Find the volume of the solid bounded by the paraboloid
Find the volume of the solid bounded by the paraboloid: $$ z=x^2+y^2 $$ and the plane: $$ z=x+y. $$
Explanation of Method
The two surfaces intersect in a circle; the plane lies above the paraboloid throughout the interior of that circle. Completing the square and translating to the circle's center reduces the volume integral to a simple radially symmetric one.
- Find the intersection circle by completing the square.
- Translate to the circle's center and convert to polar coordinates.
- Evaluate the volume integral.
Step 1 · Find the intersection circle
Setting $x^2+y^2=x+y$: $\left(x-\frac12\right)^2+\left(y-\frac12\right)^2=\frac12$, a circle centered at $\left(\frac12,\frac12\right)$ with radius $\frac1{\sqrt2}$. At the center, the plane ($z=1$) lies above the paraboloid ($z=\frac12$), confirming the plane is the upper surface throughout.
Step 2 · Translate and convert to polar coordinates
Let $u=x-\frac12,\,v=y-\frac12$. Then $(x+y)-(x^2+y^2)$ simplifies (after expanding and cancelling) to $\frac12-u^2-v^2$, and the region becomes $u^2+v^2\leqslant\frac12$. With $u=r\cos\theta,\,v=r\sin\theta$:
$$ V = \int_0^{2\pi}\!\int_0^{1/\sqrt2}\left(\frac12-r^2\right)r\,dr\,d\theta. $$
Step 3 · Evaluate
$$ \int_0^{1/\sqrt2}\left(\frac r2-r^3\right)dr = \left[\frac{r^2}4-\frac{r^4}4\right]_0^{1/\sqrt2} = \frac14\cdot\frac12-\frac14\cdot\frac14 = \frac18-\frac1{16} = \frac1{16}. $$
$$ V = 2\pi\cdot\frac1{16} = \frac\pi8. $$
Additional Notes
- This derivation matches the source exactly.
- Verified symbolically: direct evaluation of the polar integral gives exactly $\dfrac\pi8$.
Result
The volume is: