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Homework 13 - chester(crc2876 HW13 meth(91845 This...

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chester (crc2876) – HW13 – meth – (91845) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Evaluate the double integral I = i i A 3 dxdy with A = b ( x, y ) : 2 x 7 , 2 y 4 B by frst identiFying it as the volume oF a solid. 1. I = 36 2. I = 34 3. I = 28 4. I = 32 5. I = 30 correct Explanation: The value oF I is the volume oF the solid below the graph oF z = f ( x, y ) = 3 and above the region A = b ( x, y ) : 2 x 7 , 2 y 4 B . Since A is a rectangle, this solid is a box with base A and height 3. Its volume, thereFore, is given by length × width × height = (7 - 2) × (4 - 2) × 3 . Consequently, I = 30 . keywords: volume, double integral, rectangu- lar region, rectangular solid 002 10.0 points The graph oF the Function z = f ( x, y ) = 6 - x is the plane shown in z 6 x y Determine the value oF the double integral I = i i A f ( x, y ) dxdy over the region A = b ( x, y ) : 0 x 6 , 0 y 4 B in the xy -plane by frst identiFying it as the volume oF a solid below the graph oF f . 1. I = 71 cu. units 2. I = 72 cu. units correct 3. I = 70 cu. units 4. I = 68 cu. units 5. I = 69 cu. units Explanation: The double integral I = i i A f ( x, y ) dxdy is the volume oF the solid below the graph oF f having the rectangle A = b ( x, y ) : 0 x 6 , 0 y 4 B
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chester (crc2876) – HW13 – meth – (91845) 2 for its base. Thus the solid is the wedge z 6 6 x y (6 , 4) and so its volume is the area of triangular face multiplied by the thickness of the wedge. Consequently, I = 72 cu. units . keywords: double integral, linear function, volume under graph, volume, rectangular re- gion, prism, triangle 003 10.0 points Find the value of the integral I = i 2 0 f ( x, y ) dx when f ( x, y ) = x - 4 x 2 y . 1. I = 2 + 32 3 y 2 2. I = 2 - 4 y 3. I = y - 4 y 2 4. I = y - 16 y 2 5. I = 2 + 32 3 y 6. I = 2 - 32 3 y correct Explanation: Since x varies but y does not, I = i 2 0 ( x - 4 x 2 y ) dx = b 1 2 x 2 - 4 3 x 3 y B 2 0 .
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