Find the volume of the solid bounded by the plane z=0
Find the volume of the solid bounded by the plane $z=0$, the parabolic cylinder: $$ y^2=2px, $$ the plane $x=a$, and the surface: $$ z=xy^2. $$
Explanation of Method
The base region is bounded by the parabola $y^2=2px$ (so $-\sqrt{2px}\leqslant y\leqslant\sqrt{2px}$) and the plane $x=a$; the height is $z=xy^2$, which is non-negative throughout since $x\geqslant0$ and $y^2\geqslant0$.
- Identify the base region $D=\{0\leqslant x\leqslant a,\ -\sqrt{2px}\leqslant y\leqslant\sqrt{2px}\}$.
- Set up the volume integral, using symmetry in $y$.
- Evaluate the resulting single-variable integral.
Step 1 · Set up the volume integral
$$ V = \int_0^a\!\int_{-\sqrt{2px}}^{\sqrt{2px}}xy^2\,dy\,dx = 2\int_0^a x\left[\int_0^{\sqrt{2px}}y^2\,dy\right]dx. $$
Step 2 · Evaluate the inner integral
$$ \int_0^{\sqrt{2px}}y^2\,dy = \frac{(2px)^{3/2}}3, $$
so:
$$ V = \frac23\int_0^a x(2px)^{3/2}\,dx = \frac{2(2p)^{3/2}}3\int_0^a x^{5/2}\,dx. $$
Step 3 · Evaluate the outer integral
$$ \int_0^a x^{5/2}\,dx = \frac{2a^{7/2}}7, $$
so:
$$ V = \frac{2(2p)^{3/2}}3\cdot\frac{2a^{7/2}}7 = \frac{4(2p)^{3/2}a^{7/2}}{21} = \frac{8\sqrt2\,p^{3/2}a^{7/2}}{21}. $$
Additional Notes
- This derivation matches the source exactly.
- Verified symbolically: direct evaluation of the nested integral gives exactly $\dfrac{8\sqrt2\,p^{3/2}a^{7/2}}{21}$.
Result
The volume is: