I encountered this question. However, I did not understand the solution given by the books (on sketching parts)
This is the question: Find the volume of the solid using triple integral, where the solid bounded ABOVE by plane 2x+3y+z=6 and BELOW by plane z=1 in the first octant.
This is the solution: region sketch according to the solution
please help me understand how the shaded region turned out like this.
The shaded region is defined by the intersections of the planes $2x+3y+z=6$, z=1, x=0, and y=0.
The top point of the prism is found by setting $x=0$ and $y=0$, solving $2x+3y+z=6$ for the z-intercept, giving us $z=6$.
The lower left most point in the diagram is found by setting $z=1$ and $y=0$, solving $2x+3y+z=6$ for $x$, which yields $x=\frac{5}{2}$ and the point $(\frac{5}{2},0,1)$.
The right most point is similar, set $z=1$ and $x=0$ and solve $2x+3y+z=6$ for $y$ giving $y=\frac{5}{3}$ which means our next point is $(0,\frac{5}{3}, 1)$
The center rear point is the point where the plane $z=1$ intersects with the z-axis, so $(0,0,1)$