find the volume of the solid that is bounded by y^2+z^2=9, x=0,y=3x, and z=0 in

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Explanation / Answer

We need to find the volume under z = 6 - 3x - 2y in the first octant. Projecting this onto the xy-plane yields the region between 3x + 2y = 6 and x, y = 0. ==> y = 0 to y = 3 - 3x/2 with x in [0, 2]. Hence, the volume equals ?(x = 0 to 2) ?(y = 0 to 3 - 3x/2) (6 - 3x - 2y) dy dx = ?(x = 0 to 2) [(6 - 3x)y - y^2] {for y = 0 to 3 - 3x/2} dx = ?(x = 0 to 2) [(6 - 3x)(3 - 3x/2) - (3 - 3x/2)^2] dx = ?(x = 0 to 2) [(1/2)(6 - 3x)(6 - 3x) - (1/2)^2 * (6 - 3x)^2] dx = ?(x = 0 to 2) (1/4)(6 - 3x)^2 dx = ?(x = 0 to 2) (9/4)(x - 2)^2 dx = (3/4)(x - 2)^3 {for x = 0 to 2} = 6.

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