1)

The plane P : x + 2 y + 6 z = 12 intercepts x axis at A (12, 0, 0), y axis at B (0,6,0) and z axis at C (0, 0, 2).

We have the pyramid with three slanting surfaces as xy plane , y-z plane and x-z plane. The base of the pyramid in 1st octant is given by the equation.

*We look at the pyramid as : vertex at A (12,0,0) and base as the part in y-z plane*.

[ Note: We can also look at the pyramid as having base in x-y plane and apex at C(0,0,2). Similarly, we can take base on x-z plane also. ]

Base area = 1/2 * 6 * 12 = 36

Altitude or height of pyramid is = 12 (x-intercept) as y-z plane and x axis are perpendicular.

*Volume of pyramid *= 1/3 * Base * height = 1/3 * 36 * 12 = 144.

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To find the volume, we take a small area element on x-y plane and multiply with the height z of the solid at (x,y).

Let x = r cos Ф and y = r Sin Ф denote the coordinates of a tiny area element dA in the 1st quadrant of circular area, 0 <= r <= 2 and 0 <= Ф <= π/2

dA at (r, Ф) = dr * r dФ = r dr dФ

the height at (r, Ф) is z = x + y = r (Cos Ф+Sin Ф)

dV = r (Cos Ф+ SinФ) r dr dФ

Volume=

*Hence volume of the solid bounded by the given graphs is = 16/3*