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Find the volume of the pyramid bounded by the plane x+2y+6z=12 in the first octant.

Find the volume of the solid bounded by the graphs of the equations z=x+y, x^2+y^2=4 in the first octant.
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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.

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

\int\limits^{}_{} {} \, dV = \int\limits^{2}_{r=0} r^2 { \int\limits^{\pi/2}_0 {(cos\phi+Sin\phi)} \, d\phi } \, dr \\\\\int\limits^{2}_{r=0} r^2 } [Sin\phi-Cos\phi ]_0^{\pi/2} \, dr\\\\\int\limits^{2}_{r=0}2\ r^2 } \, dr\\\\2[r^3/3]_0^2=2*8/3=\frac{16}{3}\\

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

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