HW13-solutions rusin

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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,...
View Full Document

## 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.

Ask a homework question - tutors are online