Find the volume of the solid bounded by the surfaces
Find the volume of the solid bounded by the surfaces: $$ z=3+x^2+2y^2,\qquad y=2x^2-1,\qquad y=0,\qquad z=0. $$
Explanation of Method
A note on the region: the literal pair $y=2x^2-1$ (upward parabola, vertex at $(0,-1)$) and $y=0$ only encloses a bounded region where the parabola lies below the axis, i.e. for $|x|\leqslant\frac1{\sqrt2}$, giving the base region $2x^2-1\leqslant y\leqslant0$. (Reading the bound the other way, $y\geqslant2x^2-1$ together with $y\geqslant0$, would leave the region unbounded.) Since $z=3+x^2+2y^2\geqslant3>0$ throughout, the surface is entirely above $z=0$, and the volume is simply $\iint_D(3+x^2+2y^2)\,dA$.
- Identify the bounded base region $D$.
- Set up the volume as an iterated integral, using symmetry in $x$.
- Evaluate the inner integral in $y$, then the outer integral in $x$.
Step 1 · The base region
The parabola meets $y=0$ at $2x^2-1=0\Rightarrow x=\pm\frac1{\sqrt2}$. The bounded region is:
$$ D=\left\{(x,y):|x|\leqslant\frac1{\sqrt2},\ 2x^2-1\leqslant y\leqslant0\right\}. $$
Step 2 · Set up the volume integral
By symmetry in $x$:
$$ V = 2\int_0^{1/\sqrt2}\int_{2x^2-1}^0\big(3+x^2+2y^2\big)\,dy\,dx. $$
Step 3 · Evaluate the inner integral
$$ \int_{2x^2-1}^0\big(3+x^2+2y^2\big)\,dy = -\left[(3+x^2)(2x^2-1)+\frac23(2x^2-1)^3\right]. $$
Expanding (using $(2x^2-1)^3=8x^6-12x^4+6x^2-1$):
$$ = \frac{11}3+6x^4-9x^2-\frac{16x^6}3. $$
Step 4 · Evaluate the outer integral
$$ V = 2\int_0^{1/\sqrt2}\left(\frac{11}3+6x^4-9x^2-\frac{16x^6}3\right)dx = 2\left[\frac{11x}3+\frac{6x^5}5-3x^3-\frac{16x^7}{21}\right]_0^{1/\sqrt2}. $$
At $x=\frac1{\sqrt2}$, the bracket equals $\dfrac1{\sqrt2}\left(\dfrac{11}3+\dfrac3{10}-\dfrac32-\dfrac2{21}\right)$. Using the common denominator $210$: $\dfrac{770+63-315-20}{210}=\dfrac{498}{210}=\dfrac{83}{35}$. So:
$$ V = 2\cdot\frac1{\sqrt2}\cdot\frac{83}{35} = \sqrt2\cdot\frac{83}{35} = \frac{83\sqrt2}{35}. $$
Additional Notes
- A note on the source computation: the source correctly reinterpreted the bounded region (flagging and resolving the same ambiguity noted above) and correctly derived the inner integral $\frac{11}3+6x^4-9x^2-\frac{16x^6}3$, matching this page exactly. The error was in the final arithmetic: combining the four fractions $\frac{11}3,\frac3{10},\frac32,\frac2{21}$ over a common denominator of $60$ — but $21$ does not divide $60$ evenly (the correct common denominator is $210$), so the source's numerator for the $\frac2{21}$ term was wrong ($12$ instead of the correct $20$ over denominator $210$, i.e. it used $2/21\approx12/60$ when in fact $2/21\times210=20$, not $12\times3.5$). This produced the source's $\frac{34\sqrt2}{15}\approx3.2056$ instead of the correct $\frac{83\sqrt2}{35}\approx3.3537$.
- Verified by direct numerical double integration over $D$, matching $\dfrac{83\sqrt2}{35}$ to high precision.
Result
The volume is: