0 points find the volume of the solid under the graph

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Unformatted text preview: 36 ln 6 − 5 4 correct I=3 19 36 ln 6 − 5 4 013 f (x, y ) = 3(x + y ) ln x 10.0 points . Find the volume of the solid under the graph of f (x, y ) = 1 + 6x2 + 8y chester (crc2876) – HW13 – meth – (91845) and above the rectangle A= (x, y ) : 1 ≤ x ≤ 2 , 0 ≤ y ≤ 1 . 1. volume = 17 cu.units 2. volume = 19 cu.units correct 3. volume = 26 cu.units √ π 3 + 3 2 √ 3 π − 3 2 1 5. I = 8 6. I = 7 1 8 Explanation: To integrate 1 2 4. volume = 24 cu.units 1 − x2 dx y 0 5. volume = 18 cu.units Explanation: The volume is given by the double integral 1 2 (1 + 6x + 8y ) dxdy 0 with respect to x while keeping y fixed we use the substitution x = sin(u). For then dx = cos(u) du , 2 V= 1 and x=0 of f (x, y ) over the rectangular region A. Integrating each term separately, we see that 1 2 V= 1 2 1 dxdy + 0 1 0 1 6x2 dxdy =⇒ u=0 1 2 =⇒ u= x= Thus 1 2 8y dxdy = 1 + 14 + 4. 1 π /6 y 0 1 − x2 dx = y volume = 19 cu.units . 10.0 points since 1 2 I= 0 y 0 1 1. I = 2 π1 − 32 1 3. I = 8 π1 + 32 √ 3 π + 3 2 4. I = 1 2 (1 + cos(2u)) du 0 Consequently, I= √ π 3 − 3 2 1 2. I = 2 1 − x2 dx dy . 0 cos(2u) = 2 cos2 (u) − 1 . Evaluate the iterated integral 1 cos2 (u) du π /6 1 =y 2 Consequently, 014 π . 6 1 2 + 0 correct 1 2 π /6 1 0 = 1 dy sin(2u) 2 0 √ 1 3 12 π , + y 6 4 2 0 u+ y 1 2 and so 1 I= 8 √ 3 π + 3 2 ....
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This note was uploaded on 08/25/2013 for the course MATH 408M taught by Professor Kushner during the Summer '10 term at University of Texas.

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