Find the volume of the solid bounded by the surface
Find the volume of the solid bounded by the surface: $$ x^2(a-y)^2+y^2z^2=y^2(a-y)^2 \qquad(a>0). $$
Explanation of Method
Solving for $z$ shows that, at each fixed height $y\in(0,a]$, the cross-section in the $xz$-plane is a circle of radius $y$ scaled by the factor $\dfrac{a-y}y$ in the $z$-direction — an ellipse of semi-axes $y$ (in $x$) and $(a-y)$ (in $z$). The volume follows from integrating the areas of these elliptical cross-sections over $y\in[0,a]$.
- Solve for $z$ and identify the elliptical cross-section at each $y$.
- Compute the area of each cross-section.
- Integrate over $y\in[0,a]$.
Step 1 · Solve for $z$
Rearranging: $y^2z^2=(a-y)^2\big(y^2-x^2\big)$, so for $y\neq0$:
$$ z^2 = \frac{(a-y)^2}{y^2}\big(y^2-x^2\big) \quad\Longrightarrow\quad z=\pm\frac{a-y}y\sqrt{y^2-x^2}, $$
real for $|x|\leqslant y$ and $0<y\leqslant a$. At each fixed $y$, this traces an ellipse in the $xz$-plane with semi-axis $y$ along $x$ and semi-axis $\dfrac{(a-y)y}y=a-y$ along $z$ (after clearing the scale factor) — equivalently, the cross-section $\dfrac{x^2}{y^2}+\dfrac{z^2}{(a-y)^2}=1$.
Step 2 · Area of the cross-section
An ellipse with semi-axes $y$ and $(a-y)$ has area $\pi y(a-y)$. Equivalently, integrating directly:
$$ \int_{-y}^y\frac{2(a-y)}y\sqrt{y^2-x^2}\,dx = \frac{2(a-y)}y\cdot\frac{\pi y^2}2 = \pi y(a-y), $$
using $\displaystyle\int_{-y}^y\sqrt{y^2-x^2}\,dx=\frac{\pi y^2}2$ (a semicircle of radius $y$).
Step 3 · Integrate over $y$
$$ V = \pi\int_0^a y(a-y)\,dy = \pi\int_0^a\big(ay-y^2\big)\,dy = \pi\left[\frac{ay^2}2-\frac{y^3}3\right]_0^a = \pi a^3\left(\frac12-\frac13\right) = \frac{\pi a^3}6. $$
Additional Notes
- This derivation matches the source exactly (presented here via the direct ellipse-area shortcut $\pi y(a-y)$ rather than the source's explicit semicircle-integral evaluation, which gives the identical result).
- Verified symbolically: direct evaluation of $\pi\int_0^a y(a-y)\,dy$ gives exactly $\dfrac{\pi a^3}6$.
Result
The volume is: