Question

(S15.6) Find the volume of the solid bounded by the four planes x + y-3: = 3, x 3, y = 3, and Z 0 Answer:

Answer 1

Answer

The given four planes are

$x+y-3z=3$

y=3

$z=0$

Below is the graph of the given four planes.

7 7 8 -54-3-2 12 eg

The base of the solid is the shaded region in the below sketch.

4 (0, 3) (3, 3) (3,0) 4

From the above graph, we can see that x changes from 0 to 3 and y changes from 3 - x to 3.

The height of the solid is

rty- 3

Hence, the volume of the solid is

3 c3 3 dy dx 3 J3-х

一3y.dr Hry + !

3 33(3)) - (3- r)(3-2 0

$=\int_0^3 \frac{1}{3}\left (3x-\frac{9}{2} +\frac{(3-x)^2}{2}\right ) \, dx$

2-6x +9-9 d.r

3 3 0 3

2 0 3

$=\frac{1}{6}\int_0^3x^2 =\frac{1}{6}\left [ \frac{x^3}{3} \right ]_0^3 =\frac{1}{6}\left [ \frac{3^3}{3}-\frac{0^3}{3} \right ] =\frac{1}{6}\left ( 9-0 \right ) =\frac{9}{6} =\frac{3}{2}$

${\color{Blue} \boxed{\textup{Answer: }\frac{3}{2}}}$