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HW13-solutions rusin - rogers(grr459 HW13 rusin(55220 This...

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rogers (grr459) – HW13 – rusin – (55220) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. Here is a short problem set to get you think- ing about Chapter 15. It’s due ±riday night. There will be one more homework, due next ±riday night (12/7) and then you’re done! 001 10.0 points Evaluate the double integral I = i i A 4 dxdy with A = b ( x, y ) : 2 x 8 , 5 y 7 B by frst identiFying it as the volume oF a solid. 1. I = 52 2. I = 54 3. I = 46 4. I = 48 correct 5. I = 50 Explanation: The value oF I is the volume oF the solid below the graph oF z = f ( x, y ) = 4 and above the region A = b ( x, y ) : 2 x 8 , 5 y 7 B . Since A is a rectangle, this solid is a box with base A and height 4. Its volume, thereFore, is given by length × width × height = (8 - 2) × (7 - 5) × 4 . Consequently, I = 48 . 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 = 73 cu. units 2. I = 72 cu. units correct 3. I = 70 cu. units 4. I = 71 cu. units 5. I = 69 cu. units Explanation: The double integral I = i i A f ( x, y ) dxdy
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rogers (grr459) – HW13 – rusin – (55220) 2 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 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.
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