Volume of Region Bounded by x^2+2y^2=2, x+y+2z=2

[jonroberts74]

Homework Statement

find volume of the region bounded by $$x^2+2y^2=2;z=0;x+y+2z=2$$

The Attempt at a Solution

I figure "slicing" in the z=0 direction would be the easiest

the first issue I am having is the upper bound of z, it definitely seems to be 2 but but it's not making sense at the moment how to get that


when I slice with the z=0 plane I get an ellipse

y is on the interval [-1,1] and then getting x in terms of y
$$-\sqrt{2-2y^2} \le x \le \sqrt{2-2y^2}$$


so from that I get the setup

$$\int_{0}^{2}\int_{-1}^{1}\int_{-\sqrt{2-2y^2}}^{\sqrt{2-2y^2}}dxdydz$$

I feel that is correct but the part mostly bothering be is the upper bound of z.

using x+y+2z=2 I can see the intercepts for x,y,z are 2,2,1. this is where I am getting stuck

 

[HallsofIvy]

x+ y+ 2z= 2, which is equivalent to z= 1- x/2- y/2, is the upper bound. Since that involves both x and y, that should be the first integral, not the last.

 

[jonroberts74]

x+ y+ 2z= 2, which is equivalent to z= 1- x/2- y/2, is the upper bound. Since that involves both x and y, that should be the first integral, not the last.

$$\int_{-1}^{1}\int_{-\sqrt{2-2y^2}}^{\sqrt{2-2y^2}}\int_{0}^{1- x/2- y/2}dzdxdy$$

more like that then?