How is the boundary of the projection region determined for the solid bounded by $z = x^2 + y^2$ and $z = x + y$?

Conditions

• Solid is bounded above by plane $z=x+y$
• Solid is bounded below by paraboloid $z=x^2+y^2$

Reasoning, step by step

1. Set the upper and lower surface equations equal: $x^2+y^2 = x+y$.
2. Rearrange terms to group variables: $x^2-x + y^2-y = 0$.
3. Complete the square for both $x$ and $y$: $(x-1/2)^2 - 1/4 + (y-1/2)^2 - 1/4 = 0$.
4. Simplify to standard circle form: $(x-1/2)^2 + (y-1/2)^2 = 1/2$.

Example

The script states: 'To determine the integration range, we project the intersection curve onto the xy-plane by setting $x^2+y^2=x+y$. Completing the square yields ($x-1/2$)^$2+(y-1/2)^2=1/2$, defining a circular region centered at ($1/2$, $1/2$) with radius sqrt(2)/2.'

Common misconceptions

• Assuming the projection is always centered at the origin.
• Forgetting to complete the square correctly, leading to wrong center or radius.

Watch the explanation

Explore next

• Why is the plane considered the upper bound within this domain?
• What would happen if the surfaces did not intersect?