Find the volume of the solid bounded by the surfaces x² +

Find the volume of the solid bounded by the surfaces $x^2 + 4y^2 = 8z$, $x^2 + 4y^2 = 1$, and $z = 0$.

We use a scaled polar substitution tailored to the ellipse $x^2+4y^2\le1$.

Step 1: Identify the surfaces

The paraboloid $z=\dfrac{x^2+4y^2}{8}$ opens upward; the cylinder $x^2+4y^2=1$ is elliptical with semi-axes $1$ (in $x$) and $\frac12$ (in $y$); the solid sits between $z=0$ and the paraboloid, over the elliptical region $D:\ x^2+4y^2\le1$.

Step 2: Set up the double integral

\[V = \iint_D \frac{x^2+4y^2}{8}\,dA\]

Step 3: Substitute $x=r\cos\theta$, $y=\frac r2\sin\theta$

The Jacobian is:

\[J = \begin{vmatrix}\cos\theta & -r\sin\theta \\ \frac12\sin\theta & \frac r2\cos\theta\end{vmatrix} = \frac r2\cos^2\theta+\frac r2\sin^2\theta = \frac r2\]

and $x^2+4y^2 = r^2\cos^2\theta+4\cdot\frac{r^2}{4}\sin^2\theta = r^2$. The region becomes the simple disk $0\le r\le1$, $0\le\theta\le2\pi$.

Step 4: Evaluate the integral

\[V = \int_0^{2\pi}\int_0^1 \frac{r^2}{8}\cdot\frac r2\,dr\,d\theta = \int_0^{2\pi}\int_0^1\frac{r^3}{16}\,dr\,d\theta = \int_0^{2\pi}\frac{1}{64}\,d\theta = \frac{2\pi}{64} = \frac{\pi}{32}\]

\[\boxed{V = \frac{\pi}{32} \text{ cubic units}}\]

Summary

\[\boxed{V = \frac{\pi}{32} \text{ cubic units}}\]

Key insight: The scaled substitution $y=\frac r2\sin\theta$ (rather than the ordinary $y=r\sin\theta$) is exactly what's needed to turn the elliptical boundary $x^2+4y^2=1$ into the simple circle $r=1$, at the modest cost of a Jacobian factor of $\frac r2$ instead of $r$.