Find the volume of the solid delimited by the planes $x = y + 2z + 1$, $x = 0, y = 0, z = 0$ and $3y + z - 3 = 0$. Sketch it also.
I'm supposed to solve this using multivariable calculus. However that's not much of a problem, the main issue is that I can't sketch this solid without calling it infinite.
I'm facing some difficulty to make sense out of this question because the region delimited by these planes seems to be infinite. So I don't know if I'm wrong or if the question is wrongly phrased.
I found points A(1,0,0) , B(2,1,0) and C(3,0,1) on plane x = y + 2z + 1.
And I found points D(0,0,3) and (0,1,0) on plane 3y + z - 3 = 0.
Both planes are also delimited by planes yz, xz and xy.
So when I sketch it, I see that it results in an infinite solid. Is this right?
The solid should not be infinite. I have assumed that x >= 0, y >= 0 and z >= 0. Since plane 1 is parallel to the x-axis we can consider the solid in 3 sections cut by planes parallel to the y-z plane for different x values. The diagram shows the intersections of plane 2 with y-z plane at different x values. For x = 0 to 1 we have the full triangle hence the volume is 1.5 For x = 1 to 2 express the area of the small triangle at y = 0 and z = 0 in terms of x and then subtract it from the full triangle. Then integrate from x = 1 to 2. The volume should be 17/12. Do similarly for x = 2 to 7 for the shaded triangle at the top the answer should be 25/12. The total should be 5. enter image description here