Use a double integral to find the volume of the solid in the first octant bounded above by the plane (below).?

Use a double integral to find the volume of the solid in the first octant bounded above by the plane 9x + 2y + z = 20 and below by the rectangle on the xy-plane: {(x, y): 0 %26lt; x %26lt; 1, 0 %26lt; y %26lt; 2 }.








one of these could possibly be the correct response....





a. 26


b. 27


c. 28


d. 29


e. 30


f. 31


g. 32


h. 33


i. 34





or is it none of these?

Use a double integral to find the volume of the solid in the first octant bounded above by the plane (below).?
9x + 2y + z = 20


z = 20 - 9x - 2y





∫{0→1} ∫{0→2} 20 - 9x - 2y dy dx


= ∫{0→1} [20y - 9xy - y²]{0→2} dx


= ∫{0→1} [20(2) - 9x(2) - 2²] - [20(0) - 9x(0) - 0²] dx


= ∫{0→1} (40 - 18x - 4) - (0 - 0 - 0) dx


= ∫{0→1} 36 - 18x dx


= 9∫{0→1} 4 - 2x dx


= 9[4x - x²]{0→1}


= 9[[4(1) - 1²] - [4(0) - 0²]]


= 9[(4 - 1) - (0 - 0)]


= 9(3)


= 27





The answer is (b).

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