = (3x − 5y ) dydx . {4(x − 4) − 3y } dxdy when A is the region 1. I = . 10.0 points 1 10.0 points 3. I = 7 4. I = 8 3 20 20 correct 3 5. I = 7 2. I = 20 Explanation: The integral can be written as the repeated integral 3. I = 1 20 4. I = 1 correct 4 x −4 6 I= 9 5. I = 20 4 0 {4(x − 4) − 3y } dy dx. Now Explanation: After integration with respect to y we see that 1 5 3xy − y 2 2 I= 0 1 = 0 = x2 0 {4(x − 4) − 3y } dy 0 dx = 5 3 x3 − x4 d x 2 34 15 x− x 4 2 x −4 0 . x −4 0 . Consequently 6 1 3 4(x − 4)y − y 2 2 I= 4 5 (x − 4)2 dx = 2 5 (x − 4)3 6 6 4 , rogers (grr459) – HW13 – rusin – (55220) 6 y and so I= 011 20 3 4 . 10.0 points Reverse the order of integration in the integral x 2x 2 2 f (x, y ) dy dx , I= 0 0 but make no attempt to evaluate either integral. 2 4 1. I = f (x, y ) dx dy y 0 Integration is taken first with respect to y for fixed x along the dashed vertical line. To change the order of integration, now fix y and let x vary along the solid horizontal line in y y 4 2. I = f (x, y ) dx dy 0 4 2 4 2 3. I = f (x, y ) dx dy correct y/2 0 2y 2 4. I = f (x, y ) dx dy 0 0 y /2 4 5. I = x f (x, y ) dx dy 0 2 4 6. I = f (x, y ) dx dy 0 2 0 2y Explanation: The region of integration is the set of all points (x, y ) : 0 ≤ y ≤ 2x , 0 ≤ x ≤ 2 in the plane bounded by the x-axis and the graphs of y = 2x , x = 2. This is the shaded region in Since the equation of the slant line is x = y/2, integration in x is along the line from (y/2, y ) to (2, y ) for fixed y , and then from y = 0 to y = 4. Consequently, after changing the order of integration, 4 2 I= f (x, y ) dx dy 0 . y/2 keywords: double integral, reverse order integration, linear function,...
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This note was uploaded on 09/22/2013 for the course M 408 D taught by Professor Textbookanswers during the Fall '07 term at University of Texas at Austin.

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