Triple Integrals: Finding Volume of Solid S Bounded by Planes

[WMDhamnekar]

Find the volume V of the solid S bounded by the three coordinate planes, bounded above by the plane x + y + z = 2, and bounded below by the z = x + y.

How to answer this question using triple integrals? How to draw sketch of this problem here ?

 

[Prove It]

I like your idea of sketching the graphs. Maybe start by working out the intercepts with the co-ordinate axes.

E.g. for x + y + z = 2, the x intercept is x + 0 + 0 = 2 => x = 2.

With the three intercepts it's pretty easy to draw the plane.

 

[WMDhamnekar]

I like your idea of sketching the graphs. Maybe start by working out the intercepts with the co-ordinate axes.

E.g. for x + y + z = 2, the x intercept is x + 0 + 0 = 2 => x = 2.

With the three intercepts it's pretty easy to draw the plane.

I computed the answer -8 as follows but it is wrong. Where am I wrong?

$\displaystyle\int_0^2\displaystyle\int_0^2\displaystyle\int_{x+y}^{2-x-y}1 dzdydx=-8$

Even if I change the upper limit of integration of y as 2-x, the answer would be $\displaystyle\int_0^2\displaystyle\int_0^{2-x} (2-2x-2y) dydx = -\frac43$ which is also wrong.

 

[HallsofIvy]

One thing that may be causing confusion is that the problem is not stated very well. We are told the figure is bounded by the three coordinate planes, by the plane x+ y+ z= 2, and by the plane z= x+ y. But there is NO region bounded by those five planes! The plane z= x+ y "covers" the plane z= 0. I think what is intended is the region bounded by the two coordinate planes, x= 0 and y= 0, and by x+ y+ z= 2 and z= x+ y but not by the plane z= 0. Adding x+ y+ z= 2 and x+ y- z= 0 we get 2x+ 2y= 2 or x+ y= 1. The last two planes intersect at the line x+ y= 1, z= 1. The first plane, x+ y+ z, intersects the coordinate planes in a triangle with vertices (0, 2, 0), (2, 0, 0) and (0, 0, 2). The sides have equations x= 0, y+ z= 2, y= 0, x+ z= 2, and z= 0, x+ y= 2.

The second triangle passes through the origin, (0, 0, 0), and cuts the first plane in the line x+ y= 1, z= 1.
The largest value x take in the region is 1 so we must take x from 0 to 1. For every x, y goes from 0 to 1- x, and for every x, y, z goes from z= 2- x- y.

The integral can be written a $\int_{x= 0}^1\int_{y= 0}^{1- x}\int_{z= 0}^{2- x- y}dzdxdy$.

 

[WMDhamnekar]

Final answer to this question is $\displaystyle\int_{x=0}^1\displaystyle\int_{y=0}^{1-x}\displaystyle\int_{z=x+y}^{z=2-x-y} dzdydx= \frac13$

 

[HallsofIvy]

That's what I get!