Use a triple integral to find the volume of the solid in the first octant bounde

Question

Use a triple integral to find the volume of the solid in the first octant bounded by the coordinate planes, the plane 2x + y = 2, and the paraboloid z = 11 - x^2 - y^2

A 19/6

B 25/6

C 31/6

D 37/6

E 43/6

F 49/6

G 55/6

H 61/6

I none of these

Explanation / Answer

we need just a double integral. The given volume is a wedge-shaped region with the x-y plane at one end and the paraboloid at the other. The volume = integral (x = 0 ---> 1) integral (y = 0 ---> 2 - 2x) (11 - x^2 - y^2) dy dx = integral (x = 0 ---> 1) [(11 - x^2)y - y^3/3] | (y = 0 ---> 2 - 2x) dx = integral (x = 0 ---> 1) [(11 - x^2)(2 - 2x) - (2 - 2x)^3/3] dx = integral (x = 0 ---> 1) (14x^3/3 - 10x^2 - 14x + 58/3) dx = (7x^4/6 - 10x^3/3 - 7x^2 + 58x/3) | (x = 0 ---> 1) = 7/6 - 10/3 - 7 + 58/3 = 61/6 The answer is H

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