# Surface area bounded by 2 different planes

1. Oct 24, 2016

1. The problem statement, all variables and given/known data
Find the surface area of portion of plane x + y + z = 3 that lies above the disc (x^2) + (y^2) < 2 in the first octant ...

2. Relevant equations

3. The attempt at a solution
Here's the solution provided by the author .....
I think it's wrong .... I think it should be the green coloured area + the black area ....

If it's only the black area , then the problem is find the surface area of portion of plane x + y + z = 3 that lies above the cylinder (x^2) + (y^2) < 2 in the first octant..

Last edited by a moderator: Oct 24, 2016
2. Oct 24, 2016

Image here

#### Attached Files:

• ###### 397.jpg
File size:
65.7 KB
Views:
34
3. Oct 24, 2016

### Staff: Mentor

No, it's just the black area.
That's not what they wrote. The disc they described consists of all the points in the x-y plane that lie inside the circle $x^2 + y^2 = 2$.

4. Oct 25, 2016

The disc here refers to the circle with radius 2 lie on the xy plane where z = 0???

5. Oct 25, 2016

### Staff: Mentor

Almost -- the radius is $\sqrt{2}$. And yes, the disc is in the x-y plane.