Find the volume of the solid bounded by the prism whose
Find the volume of the solid bounded by the prism whose base is the triangle with vertices: $$ O(0,0,0),\quad A(1,0,0),\quad B\!\left(0,\frac12,0\right), $$ with edges parallel to the $z$-axis, and the surface: $$ z=x^2+y+1 \qquad(z>0). $$
Explanation of Method
The prism's base is a right triangle in the $xy$-plane; the top is capped by the surface $z=x^2+y+1$ (which is manifestly positive throughout the triangle, since $x,y\geqslant0$), so the volume is a straightforward double integral of the height over the triangle.
- Find the equation of the hypotenuse $AB$ and describe the triangle.
- Set up the volume integral.
- Evaluate the resulting polynomial integral.
Step 1 · Describe the triangle
The line through $A(1,0)$ and $B\left(0,\frac12\right)$ is $x+2y=1$, i.e. $y=\dfrac{1-x}2$. So the base region is:
$$ D = \left\{(x,y):0\leqslant x\leqslant1,\ 0\leqslant y\leqslant\frac{1-x}2\right\}. $$
Step 2 · Set up the volume integral
$$ V = \int_0^1\!\int_0^{(1-x)/2}\big(x^2+y+1\big)\,dy\,dx. $$
Step 3 · Evaluate the inner integral
$$ \int_0^{(1-x)/2}(x^2+y+1)\,dy = \left[x^2y+\frac{y^2}2+y\right]_0^{(1-x)/2} = \frac{x^2(1-x)}2+\frac{(1-x)^2}8+\frac{1-x}2. $$
Factoring out $\dfrac{1-x}8$: this equals $\dfrac{1-x}8\big(4x^2-x+5\big)$.
Step 4 · Evaluate the outer integral
Expanding $(1-x)(4x^2-x+5)=-4x^3+5x^2-6x+5$:
$$ V = \frac18\int_0^1\big(-4x^3+5x^2-6x+5\big)\,dx = \frac18\left[-x^4+\frac{5x^3}3-3x^2+5x\right]_0^1. $$
$$ = \frac18\left[-1+\frac53-3+5\right] = \frac18\left[1+\frac53\right] = \frac18\cdot\frac83 = \frac13. $$
Additional Notes
- This derivation matches the source exactly (the source's own brief self-correction mid-arithmetic, "Wait, let me recalculate," resolved to the same correct value).
- Verified symbolically: direct evaluation of the nested integral gives exactly $\dfrac13$.
Result
The volume is: