Find the volume of the solid bounded by the paraboloid

Find the volume of the solid bounded by the paraboloid of revolution: $$ z=x^2+y^2, $$ the coordinate planes, and the plane $x+y=1$.

Step 1 · Set up the volume integral

$$ V = \int_0^1\!\int_0^{1-x}\big(x^2+y^2\big)\,dy\,dx. $$

Step 2 · Evaluate the inner integral

$$ \int_0^{1-x}(x^2+y^2)\,dy = \left[x^2y+\frac{y^3}3\right]_0^{1-x} = x^2(1-x)+\frac{(1-x)^3}3. $$

Step 3 · Evaluate the outer integral

$$ V = \int_0^1\left[x^2-x^3+\frac{(1-x)^3}3\right]dx = \int_0^1x^2\,dx-\int_0^1x^3\,dx+\frac13\int_0^1(1-x)^3\,dx. $$

With $\displaystyle\int_0^1x^2\,dx=\frac13$, $\displaystyle\int_0^1x^3\,dx=\frac14$, and $\displaystyle\int_0^1(1-x)^3\,dx=\frac14$:

$$ V = \frac13-\frac14+\frac13\cdot\frac14 = \frac13-\frac14+\frac1{12} = \frac{4-3+1}{12} = \frac2{12} = \frac16. $$

Additional Notes

• This derivation matches the source exactly. By the symmetry of the region and the integrand $x^2+y^2$, integrating in the opposite order ($x$ first, then $y$) gives the same result.
• Verified symbolically: direct evaluation of the nested integral gives exactly $\dfrac16$.