Find the volume of the solid bounded by the surfaces

Find the volume of the solid bounded by the surfaces: $$ z=\left(\frac{x^2}4+\frac{y^2}9\right)^2,\qquad\frac{x^2}4+\frac{y^2}9=x,\qquad z=0. $$

Step 1 · Identify the boundary ellipse

Completing the square in $\dfrac{x^2}4+\dfrac{y^2}9=x$ gives $\dfrac{(x-2)^2}4+\dfrac{y^2}9=1$: an ellipse centered at $(2,0)$ with semi-axes $2$ (in $x$) and $3$ (in $y$).

Step 2 · Re-parametrize

Let $x=2+2r\cos\theta,\,y=3r\sin\theta$, $r\in[0,1],\,\theta\in[0,2\pi]$, with Jacobian:

$$ J=\begin{vmatrix}2\cos\theta&-2r\sin\theta\\3\sin\theta&3r\cos\theta\end{vmatrix}=6r. $$

Then:

$$ \frac{x^2}4+\frac{y^2}9 = \frac{(2+2r\cos\theta)^2}4+r^2\sin^2\theta = 1+2r\cos\theta+r^2\cos^2\theta+r^2\sin^2\theta = 1+2r\cos\theta+r^2. $$

Step 3 · Set up and expand the volume integral

$$ V=\int_0^{2\pi}\int_0^1\big(1+2r\cos\theta+r^2\big)^2\cdot6r\,dr\,d\theta. $$

Expanding $(1+2r\cos\theta+r^2)^2=1+4r\cos\theta+2r^2+4r^2\cos^2\theta+4r^3\cos\theta+r^4$, and multiplying by $r$:

$$ r(1+2r\cos\theta+r^2)^2 = r+4r^2\cos\theta+2r^3+4r^3\cos^2\theta+4r^4\cos\theta+r^5. $$

Step 4 · Integrate over $\theta$

Terms with $\cos\theta$ vanish over $[0,2\pi]$; using $\int_0^{2\pi}\cos^2\theta\,d\theta=\pi$:

$$ \int_0^{2\pi}r(1+2r\cos\theta+r^2)^2\,d\theta = 2\pi r+4r^3\pi+4\pi r^3+2\pi r^5 = 2\pi\big(r+4r^3+r^5\big). $$

Step 5 · Integrate over $r$

$$ V = 6\cdot2\pi\int_0^1\big(r+4r^3+r^5\big)\,dr = 12\pi\left[\frac{r^2}2+r^4+\frac{r^6}6\right]_0^1 = 12\pi\left(\frac12+1+\frac16\right) = 12\pi\cdot\frac{10}6 = 20\pi. $$

Additional Notes

• This derivation matches the source exactly.
• Verified two ways: a direct Cartesian double integral over the boundary ellipse, and an independent evaluation of the $(r,\theta)$ integral, both giving exactly $20\pi$.